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Page 1
CHAPTER 7
Trigonometric Identities and Equations
Getting Started, p. 386
1. a) Do the following to isolate and solve for x:
3x 2 7 5 5 2 9x
3x 2 7 1 9x 5 5 2 9x 1 9x
12x 2 7 5 5
12x 2 7 1 7 5 5 1 7
12x 5 12
12
12x
5
12
12
x51
b) Do the following to isolate and solve for x:
x
1
2(x 1 3) 2 5
4
2
x
1
2x 1 6 2 5
4
2
x
1
2x 1 6 2 2 6 5 2 6
4
2
x
11
2x 2 5 2
4
2
7x
11
52
4
2
7x
4
11
4
3 52 3
4
7
2
7
44
x52
14
22
x52
7
c) Factor the left side of the equation and solve for
x as follows:
x2 2 5x 2 24 5 0
(x 2 8)(x 1 3) 5 0
x 2 8 5 0 or x 1 3 5 0
x 2 8 1 8 5 0 1 8 or x 1 3 2 3 5 0 2 3
x 5 8 or x 5 23
d) Move all terms to the left side of the equation,
factor, and solve for x as follows:
6x2 1 11x 5 10
2
6x 1 11x 2 10 5 10 2 10
6x2 1 11x 2 10 5 0
(3x 2 2)(2x 1 5) 5 0
3x 2 2 5 0
2x 1 5 5 0
or
3x 2 2 1 2 5 0 1 2 or 2x 1 5 2 5 5 0 2 5
Advanced Functions Solutions Manual
3x 5 2 or 2x 5 25
2x
3x
2
5
5 or
52
3
3
2
2
5
2
x 5 or x 5 2
3
2
e) Use the quadratic formula to solve for x as
follows:
x2 1 2x 2 1 5 0
x5
x5
x5
x5
x5
2b 6 "b2 2 4ac
2a
22 6 "(2)2 2 4(1)(21)
2(1)
22 6 !4 1 4
2
22 6 !8
2
22 6 2 !2
2
x 5 21 6 !2
f) Move all terms to the left side of the equation and
use the quadratic formula to solve for x as follows:
3x2 5 3x 1 1
3x2 2 3x 2 1 5 3x 1 1 2 3x 2 1
3x2 2 3x 2 1 5 0
x5
x5
x5
2b 6 "b2 2 4ac
2a
3 6 "(23)2 2 4(3)(21)
2(3)
3 6 !9 1 12
6
3 6 !21
6
2. To show that AB 5 CD, the distance formula
should be used as follows:
First for AB:
x5
d 5 "(x2 2 x1 )2 1 (y2 2 y1 )2
d5
2
1
(2 2 1)2 1 a 2 0b
Å
2
7-1
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d5
d5
d5
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1 2
(1)2 1 a b
Å
2
Ä
11
1
4
5
Ä4
!5
2
Now for CD:
d5
d 5 "(x2 2 x1 )2 1 (y2 2 y1 )2
d5
1 2
a0 2 a2 bb 1 (6 2 5)2
Å
2
c) From the figure:
15
sin B 5
17
sin21 (sin B) 5 sin21 a
15
b
17
15
m/B 5 sin21 a b
17
m/B 5 61.9 °
4. a)
y
4
2 u
x
P(–2, 2)
0
–4 –2
2 4
–2
–4
d5
1 2
a b 1 (1)2
Å 2
d5
1
11
Ä4
b) Drawing a segment from (22, 2) down to the
negative x-axis forms a right triangle with two legs
of length 2. The tangent of the related acute angle is
2
2 or 1.
d5
5
Ä4
The measure in radians of the acute angle that has a
p
tangent equal to 1 is , so the value of the related
!5
2
So, AB 5 CD.
8
opposite side
3. a) sin A 5
5 ;
hypotenuse
17
15
adjacent side
5 ;
cos A 5
hypotenuse
17
8
opposite side
5 ;
tan A 5
adjacent side
15
17
hypotenuse
5 ;
csc A 5
opposite side
8
17
hypotenuse
5 ;
sec A 5
adjacent side
15
15
adjacent side
5
cot A 5
opposite side
8
b) To determine the measure of /A in radians, any
of the ratios found in part (a) can be used. In this
case, sin A will be used:
8
sin A 5
17
8
sin21 (sin A) 5 sin21
17
8
m/A 5 sin21
17
m/A 5 0.5 radians
d5
7-2
4
p
acute angle is radians.
4
c) Since the point at (22, 2) is in the second
quadrant, the terminal arm of the angle u must also
be in the second quadrant. Since the angle with a
terminal arm in the second quadrant and a tangent
3p
3p
of 21 measures
radians, u 5
radians.
4
4
p
4
5. a) The x-coordinate of point A is cos , while the
p
4
y-coordinate is sin . Therefore, the coordinates of
!2
R
point A are Q !2
2 , 2 . The x-coordinate of point B
p
p
is cos , while the y-coordinate is sin . Therefore,
3
!3
R
2 .
the coordinates of point B are Q 12,
x-coordinate of point C is cos
y-coordinate is sin
point C are Q 2 12,
2p
.
3
!3
R
2 .
2p
,
3
3
The
while the
Therefore, the coordinates of
The x-coordinate of point D the
5p
5p
is cos , while the y-coordinate is sin . Therefore,
6
6
1
R
coordinates of point D are Q 2 !3
2 , 2 . The x-coordinate
of point E is cos
7p
,
6
while the y-coordinate is sin
7p
.
6
1
R
Therefore, the coordinates of point E are Q 2 !3
2 , 22 .
The x-coordinate of point F is cos
5p
, while the
4
Chapter 7: Trigonometric Identities and Equations
08-035_07_AFSM_C07_001-060.qxd
y-coordinate is sin
7/23/08
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5p
. Therefore, the
4
Page 3
coordinates of
!2
R
point F are Q 2 !2
2 , 2 2 . The x-coordinate of point
G is cos
4p
,
3
while the y-coordinate is sin
4p
.
3
R
Therefore, the coordinates of point G are Q 2 12, 2 !3
2 .
The x-coordinate of point H is cos
y-coordinate is sin
5p
.
3
5p
,
3
while the
Therefore, the coordinates of
R
point H are Q 12, 2 !3
2 . The x-coordinate of point I is
cos
7p
,
4
while the y-coordinate is sin
7p
.
4
Therefore,
!2
R
the coordinates of point I are Q !2
2 , 2 2 . The
x-coordinate of point J is cos
y-coordinate is
of point J are
b) i) cos
at Q 2 !2
2 ,
3p
4
11p
.
sin
6
Q !3
2 ,
while the
Therefore, the coordinates
2 12 R .
is equal to the x-coordinate of the point
!2
R
2 ,
so cos
11p
is equal
6
11p
5 2 12.
sin
6
ii) sin
11p
,
6
3p
5 2 !2
2 .
4
to the y-coordinate of point J,
so
iii) cos p is equal to the x-coordinate of the point at
(21, 0), so cos p 5 21.
p
iv) csc is equal to the reciprocal of the y-coordinate
6
1
p
1
R
of the point at Q !3
5 1 5 2.
2 , 2 , so csc
6
2 34,
2
6. a) Since tan x 5
the leg opposite angle x in a
right triangle has a length of 3, while the leg adjacent
to angle x has a length of 4. For this reason, the length
of the hypotenuse can be calculated as follows:
32 1 42 5 z 2
9 1 16 5 z 2
25 5 z 2
z55
Therefore, if the angle x is in the second quadrant,
the other five trigonometric ratios are as follows:
3
opposite leg
5 ;
sin x 5
hypotenuse
5
4
adjacent leg
52 ;
cos x 5
hypotenuse
5
5
hypotenuse
5 ;
csc x 5
opposite leg
3
5
hypotenuse
52 ;
sec x 5
adjacent leg
4
Advanced Functions Solutions Manual
adjacent leg
4
52
opposite leg
3
If the angle x is in the fourth quadrant, the other
five trigonometric ratios are as follows:
3
opposite leg
52 ;
sin x 5
hypotenuse
5
4
adjacent leg
5 ;
cos x 5
hypotenuse
5
5
hypotenuse
52 ;
csc x 5
opposite leg
3
5
hypotenuse
5 ;
sec x 5
adjacent leg
4
4
adjacent leg
52
cot x 5
opposite leg
3
cot x 5
b) To determine the value of the angle x, any of the
ratios found in part (a) can be used. If x is in the
second quadrant, sin x 5 35, so x can be solved for as
follows:
3
sin x 5
5
3
sin21 (sin x) 5 sin21 a b
5
3
x 5 sin21 a b
5
x 5 2.5
If x is in the fourth quadrant, sin x 5 2 35, so x can
be solved for as follows:
3
sin x 5 2
5
3
sin21 (sin x) 5 sin21 a2 b
5
3
x 5 sin21 a2 b
5
x 5 5.6
sin u
, cos u 2 0 is true.
7. a) tan u 5
cos u
b) sin2 u 1 cos2 u 5 1 is true.
1
c) sec u 5
, sin u 2 0 is false, since sec u is the
sin u
reciprocal of cos u, not sin u.
d) cos2 u 5 sin2 u 2 1 is false. This can be shown
by performing the following operations:
cos2 u 5 sin2 u 2 1
2
cos u 1 1 5 sin2 u 2 1 1 1
cos2 u 1 1 5 sin2 u
cos2 u 1 1 2 cos2 u 5 sin2 u 2 cos2 u
sin2 u 2 cos2 u 5 1
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Since sin2 u 1 cos2 u 5 1 is true,
sin2 u 2 cos2 u 5 1 must be false.
e) 1 1 tan2 u 5 sec2 u is true
f)
cos u
, sin u 2 0
cot u 5
sin u
is true.
8. The sinusoidal function y 5 sin x is transformed
into the function y 5 a sin k(x 2 d) 1 c by
vertically stretching or compressing the function
y 5 sin x by a factor of a, horizontally stretching or
1
compressing it by a factor of , reflecting it in the
k
x-axis if a , 0, reflecting it in the y-axis is k , 0,
vertically translating it c units up or down, and
horizontally translating it d units to the right or the
left. Therefore, an appropriate flow chart would be
as follows:
Perform a vertical stretch> compression by a
factor of 0 a 0 .
Use ` ` to determine the horizontal
1
k
stretch> compression.
Use a and k to determine whether the function is
reflected in the y-axis or the x-axis.
Perform a vertical translation of
c units up or down.
Perform a horizontal translation of d units to
the right or the left.
7.1 Exploring Equivalent
Trigonometric Functions, pp. 392–393
1. a) Answers may vary. For example: The graph is
that of the trigonometric function y 5 cos u. Since
the period of y 5 cos u is 2p radians, the graph
repeats itself every 2p radians. Therefore, three
possible equivalent expressions for the graph are
y 5 cos (u 1 2p), y 5 cos (u 1 4p), and
y 5 cos (u 2 2p).
7-4
b) The trigonometric functions y 5 cos u and
p
y 5 sin Q u 1 R are equivalent. Since the period
2
of y 5 sin Q u 1 R is 2p radians, the graph repeats
2
itself every 2p radians. Therefore, three possible
equivalent expressions for the graph using the sine
p
3p
function are y 5 sin Q u 1 R , y 5 sin Q u 2 R ,
and
p
5p
y 5 sin Q u 1 R .
2
2
2
2. a) The graph of the trigonometric function
y 5 csc u is symmetric with respect to the origin,
so y 5 csc u is an odd function. Therefore,
csc (2u) 5 2csc u. The graph of the trigonometric
function y 5 sec u is symmetric with respect to the
y-axis, so y 5 sec u is an even function. Therefore,
sec (2u) 5 sec u. The graph of the trigonometric
function y 5 cot u is symmetric with respect to the
origin, so y 5 cot u is an odd function. Therefore,
cot (2u) 5 2cot u.
b) If the graph of the trigonometric function
y 5 csc u is reflected in the y-axis, the equation of
the resulting graph is y 5 csc (2u). Also, if the
graph of y 5 csc u is reflected in the x-axis, the
equation of the resulting graph is y 5 2csc u. Since
the graph of y 5 csc u is symmetric with respect to
the origin, a reflection in the y-axis is the same as a
reflection in the x-axis. Therefore, the equation
csc (2u) 5 2csc u must be true. If the graph of the
trigonometric function y 5 sec u is reflected in the
y-axis, the equation of the resulting graph is
y 5 sec (2u). Since the graph of y 5 sec u is
symmetric with respect to the y-axis, a reflection of
the function in the y-axis results in the function itself.
Therefore, the equation sec (2u) 5 sec u must be
true. If the graph of the trigonometric function
y 5 cot u is reflected in the y-axis, the equation of
the resulting graph is y 5 cot (2u). Also, if the
graph of y 5 cot u is reflected in the x-axis, the
equation of the resulting graph is y 5 2cot u. Since
the graph of y 5 cot u is symmetric with respect to
the origin, a reflection in the y-axis is the same as a
reflection in the x-axis. Therefore, the equation
cot (2u) 5 2cot u must be true.
p
3. a) Since sin u 5 cos a 2 ub,
2
3p
p
p
p
p
sin 5 cos a 2 b 5 cos a
2 b
6
2
6
6
6
2p
p
5 cos
5 cos
6
3
Chapter 7: Trigonometric Identities and Equations
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b) Since cos u 5 sin a
p
2 ub,
2
p
5p
5p
cos
5 sin a 2
b
12
2
12
6p
5p
p
5 sin a
2
b 5 sin
12
12
12
p
c) Since tan u 5 cot a 2 ub,
2
p
3p
3p
tan
5 cot a 2
b
8
2
8
4p
3p
p
5 cot a
2
b 5 cot
8
8
8
p
d) Since cos u 5 sin a 2 ub,
2
p
5p
5p
cos
5 sin a 2
b
16
2
16
8p
5p
3p
5 sin a
2
b 5 sin
16
16
16
p
e) Since sin u 5 cos a 2 ub,
2
p
p
p
sin 5 cos a 2 b
8
2
8
4p
p
5 cos a
2 b
8
8
3p
5 cos
8
p
f) Since tan u 5 cot a 2 ub,
2
3p
p
p
p
p
tan 5 cot a 2 b 5 cot a
2 b
6
2
6
6
6
2p
p
5 cot
5 cot
6
3
p
4. a) Since sin u 5 cos a 2 ub,
2
1
1
5
p
sin u
cos Q 2 u R
2
Therefore, csc u 5 sec a
p
2 ub.
2
p
1
1
.
5
Since cos u 5 sin a 2 ub,
p
2
cos u
sin Q 2 u R
2
Therefore, sec u 5 csc a
p
2 ub.
2
Advanced Functions Solutions Manual
1
5
Since tan u 5 cot Q 2 2 u R ,
tan u
p
Therefore, cot u 5 tan Q 2 2 u R .
p
1
p
cot Q 2 2 u R
.
b) The trigonometric function y 5 sec Q 2 u R
2
p .
can also be written as y 5 sec Q2 Q u 2 RR If the
p
2
graph of the trigonometric function
y 5 sec Q2 Q u 2 RR is reflected in the y-axis, the
2
p
equation of the resulting graph is y 5 sec Q u 2 R .
p
2
p
y 5 sec Q2 Q u 2 RR
2
Since the graph of
is symmetric
with respect to the y-axis, a reflection of the function
in the y-axis results in the function itself. For this
p
p
reason, sec Q 2 u R 5 sec Q u 2 R , so
2
2
1
p
1
csc u 5 sec Q u 2 R . Therefore,
5
p ,
2
csc u
sec Q u 2 2 R
or sin u 5 cos Q u 2 R . This is a known identity, so
2
p
the cofunction identity csc u 5 sec Q 2 u R must be
2
p
true. The trigonometric function y 5 csc Q 2 u R can
p
2
also be written as y 5 csc Q2 Q u 2 RR . If the graph
2
p
of the trigonometric function y 5 csc Q2 Q u 2 RR
2
is reflected in the y-axis, the equation of the resulting
p
resulting graph is y 5 csc Q u 2 R . Since the graph
p
2
of y 5 csc Q2 Q u 2 RR is symmetric with respect to
2
the origin, a reflection in the y-axis is the same
as a reflection in the x-axis. For this reason,
p
p
2csc a 2 ub 5 csc au 2 b, or
2
2
p
p
csc a 2 ub 5 csc au 2 1 pb
2
2
p
5 csc au 1 b, so
2
p
sec u 5 csc au 1 b
2
1
1
,
5
Therefore,
p or
sec u
csc Q u 1 R
p
2
cos u 5 sin au 1
p
b
2
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This is a known identity, so the cofunction identity
p
sec u 5 csc Q 2 u R
2
must be true.
The trigonometric function y 5 tan Q 2 u R can also
2
p
be written as y 5 tan Q2 Q u 2 RR . If the graph of
2
p
the trigonometric function y 5 tan Q2 Qu 2 RR is
2
reflected in the y-axis, the equation of the resulting
p
graph is y 5 tan Q2 Q u 2 RR .
p
2
Since the graph of y 5 tan Q2 Q u 2 RR is symmetric
2
with respect to the origin, a reflection in the
y-axis is the same as a reflection in the x-axis. For
p
p
this reason, 2tan Q 2 u R 5 tan Q u 2 R , or
2
p
2
tan Q 2 u R 5 tan Q u 2 1 p R 5 tan Q u 1 R ,
2
2
2
p
so cot u 5 tan Q u 1 R . Therefore,
p
p
p
2
1
1
,
5
cot u
tan Q u 1 p R
2
or tan u 5 cot Q u 1 R . This
2
p
is a known identity, so the cofunction identity
p
cot u 5 tan Q 2 u R must be true.
2
5. a) Since sin u 5 sin (p 2 u),
7p
7p
p
sin
5 sin ap 2
b 5 sin
8
8
8
b) Since cos u 5 2cos (p 1 u),
13p
13p
5 2cos ap 1
b
12
12
13p
25p
12p
1
b 5 2cos
5 2cos a
12
12
12
25p
2 2pb
5 2cos a
12
24p
p
25p
2
b 5 2cos
5 2cos a
12
12
12
c) Since tan u 5 tan (p 1 u),
9p
5p
4p
5p
tan
5 tan a
1
b 5 tan
4
4
4
4
9p
5 tan a
2 2pb
4
9p
8p
p
5 tan a
2
b 5 tan
4
4
4
d) Since cos u 5 cos (2p 2 u),
11p
11p
cos
5 cos a2p 2
b
6
6
12p
11p
p
5 cos a
2
b 5 cos
6
6
6
cos
7-6
e) Since sin u 5 2sin (2p 2 u),
13p
13p
sin
5 2sin a2p 2
b
8
8
16p
13p
3p
5 2sin a
2
b 5 2sin
8
8
8
f) Since tan u 5 2tan (2p 2 u),
5p
5p
tan
5 2tan a2p 2
b
3
3
p
6p
5p
5 2tan a
2
b 5 2tan
3
3
3
6. a) Assume the circle is a unit circle. Let the
coordinates of Q be (x, y). Since P and Q are
reflections of each other in the line y 5 x, the
coordinates of P are (y, x). Draw a line from P to
the positive x-axis. The hypotenuse of the new right
p
triangle makes an angle of Q 2 u R with the
2
positive x-axis. Since the x-coordinate of P is y,
p
cos Q 2 u R 5 y. Also, since the y-coordinate of
2
Q is y, sin u 5 y. Therefore, cos Q 2 u R 5 sin u.
2
b) Assume the circle is a unit circle. Let the
coordinates of the vertex on the circle of the right
triangle in the first quadrant be (x, y). Then
sin u 5 y, so 2sin u 5 2y. The point on the circle
p
that results from rotating the vertex by
2
counterclockwise about the origin has coordinates
p
(2y, x), so cos Q 1 u R 5 2y. Therefore,
p
2
p
cos Q 1 u R 5 2sin u.
2
7. a) true; the period of cosine is 2p
p
b) false; Answers may vary. For example: Let u 5 .
2
p
2
Then the left side is sin , or 1. The right side is
2sin
p
2
or 21.
c) false; Answers may vary. For example: Let u 5 p.
Then the left side is cos p, or 21. The right side is
2cos 5p, or 1.
p
d) false; Answers may vary. For example: Let u 5 .
Then the left side is
3p
tan ,
4
or
2 !2
2 .
4
The right side
is tan , or !2
2 .
4
e) false; Answers may vary. For example: Let u 5 p.
p
Then the left side is cot
p
tan ,
4
3p
,
4
or –1. The right side is
or 1.
Chapter 7: Trigonometric Identities and Equations
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Page 7
f) false; Answers may vary. For example: Let
p
5p
u 5 . Then the left side is sin , or 1. The right
2
side is
p
sin a2 b,
2
2
or –1.
5a
1. a) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
sin a cos 2a 1 cos a sin 2a 5 sin (a 1 2a) 5 sin 3a.
b) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
cos 4x cos 3x 2 sin 4x sin 3x 5 cos (4x 1 3x)
5 cos 7x
tan a 2 tan b
,
2. a) Since tan (a 2 b) 5
1 1 tan a tan b
tan 170° 2 tan 110°
5 tan (170° 2 110°)
1 1 tan 170° tan 110°
5 tan 60° 5 !3
b) Since cos (a 2 b) 5 cos a cos b 1 sin a sin b,
5p
p
5p
p
p
5p
cos
cos
1 sin
sin
2 b
5 cos a
12
12
12
12
12
12
4p
p
1
5 cos
5 cos
5
12
3
2
3. a) Two angles from the special triangles are 30°
and 45°, so 75° 5 30° 1 45°.
b) Two angles from the special triangles are 30° and
45°, so 215° 5 30° 2 45°.
p
c) Two angles from the special triangles are and
6
p
p
p
2p
p
p
, so 2 5 2
5 2 .
6
6
6
6
3
p
p
d) Two angles from the special triangles are and ,
6
4
p
3p
2p
p
p
2
5 2 .
so 5
12
12
12
4
6
e) Two angles from the special triangles are 45° and
60°, so 105° 5 45° 1 60°.
p
p
f) Two angles from the special triangles are and ,
3
2
5p
2p
3p
p
p
5
1
5 1 .
so
6
6
6
3
2
4. a) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
sin 75° 5 sin (30° 1 45°)
5 sin 30° cos 45° 1 cos 30° sin 45°
1 !2
!3 !2
5 a ba
b1a
ba
b
2
2
2
2
!2
!6
5
1
4
4
!2 1 !6
5
4
Advanced Functions Solutions Manual
!2 !3
!2 1
ba
b1a
ba b
2
2
2
2
!2
!6
1
4
4
!2 1 !6
5
4
5
7.2 Compound Angle Formulas,
pp. 400–401
3
b) Since cos (a 2 b) 5 cos a cos b 1 sin a sin b,
cos 15° 5 cos (45° 2 30°)
5 cos 45° cos 30° 1 sin 45° sin 30°
tan a 1 tan b
,
1 2 tan a tan b
2p
3p
p
p
5p
5 tan a
1
b 5 tan a 1 b
tan
12
12
12
6
4
c) Since tan (a 1 b) 5
p
p
!3
1 tan
6
4
3 1 1
5
p
p 5
1 2 tan tan
R
1 2 Q !3
6
4
3 (1)
tan
5
5
5
5
!3
3
3 1 3
3
!3
3 2 3
5
!3 1 3
3
3 2 !3
3
!3 1 3
!3 1 3
3
3
5
3
3 2 !3
3 2 !3
( !3 1 3)( 3 1 !3)
( 3 2 !3)( 3 1 !3)
3 !3 1 9 1 3 1 3!3
9 2 3 !3 1 3 !3 2 3
12 1 6!3
5 2 1 !3
6
d) Since sin (a 2 b) 5 sin a cos b 2 cos a sin b,
p
3p
4p
p
p
sin a2 b 5 sin a
2
b 5 sin a 2 b
12
12
12
4
3
p
p
p
p
5 sin cos 2 cos sin
4
3
4
3
5
5a
5
!2 1
!2 !3
ba b 2 a
ba
b
2
2
2
2
!2
!6
2
4
4
!2 2 !6
4
e) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
cos 105° 5 cos (45° 1 60°)
5 cos 45° cos 60° 2 sin 45° sin 60°
5
5a
!2 1
!2 !3
ba b 2 a
ba
b
2
2
2
2
7-7
08-035_07_AFSM_C07_001-060.qxd
5
5
7/23/08
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Page 8
!2
!6
2
4
4
!2 2 !6
4
tan a 1 tan b
,
1 2 tan a tan b
23p
9p
14p
3p
7p
tan
5 tan a
1
b 5 tan a
1
b
12
12
12
4
6
f) Since tan (a 1 b) 5
3p
7p
1 tan
21 1 !3
4
6
3
5
5
3p
7p
1 2 tan
tan
R
1 2 (21) Q !3
3
4
6
tan
2 33 1
5
3
3
5
5
5
!3
3
R
2 Q 2 !3
3
110
1 2 (1)(0)
110
5
120
1
5 51
1
d) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
p
p
p
p
sin a2 1 b 5 sin a2 b cos
2
3
2
3
p
p
1 cos a2 b sin
2
3
!3
1
b
5 (21)a b 1 (0)a
2
2
1
52 10
2
1
52
2
tan a 2 tan b
,
e) Since tan (a 2 b) 5
1 1 tan a tan b
p
p
tan 2 tan
p
p
3
6
tan a 2 b 5
p
p
3
6
1 1 tan tan
5
5
!3 2 3
3
3 1 !3
3
!3 2 3
3
!3 2 3
3
5
3
3 1 !3
3 1 !3
( !3 2 3)( 3 2 !3)
( 3 1 !3)( 3 2 !3)
3 !3 2 9 2 3 1 3 !3
9 1 3!3 2 3!3 2 3
212 1 6!3
6
5 22 1 !3
5. a) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
p
p
p
sin ap 1 b 5 sin p cos 1 cos p sin
6
6
6
!3
1
5 (0)a
b 1 (21)a b
2
2
1
1
5 0 1 a2 b 5 2
2
2
b) Since cos (a 2 b) 5 cos a cos b 1 sin a sin b,
p
p
p
cos ap 2 b 5 cos p cos 1 sin p sin
4
4
4
!2
!2
5 (21)a
b 1 (0)a
b
2
2
!2
10
52
2
!2
52
2
tan a 1 tan b
c) Since tan (a 1 b) 5
,
1 2 tan a tan b
p
tan 1 tan p
p
4
tan a 1 pb 5
p
4
1 2 tan tan p
5
3
5
5
5
5
!3 2
!3
3
!3
3
1 1 ( !3)(
3 !3
3
2 !3
3
1 1 33
6
)
3 !3 2 !3
3
111
2 !3
3
2
1
2!3
3
5
3
2
2 !3
5
6
!3
5
3
f) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
p
p
p
p
p
p
cos a 1 b 5 cos cos 2 sin sin
2
3
2
3
2
3
1
!3
5 (0)a b 2 (1)a
b
2
2
4
7-8
Chapter 7: Trigonometric Identities and Equations
08-035_07_AFSM_C07_001-060.qxd
7/23/08
2:09 PM
Page 9
!3
2
!3
52
2
6. a) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
sin (p 1 x) 5 sin p cos x 1 cos p sin x
5 (0)(cos x) 1 (21)(sin x)
5 0 1 (2sin x)
5 0 2 sin x
5 2sin x
b) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
3p
3p
3p
b 5 cos x cos
2 sin x sin
cos ax 1
2
2
2
5 (cos x)(0) 2 (sin x)(21)
5 0 2 (2sin x) 5 sin x
c) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
p
p
p
cos ax 1 b 5 cos x cos 2 sin x sin
2
2
2
5 (cos x)(0) 2 (sin x)(1)
5 0 2 sin x
5 2sin x
tan a 1 tan b
,
d) Since tan (a 1 b) 5
1 2 tan a tan b
tan x 1 tan p
tan (x 1 p) 5
1 2 tan x tan p
tan x 1 0
5
1 2 (tan x)(0)
tan x 1 0
5
120
tan x
5
1
5 tan x
e) Since sin (a 2 b) 5 sin a cos b 2 cos a sin b,
sin (x 2 p) 5 sin x cos p 2 cos x sin p
5 (sin x)(21) 2 (cos x)(0)
5 2sin x 2 0
5 2sin x
tan a 2 tan b ,
f) Since tan (a 2 b) 5
1 1 tan a tan b
tan 2p 2 tan x
tan (2p 2 x) 5
1 1 tan 2p tan x
0 2 tan x
5
1 1 (0)(tan x)
2tan x
5
110
2tan x
5
1
5 2tan x
502
Advanced Functions Solutions Manual
7. a) Since sin (p 1 x) 5 sin (x 1 p), sin (p 1 x)
is equivalent to sin x translated p units to the left,
which is equivalent to 2sin x.
b) cos Qx 1
3p
2
3p
R
2
is equivalent to cos x translated
units to the left, which is equivalent to sin x.
c) cos Q x 1 R is equivalent to cos x translated
2
p
p
2
units to the left, which is equivalent to 2sin x.
d) tan (x 1 p) is equivalent to tan x translated
p units to the left, which is equivalent to tan x.
e) sin (x 2 p) is equivalent to sin x translated
p units to the right, which is equivalent to 2sin x.
f) Since tan (2p 2 x) 5 tan (2x 1 2p), and
since the period of the tangent function is 2p,
tan (2p 2 x) is equivalent to tan (2x), which is
equivalent to tan x reflected in the y-axis, which is
equivalent to 2tan x.
8. a) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
cos 75° 5 cos (30° 1 45°)
5 cos 30° cos 45° 2 sin 30° sin 45°
!3 !2
1 !2
5a
ba
b 2 a ba
b
2
2
2
2
!6
!2
5
2
4
4
!6 2 !2
5
4
tan a 2 tan b
,
b) Since tan (a 2 b) 5
1 1 tan a tan b
tan (215°) 5 tan (30° 2 45°)
tan 30° 2 tan 45°
5
1 1 tan 30° tan 45°
!3
!3
3
3 2 1
3 2 3
5
5
3
!3
1 1 ( !3
3 1 3
3 ) (1)
5
5
5
5
!3 2 3
3
3 1 !3
3
5
!3 2 3
3
3
3
3 1 !3
( !3 2 3)( 3 2 !3)
!3 2 3
5
3 1 !3
( 3 1 !3)( 3 2 !3)
3 !3 2 9 2 3 1 3!3
9 1 3 !3 2 3 !3 2 3
212 1 6 !3
6
5 22 1 !3
7-9
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Page 10
c) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
2p
9p
11p
5 cos a
1
b
cos
12
12
12
3p
p
b
5 cos a 1
6
4
3p
p
3p
p
2 sin sin
5 cos cos
6
4
6
4
!2
1 !2
!3
b a2
b 2 a ba
b
5a
2
2
2
2
!6
!2
52
2
4
4
2 !6 2 !2
5
4
d) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
13p
3p
10p
sin
5 sin a
1
b
12
12
12
p
5p
5 sin a 1
b
4
6
p
5p
p
5p
5 sin cos
1 cos sin
4
6
4
6
!3
!2 1
!2
b a2
b1a
ba b
2
2
2
2
!2
!6
1
52
4
4
5a
5
!2 2 !6
4
tan a 1 tan b
,
1 2 tan a tan b
7p
3p
4p
tan
5 tan a
1
b
12
12
12
p
p
5 tan a 1 b
4
3
e) Since tan (a 1 b) 5
p
p
1 tan
4
3
5
p
p
1 2 tan tan
4
3
tan
5
5
5
7-10
1 1 !3
1 2 (1)( !3)
1 1 !3
1 2 !3
( 1 1 !3)( 1 1 !3)
( 1 2 !3)( 1 1 !3)
5
1 1 !3 1 !3 1 3
1 2 !3 1 !3 2 3
4 1 2 !3
5 22 2 !3
22
tan a 2 tan b
,
f) Since tan (a 2 b) 5
1 1 tan a tan b
5p
4p
9p
3p
p
tan a2 b 5 tan a
2
b 5 tan a 2
b
12
12
12
3
4
5
p
3p
2 tan
3
4
5
p
3p
1 1 tan tan
3
4
tan
5
5
5
5
5
!3 2 (21)
1 1 ( !3) (21)
!3 1 1
1 2 !3
1 1 !3
1 2 !3
( 1 1 !3)( 1 1 !3)
( 1 2 !3)( 1 1 !3)
1 1 !3 1 !3 1 3
1 2 !3 1 !3 2 3
4 1 2 !3
5 22 2 !3
22
9. a) Since sin x 5 45, the leg opposite the angle x in a
right triangle has a length of 4, while the hypotenuse
of the right triangle has a length of 5. For this reason,
the other leg of the right triangle can be calculated as
follows:
x 2 1 42 5 52
x 2 1 16 5 25
2
x 1 16 2 16 5 25 2 16
x2 5 9
x 5 3, in quadrant I
5
Since cos x 5
adjacent leg
, cos x 5 35.
hypotenuse
In addition,
since sin y 5 2 12
13 , the leg opposite the angle y in a
right triangle has a length of 12, while the hypotenuse
of the right triangle has a length of 13. For this
reason, the other leg of the right triangle can be
calculated as follows:
x 2 1 122 5 132
x 2 1 144 5 169
x 2 1 144 2 144 5 169 2 144
Chapter 7: Trigonometric Identities and Equations
08-035_07_AFSM_C07_001-060.qxd
7/23/08
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Page 11
x 2 5 25
x 5 5, in quadrant IV
Since cos y 5
adjacent leg
, cos y 5
hypotenuse
5
13 .
Therefore,
since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
cos (x 1 y) 5 cos x cos y 2 sin x sin y
5
4
12
3
5 a b a b 2 a b a2 b
5 13
5
13
48
15
2 a2 b
5
65
65
48
15
1
5
65
65
63
5
65
b) Since sin x 5 45, the leg opposite the angle x in a
right triangle has a length of 4, while the hypotenuse
of the right triangle has a length of 5. For this reason,
the other leg of the right triangle can be calculated
as follows:
x 2 1 42 5 52
x 2 1 16 5 25
2
x 1 16 2 16 5 25 2 16
x2 5 9
x 5 3, in quadrant I
Since cos x 5
adjacent leg
, cos x 5 35.
hypotenuse
In addition,
since sin y 5 2 12
13 , the leg opposite the angle y in a
right triangle has a length of 12, while the hypotenuse
of the right triangle has a length of 13. For this reason,
the other leg of the right triangle can be calculated as
follows:
x 2 1 122 5 132
x 2 1 144 5 169
2
x 1 144 2 144 5 169 2 144
x 2 5 25
x 5 5, in quadrant IV
Since cos y 5
adjacent leg
, cos y 5
hypotenuse
5
13 .
Therefore,
since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
sin (x 1 y) 5 sin x cos y 1 cos x sin y
4
5
3
12
5 a b a b 1 a b a2 b
5 13
5
13
36
20
1 a2 b
5
65
65
36
20
2
5
65
65
16
52
65
Advanced Functions Solutions Manual
c) Since sin x 5 45, the leg opposite the angle x in a
right triangle has a length of 4, while the hypotenuse
of the right triangle has a length of 5. For this
reason, the other leg of the right triangle can be
calculated as follows:
x 2 1 42 5 52
x 2 1 16 5 25
2
x 1 16 2 16 5 25 2 16
x2 5 9
x 5 3, in quadrant I
Since cos x 5
adjacent leg
, cos x 5 35.
hypotenuse
In addition,
since sin y 5 2 12
13 , the leg opposite the angle y
in a right triangle has a length of 12, while the
hypotenuse of the right triangle has a length of 13.
For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 122 5 132
x 2 1 144 5 169
2
x 1 144 2 144 5 169 2 144
x 2 5 25
x 5 5, in quadrant IV
adjacent leg
Since cos y 5
, cos y 5 135 . Therefore,
hypotenuse
since cos (a 2 b) 5 cos a cos b 1 sin a sin b,
cos (x 2 y) 5 cos x cos y 1 sin x sin y
5
4
12
15
48
3
1 a2 b
5 a b a b 1 a b a2 b 5
5 13
5
13
65
65
15
48
33
2
52
5
65
65
65
d) Since sin x 5 45, the leg opposite the angle x in a
right triangle has a length of 4, while the hypotenuse
of the right triangle has a length of 5. For this reason,
the other leg of the right triangle can be calculated as
follows:
x 2 1 42 5 52
x 2 1 16 5 25
x 2 1 16 2 16 5 25 2 16
x2 5 9
x 5 3, in quadrant I
Since cos x 5
adjacent leg
, cos x 5 35.
hypotenuse
In addition,
since sin y 5 2 12
13 , the leg opposite the angle y
in a right triangle has a length of 12, while the
hypotenuse of the right triangle has a length of 13.
For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 122 5 132
x 2 1 144 5 169
7-11
08-035_07_AFSM_C07_001-060.qxd
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Page 12
x 2 1 144 2 144 5 169 2 144
x 2 5 25
x 5 5, in quadrant IV
Since cos y 5
adjacent leg
, cos y 5
hypotenuse
5
13 .
Therefore,
since sin (a 2 b) 5 sin a cos b 2 cos a sin b,
sin (x 2 y) 5 sin x cos y 2 cos x sin y
3
5
3
12
20
36
5 a b a b 2 a b a2 b 5
2 a2 b
5 13
5
13
65
65
36
56
20
1
5
5
65
65
65
e) Since sin x 5 45, the leg opposite the angle x in a
right triangle has a length of 4, while the hypotenuse
of the right triangle has a length of 5. For this reason,
the other leg of the right triangle can be calculated as
follows:
x 2 1 42 5 52
x 2 1 16 5 25
x 2 1 16 2 16 5 25 2 16
x2 5 9
x 5 3, in quadrant I
Since tan x 5
opposite leg
, tan x 5 43.
adjacent leg
In addition,
2 12
13 ,
since sin y 5
the leg opposite the angle y in a
right triangle has a length of 12, while the hypotenuse
of the right triangle has a length of 13. For this reason,
the other leg of the right triangle can be calculated as
follows:
x2 1 122 5 132
x2 1 144 5 169
x2 1 144 2 144 5 169 2 144
x2 5 25
x 5 5, in quadrant IV
opposite leg
Since tan y 5
, tan y 5 2 125. (The
adjacent leg
reason the sign is negative is because angle y is in
the fourth quadrant.) Therefore, since
tan a 1 tan b
,
tan (a 1 b) 5
1 2 tan a tan b
tan x 1 tan y
tan (x 1 y) 5
1 2 tan x tan y
4
12
20
36
1
2
(
)
3
5
15 1 ( 2 15 )
5
5
1 2 ( 43)( 2 125)
1 2 ( 2 48
15 )
2 36
2 16
15
15
48 5 63
1 1 15
15
16
15
16
52 3
52
15
63
63
5
20
15
7-12
f) Since sin x 5 45, the leg opposite the angle x in
a right triangle has a length of 4, while the hypotenuse
of the right triangle has a length of 5. For this reason,
the other leg of the right triangle can be calculated as
follows:
x2 1 42 5 52
x2 1 16 5 25
2
x 1 16 2 16 5 25 2 16
x2 5 9
x 5 3, in quadrant I
Since tan x 5
opposite leg
, tan x 5 43.
adjacent leg
In addition,
2 12
13 ,
since sin y 5
the leg opposite the angle y in a
right triangle has a length of 12, while the hypotenuse
of the right triangle has a length of 13. For this reason,
the other leg of the right triangle can be calculated as
follows:
x2 1 122 5 132
x2 1 144 5 169
2
x 1 144 2 144 5 169 2 144
x 2 5 25
x 5 5, in quadrant IV
Since tan y 5
opposite leg
, tan y 5 2 125.
adjacent leg
(The
reason the sign is negative is because angle y is in
the fourth quadrant.) Therefore, since
tan a 2 tan b
,
tan (a 2 b) 5
1 1 tan a tan b
tan x 2 tan y
tan (x 2 y) 5
1 1 tan x tan y
5
4
3
2 ( 2 125)
1 1 ( 43)( 2 125)
5
20
15
1 36
15
1 1 ( 2 48
15 )
5
56
15
2 33
15
56
15
56
3 a2 b 5 2
15
33
33
7
10. Since sin a 5 25, the leg opposite the angle a in
a right triangle has a length of 7, while the hypotenuse
of the right triangle has a length of 25. For this reason,
the other leg of the right triangle can be calculated as
follows:
x 2 1 72 5 252
x 2 1 49 5 625
2
x 1 49 2 49 5 625 2 49
x 2 5 576
x 5 24, in quadrant I
5
adjacent leg
, cos a 5 24
25 . Also, since
hypotenuse
opposite leg
tan a 5
, tan a 5 247 . In addition, since
adjacent leg
Since cos a 5
Chapter 7: Trigonometric Identities and Equations
08-035_07_AFSM_C07_001-060.qxd
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Page 13
cos b 5 135 , the leg adjacent to the angle b in a right
triangle has a length of 5, while the hypotenuse of
the right triangle has a length of 13. For this reason,
the other leg of the right triangle can be calculated
as follows:
52 1 y 2 5 132
25 1 y 2 5 169
25 1 y 2 2 25 5 169 2 25
y 2 5 144
x 5 12, in quadrant I
opposite leg
, sin b 5 12
13 . Also,
hypotenuse
opposite leg
tan b 5
, tan b 5 125. Therefore,
adjacent leg
Since sin b 5
since
since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
sin(a 1 b) 5 sin a cos b 1 cos a sin b
5
24 12
7
5 a ba b 1 a ba b
25 13
25 13
288
323
35
1
5
5
325
325
325
tan a 1 tan b
Also, since tan (a 1 b) 5
,
1 2 tan a tan b
tan a 1 tan b
tan (a 1 b) 5
1 2 tan a tan b
7
1 12
5 24 7 5 12
1 2 ( 24)( 5 )
35
120
1 288
120
84 5
1 2 120
5 sin p cos x 1 cos p sin x 1 sin p cos x
2 cos p sin x
5 2 sin p cos x 5 (2)(0)(sin x) 5 0
b) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
and since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
the expression cos Q x 1 R 2 sin Q x 1 R can be
3
6
p
p
simplified as follows:
p
p
cos x cos 2 sin x sin
3
3
p
p
2 asin x cos 1 cos x sin b
6
6
p
p
p
5 cos x cos 2 sin x sin 2 sin x cos
3
3
6
p
2 cos x sin
6
!3
1
5 a b (cos x) 2 a
b (sin x)
2
2
!3
1
2a
b (sin x) 2 a b (cos x)
2
2
5 2 !3 sin x
13. Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
since sin (a 2 b) 5 sin a cos b 2 cos a sin b, since
cos (a 1 b) 5 cos a cos b 2 sin a sin b, and since
cos (a 2 b) 5 cos a cos b 1 sin a sin b, the
expression
sin (f 1 g) 1 sin (f 2 g)
cos (f 1 g) 1 cos (f 2 g)
can be
323
120
36
120
simplified as follows:
sin (f 1 g) 1 sin (f 2 g)
cos (f 1 g) 1 cos (f 2 g)
120
323
323
3
5
5
sin f cos g 1 cos f sin g 1 sin f cos g 2 cos f sin g
120
36
36
5
cos f cos g 2 sin f sin g 1 cos f cos g 1 sin f sin g
2 sin f cos g
sin f
11. a) Since cos (a 2 b) 5 cos a cos b 1 sin a sin b,
5
5
2 cos f cos g
cos f
p
p
p
cos a 2 xb 5 cos cos x 1 sin sin x
5
tan
f,
cos
f
2
0,
cos g 2 0
2
2
2
5
5 (0)(cos x) 1 (1)(sin x)
5 0 1 sin x 5 sin x
b) Since sin (a 2 b) 5 sin a cos b 2 cos a sin b,
p
p
p
sin a 2 xb 5 sin cos x 2 cos sin x
2
2
2
5 (1)(cos x) 2 (0)(sin x)
5 cos x 2 0 5 cos x
12. a) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
and since sin (a 2 b) 5 sin a cos b 2 cos a sin b,
the expression sin (p 1 x) 1 sin (p 2 x) can be
simplified as follows:
sin (p 1 x) 1 sin (p 2 x)
Advanced Functions Solutions Manual
y
14. If sin a and sin b are written as r , where y is
the side opposite angles a and b in a right triangle,
and r is the radius of the right triangle, the side x
adjacent to angles a and b can be found with the
formula x 2 1 y 2 5 r 2. Once x is determined, cos a
and cos b can be written as rx, and since the
terminal arms of angles a and b lie in the first
quadrant, cos a and cos b are positive. With
sin a, sin b, cos a, and cos b known, cos (a 1 b)
can be found with the formula
cos (a 1 b) 5 cos a cos b 2 sin a sin b.
7-13
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Page 14
Therefore, an appropriate flow chart would be as
follows:
y
Write sin a in terms of r .
Solve for x using the Pythagorean
theorem, x 2 1 y 2 5 r 2.
p
Since aP S 0, T , choose the positive
2
value of x and determine cos a.
y
Write sin b in terms of r .
Solve for x using the Pythagorean
theorem, x 2 1 y 2 5 r 2.
p
Since bP S 0, T , choose the positive
2
value of x and determine cos b.
Use the formula
cos (a 1 b) 5 cos a cos b 2 sin a sin b
to evaluate cos (a 1 b).
15. The compound angle formulas used in this
lesson are as follows:
sin (x 1 y) 5 sin x cos y 1 cos x sin y
sin (x 2 y) 5 sin x cos y 2 cos x sin y
cos (x 1 y) 5 cos x cos y 2 sin x sin y
cos (x 2 y) 5 cos x cos y 1 sin x sin y
tan x 1 tan y
tan (x 1 y) 5
1 2 tan x tan y
tan x 2 tan y
tan (x 2 y) 5
1 1 tan x tan y
The two sine formulas are the same, except for the
operators. The operator for sin (x 1 y) is 1, while the
operator for sin (x 2 y) is 2. Remembering that the
same operator is used on both the left and right sides in
both equations will help you remember the formulas.
Similarly, the two cosine formulas are the same, except
for the operators. The operator for cos (x 1 y) is 2,
while the operator for cos (x 2 y) is 1.
Remembering that the operator on the left side is the
opposite of the operator on the right side in both
7-14
equations will help you remember the formulas. The
two tangent formulas are the same, except for the
operators in the numerator and the denominator on the
right side. The operators for tan (x 1 y) are 1 in the
numerator and – in the denominator, while the
operators for tan (x 2 y) are 2 in the numerator
and 1 in the denominator. Remembering that the
operators in the numerator and the denominator are
opposite in both equations, and that the operator in the
numerator is the same as the operator on the left side,
will help you remember the formulas.
16. Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
and since cos (a 2 b) 5 cos a cos b 1 sin a sin b,
the formula
sin C 1 sin D 5 2 sin Q
C1D
C2D
R cos Q
R
2
2
can
be developed as follows:
C2D
C1D
b cos a
b
2 sin a
2
2
D
C
D
C
5 (2)aasin b acos b 1 acos b asin bb
2
2
2
2
D
C
D
C
3 aacos b acos b 1 asin b asin bb
2
2
2
2
C
D
C
D
5 (2) aasin b acos b acos2 b 1 asin b
2
2
2
2
3 acos
D
C
D
D
b acos2 b 1 asin b acos b
2
2
2
2
3 asin2
C
C
D
C
b 1 asin b acos b asin2 bb
2
2
2
2
C
D
D
C
5 2asin b acos b acos2 1 sin2 b
2
2
2
2
D
D
C
C
1 2asin b acos b acos2 1 sin2 b
2
2
2
2
C
C
D
D
5 2asin b acos b 1 2asin b acos b
2
2
2
2
C
D
5 sin a2a bb 1 sin a2a bb
2
2
5 sin C 1 sin D
tan a 1 tan b
,
17. Since tan (a 1 b) 5
1 2 tan a tan b
cot (x 1 y) in terms of cot x and cot y can be
determined as follows:
cot (x 1 y)
1
1
1 2 tan x tan y
5
5 tan x 1 tan y 5
tan (x 1 y)
tan x 1 tan y
1 2 tan x tan y
Chapter 7: Trigonometric Identities and Equations
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Page 15
cot x cot y
1
cot x cot y 2 cot x cot y
5 cot y
5 1
cot x
1
cot x cot y 1 cot x cot y
cot x 1 cot y
cot x cot y 2 1
cot x cot y 2 1
cot x cot y
cot x cot y
5 cot x 1 cot y 5
3
cot x cot y
cot x 1 cot y
cot x cot y
1
1 2 cot x cot y
cot x cot y 2 1
cot x 1 cot y
18. Let C 5 x 1 y and let D 5 x 2 y.
cos C 1 cos D 5 cos (x 1 y) 1 cos (x 2 y)
5 cos x cos y 2 sin x sin y
1 cos x cos y 1 sin x sin y 5 2 cos x cos y
x1y1x2y
C1D
5
5x
2
2
x1y2x1y
C2D
5
5y
2
2
C2D
C1D
bcos a
b.
So, cos C 1 cos D 5 2 cos a
2
2
19. Let C 5 x 1 y and let D 5 x 2 y.
cos C 2 cos D 5 cos (x 1 y) 2 cos (x 2 y)
5 cos x cos y 2 sin x sin y
2 (cos x cos y 2 sin x sin y) 5 22 sin x sin y
x1y1x2y
C1D
5
5x
2
2
x1y2x1y
C2D
5
5y
2
2
C1D
C2D
bsin a
b.
So, cos C 2 cos D 5 22 sin a
2
2
5
7.3 Double Angle Formulas,
pp. 407–408
1. a) Since sin 2u 5 2 sin u cos u,
2 sin 5x cos 5x 5 sin 2(5x) 5 sin 10x.
b) Since cos 2u 5 cos2 u 2 sin2 u,
cos2 u 2 sin2 u 5 cos 2u.
c) Since cos 2u 5 1 2 2 sin2 u,
1 2 2 sin2 3x 5 cos 2(3x) 5 cos 6x.
2 tan u
,
d) Since tan 2u 5
1 2 tan2 u
2 tan 4x
5 tan 2(4x) 5 tan 8x.
1 2 tan2 4x
e) Since sin 2u 5 2 sin u cos u,
4 sin u cos u 5 (2)(2 sin u cos u) 5 2 sin 2u.
f) Since cos 2u 5 2 cos2 u 2 1,
u
u
2 cos2 2 1 5 cos 2a b 5 cos u.
2
2
Advanced Functions Solutions Manual
2. a) Since sin 2u 5 2 sin u cos u,
2 sin 45° cos 45° 5 sin 2(45°) 5 sin 90° 5 1
b) Since cos 2u 5 cos2 u 2 sin2 u,
cos2 30° 2 sin2 30° 5 cos 2(30°) 5 cos 60° 5
1
2
c) Since sin 2u 5 2 sin u cos u,
p
p
p
p
1
2 sin
cos
5 sin 2a b 5 sin 5
12
12
12
6
2
d) Since cos 2u 5 cos2 u 2 sin2 u,
p
p
p
p
!3
2 sin2
5 cos 2a b 5 cos 5
cos2
12
12
12
6
2
2
e) Since cos 2u 5 1 2 2 sin u,
3p
6p
3p
1 2 2 sin2
5 cos 2a b 5 cos
8
8
8
3p
!2
5 cos
52
4
2
f) Since sin 2u 5 2 sin u cos u,
sin 60°
2 tan 60° cos2 60° 5 (2)a
b (cos2 60°)
cos 60°
5 2 sin 60° cos 60° 5 sin 2(60°)
!3
2
3. a) Since sin 2u 5 2 sin u cos u,
5 sin 120° 5
sin 4u 5 2 sin 2u cos 2u
b) Since cos 2u 5 2 cos2 u 2 1,
cos 3x 5 2 cos2 (1.5x) 2 1
2 tan u
,
c) Since tan 2u 5
1 2 tan2 u
2 tan (0.5x)
tan x 5
1 2 tan2 (0.5x)
d) Since cos 2u 5 cos2 u 2 sin2 u,
cos 6u 5 cos2 3u 2 sin2 3u
e) Since sin 2u 5 2 sin u cos u,
sin x 5 2 sin (0.5x) cos (0.5x)
2 tan u
,
f) Since tan 2u 5
1 2 tan2 u
2 tan (2.5u)
tan 5u 5
1 2 tan2 (2.5u)
4. Since cos u 5 35, the leg adjacent to the angle u
in a right triangle has a length of 3, while the
hypotenuse of the right triangle has a length of 5.
For this reason, the other leg of the right triangle
can be calculated as follows:
32 1 y 2 5 52
9 1 y 2 5 25
9 1 y 2 2 9 5 25 2 9
7-15
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Page 16
y 2 5 16
y 5 4, in quadrant I
Since sin u 5
opposite leg
, sin u 5 45.
hypotenuse
Therefore, since sin 2u 5 2 sin u cos u,
4 3
24
sin 2u 5 (2)a b a b 5 .
5 5
25
Also, since cos 2u 5 cos2 u 2 sin2 u,
3 2
4 2
cos 2u 5 a b 2 a b
5
5
9
16
7
5
2
52 .
25 25
25
Finally, since tan u 5
opposite leg
, tan u 5 43.
adjacent leg
2 tan u
,
Since tan 2u 5
1 2 tan2 u
(2)( 43 )
tan 2u 5
1 2 ( 43 )2
5
8
3
16
9
5
8
3
9
9
16
9
5
8
3
12
2
2 79
8
9
24
5 3 2 52
3
7
7
7
5. Since tan u 5 2 24, the leg opposite the angle u
in a right triangle has a length of 7, while the leg
adjacent to the angle u has a length of 24. For this
reason, the hypotenuse of the right triangle can be
calculated as follows:
72 1 242 5 c 2
49 1 576 5 c 2
625 5 c 2
c 5 25, in quadrant II
opposite leg
Since sin u 5
, sin u 5 257 , and since
hypotenuse
adjacent leg
cos u 5
, cos u 5 2 24
25 . (The reason the
hypotenuse
sign is negative is because angle u is in the second
quadrant.) Therefore, since sin 2u 5 2 sin u cos u,
336
sin 2u 5 (2)( 257 ) (2 24
25 ) 5 2 625 . Also, since
cos 2u 5 cos2 u 2 sin2 u,
cos 2u 5 a2
7 2
24 2
b 2a b
25
25
576
49
527
5
2
5
625
625
625
Finally, since
2 tan u
tan 2u 5
,
1 2 tan2 u
(2)(2 247 )
2 127
tan 2u 5
7 2 5
49
1 2 (2 24 )
1 2 576
7-16
2 127
2 127
5
576
49
527
576 2 576
576
7
576
336
52 3
52
12
527
527
,
6. Since sin u 5 2 12
the
leg
opposite the angle u
13
in a right triangle has a length of 12, while the
hypotenuse of the right triangle has a length of 13.
For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 122 5 132
x 2 1 144 5 169
2
x 1 144 2 144 5 169 2 144
x 2 5 25
x 5 5, in quadrant IV
5
Since cos u 5
adjacent leg
, cos u 5
hypotenuse
5
13 .
Therefore,
since sin 2u 5 2 sin u cos u,
5
120
sin 2u 5 (2)( 2 12
13 )( 13 ) 5 2 169 . Also, since
2
2
cos 2u 5 cos u 2 sin u,
2
25
144
119
cos 2u 5 ( 135 )2 2 ( 2 12
13 ) 5 169 2 169 5 2 169 . Finally,
since tan u 5
opposite leg
, tan u 5 2 125.
adjacent leg
(The reason
the sign is negative is because angle u is in the fourth
quadrant.) Since
2 tan u
,
tan 2u 5
1 2 tan2 u
2 245
(2)(2 125 )
2 245
2 245
5
tan 2u 5
5
5
25
144
1 2 (2 125 )2
1 2 144
2 119
25
25 2 25
25
5
24
25
120
2 245
5 2 32
5
2 119
5
119
119
25
7. Since cos u 5 2 45, the leg adjacent to the angle u
in a right triangle has a length of 4, while the
hypotenuse of the right triangle has a length of 5.
For this reason, the other leg of the right triangle
can be calculated as follows:
42 1 y 2 5 52
16 1 y 2 5 25
16 1 y 2 2 16 5 25 2 16
y2 5 9
y 5 3, in quadrant II
Since sin u 5
opposite leg
, sin u 5 35.
hypotenuse
Therefore, since sin 2u 5 2 sin u cos u,
sin 2u 5 (2)( 35)( 2 45) 5 2 24
25 .
Also, since cos 2u 5 cos2 u 2 sin2 u,
Chapter 7: Trigonometric Identities and Equations
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Page 17
4 2
3 2
cos 2u 5 a2 b 2 a b
5
5
9
7
16
2
5
5
25
25
25
Finally, since tan u 5
opposite leg
, tan u 5 2 34.
adjacent leg
2 tan u
tan 2u 5
,
1 2 tan2 u
sin
(2)(2 34 )
2 32
2 32
5
5
16
9
1 2 (2 34 )2
1 2 169
16 2 16
3
16
24
23
52
5 72 5 2 3
2
7
7
16
tan 2u 5
8. To determine the value of a, first rearrange the
equation as follows:
2 tan x 2 tan 2x 1 2a 5 1 2 tan 2x tan2 x
2 tan x 2 tan 2x 1 2a 1 tan 2x
5 1 2 tan 2x tan2 x 1 tan 2x
2 tan x 1 2a 5 1 2 tan 2x tan2 x 1 tan 2x
2 tan x 1 2a 2 1 5 1 2 tan 2x tan2 x 1 tan 2x 2 1
2 tan x 1 2a 2 1 5 tan 2x 2 tan 2x tan2 x
2 tan x 1 2a 2 1 5 (tan 2x)(1 2 tan2 x)
2 tan x 1 2a 2 1
(tan 2x)(1 2 tan2 x)
5
1 2 tan2 x
1 2 tan2 x
2 tan x 1 2a 2 1
tan 2x 5
1 2 tan2 x
2 tan u
Since tan 2u 5
, the value of 2a 2 1 must
1 2 tan2 u
equal 0. Therefore, a can be solved for as follows:
2a 2 1 5 0
2a 2 1 1 1 5 0 1 1
2a 5 1
2a
1
5
2
2
1
a5
2
p
9. Jim can find the sine of by using the formula
8
cos 2x 5 1 2 2 sin2 x and isolating sin x on one side
of the equation as follows:
cos 2x 5 1 2 2 sin2 x
cos 2x 1 2 sin2 x 5 1 2 2 sin2 x 1 2 sin2 x
cos 2x 1 2 sin2 x 5 1
cos 2x 1 2 sin2 x 2 cos 2x 5 1 2 cos 2x
2 sin2 x 5 1 2 cos 2x
1 2 cos 2x
2 sin2 x
5
2
2
is
!2
2 ,
so
p
1 2 cos ((2)( p8 ))
56
8
Å
2
56
56
56
1 2 cos p4
1 2 "2
2
5 6
Å
2
Å 2
2 2 "2
2 "2
2
2
5
6
Å 2
Å 2
2
2
1
2 2 !2
3
2
2
Å
2 2 !2
"2 2 !2
56
Å
4
2
p
p
Since is in the first quadrant, the sign of sin is
56
8
positive. Therefore, sin
p
5
8
8
"2 2 !2
.
2
10. Marion can find the cosine of
2
p
12
by using the
formula cos 2x 5 2 cos x 2 1 and isolating cos x
on one side of the equation as follows:
cos 2x 5 2 cos2 x 2 1
cos 2x 1 1 5 2 cos2 x 2 1 1 1
cos 2x 1 1 5 2 cos2 x
2 cos2 x
cos 2x 1 1
5
2
2
1 1 cos 2x
cos2 x 5
2
1 1 cos 2x
cos x 5 6
Å
2
The cosine of
cos
p
6
is
!3
,
2
so
p
p
1 1 cos ((2)( 12
))
56
12
Å
2
56
1 1 cos p6
11
56
2
Å
Å 2
!3
2
2 1 !3
1 !3
2
2
5
6
Å 2
Å 2
1
2 1 !3
3
56
2
2
Å
56
56
Advanced Functions Solutions Manual
p
4
The cosine of
(The reason the sign is negative is because angle u is
in the second quadrant.) Since
1 2 cos 2x
2
1 2 cos 2x
sin x 5 6
2
Å
sin2 x 5
2
2
2 1 !3
"2 1 !3
56
Å
4
2
7-17
08-035_07_AFSM_C07_001-060.qxd
Since
p
12
7/18/08
12:55 PM
Page 18
is in the first quadrant, the sign of cos
is positive. Therefore, cos
"2 1 !3
p
5
.
12
2
p
12
11. a) Since sin 2u 5 2 sin u cos u,
sin 4x 5 2 sin 2x cos 2x
5 (2)(2 sin x cos x)(cos 2x). At this point,
either the formula cos 2u 5 2 cos2 u 2 1,
cos 2u 5 cos2 u 2 sin2 u, or cos 2u 5 1 2 2 sin2 u
can be used to simplify for cos 2x. If the formula
cos 2u 5 2 cos2 u 2 1 is used, the formula for sin 4x
can be developed as follows:
sin 4x 5 (2)(2 sin x cos x)(cos 2x)
5 (2)(2 sin x cos x)(2 cos2 x 2 1)
5 (4 sin x cos x)(2 cos2 x 2 1)
5 8 cos3 x sin x 2 4 sin x cos x
If the formula cos 2u 5 cos2 u 2 sin2 u is used, the
formula for sin 4x can be developed as follows:
sin 4x 5 (2)(2 sin x cos x)(cos 2x)
5 (2)(2 sin x cos x)(cos2 x 2 sin2 x)
5 (4 sin x cos x)(cos2 x 2 sin2 x)
5 4 cos3 x sin x 2 4 sin3 x cos x
If the formula cos 2u 5 1 2 2 sin2 u is used, the
formula for sin 4x can be developed as follows:
sin 4x 5 (2)(2 sin x cos x)(cos 2x)
5 (2)(2 sin x cos x)(1 2 2 sin2 x)
5 (4 sin x cos x)(1 2 2 sin2 x)
5 4 sin x cos x 2 8 sin3 x cos x
2p
!3
b) The value of sin
is
. Using the formula
3
2
sin 4x 5 4 sin x cos x 2 8 sin3 x cos x,
sin
2p
8p
5 sin
3
3
can be verified as follows:
sin 4x 5 4 sin x cos x 2 8 sin3 x cos x
2p
2p
2p
2p
2p
sin 4a b 5 4 sin
cos
2 8 sin3
cos
3
3
3
3
3
3
!3
1
!3
8p
1
5 (4)a
b a2 b 2 (8)a
b a2 b
sin
3
2
2
2
2
sin
4!3
3 !3
8p
52
2 (24)a
b
3
4
8
sin
4!3
3!3
8p
52
2 a2
b
3
4
2
sin
4!3
6!3
8p
52
2 a2
b
3
4
4
sin
7-18
4!3
6 !3
8p
52
1
3
4
4
8p
2 !3
5
3
4
!3
8p
5
sin
3
2
sin
12. a) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
sin 3u 5 sin (2u 1 u) 5 sin 2u cos u 1 cos 2u sin u.
Since sin 2u 5 2 sin u cos u,
sin 3u 5 (2 sin u cos u) (cos u) 1 cos 2u sin u
5 2 cos2 u sin u 1 cos 2u sin u. At this point, either
the formula cos 2u 5 2 cos2 u 2 1,
cos 2u 5 cos2 u 2 sin2 u, or cos 2u 5 1 2 2 sin2 u
can be used to simplify for cos 2u. If the formula
cos 2u 5 2 cos2 u 2 1 is used, the formula for
sin 3u can be developed as follows:
sin 3u 5 2 cos2 u sin u 1 cos 2u sin u
sin 3u 5 2 cos2 u sin u 1 (2 cos2 u 2 1)(sin u)
sin 3u 5 2 cos2 u sin u 1 2 cos2 u sin u 2 sin u
sin 3u 5 4 cos2 u sin u 2 sin u
If the formula cos 2u 5 cos2 u 2 sin2 u is used, the
formula for sin 3u can be developed as follows:
sin 3u 5 2 cos2 u sin u 1 cos 2u sin u
sin 3u 5 2 cos2 u sin u 1 (cos2 u 2 sin2 u)(sin u)
sin 3u 5 2 cos2 u sin u 1 cos2 u sin u 2 sin3 u
sin 3u 5 3 cos2 u sin u 2 sin3 u
If the formula cos 2u 5 1 2 2 sin2 u is used, the
formula for sin 3u can be developed as follows:
sin 3u 5 2 cos2 u sin u 1 cos 2u sin u
sin 3u 5 2 cos2 u sin u 1 (1 2 2 sin2 u)(sin u)
sin 3u 5 2 cos2 u sin u 1 sin u 2 2 sin3 u
b) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
cos 3u 5 cos (2u 1 u) 5 cos 2u cos u 2 sin u sin 2u.
Since sin 2u 5 2 sin u cos u,
cos 3u 5 cos 2u cos u 2 (sin u)(2 sin u cos u)
5 cos 2u cos u 2 2 sin2 u cos u. At this point, either
the formula cos 2u 5 2 cos2 u 2 1,
cos 2u 5 cos2 u 2 sin2 u, or cos 2u 5 1 2 2 sin2 u
can be used to simplify for cos 2u. If the formula
cos 2u 5 2 cos2 u 2 1 is used, the formula for
cos 3u can be developed as follows:
cos 3u 5 cos 2u cos u 2 2 sin2 u cos u
cos 3u 5 (2 cos2 u 2 1)(cos u) 2 2 sin2 u cos u
cos 3u 5 2 cos3 u 2 cos u 2 2 sin2 u cos u
If the formula cos 2u 5 cos2 u 2 sin2 u is used, the
formula for cos 3u can be developed as follows:
cos 3u 5 cos 2u cos u 2 2 sin2 u cos u
cos 3u 5 (cos2 u 2 sin2 u)(cos u) 2 2 sin2 u cos u
cos 3u 5 cos3 u 2 sin2 u cos u 2 2 sin2 u cos u
cos 3u 5 cos3 u 2 3 sin2 u cos u
Chapter 7: Trigonometric Identities and Equations
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Page 19
If the formula cos 2u 5 1 2 2 sin2 u is used, the
formula for cos 3u can be developed as follows:
cos 3u 5 cos 2u cos u 2 2 sin2 u cos u
cos 3u 5 (1 2 2 sin2 u)(cos u) 2 2 sin2 u cos u
cos 3u 5 cos u 2 2 sin2 u cos u 2 2 sin2 u cos u
cos 3u 5 cos u 2 4 sin2 u cos u
tan a 1 tan b
c) Since tan (a 1 b) 5
,
1 2 tan a tan b
tan 2u 1 tan u
tan 3u 5 tan (2u 1 u) 5
1 2 tan 2u tan u
2 tan u
Since tan 2u 5
,
1 2 tan2 u
tan 3u 5
5
5
5
5
2 tan u
1 tan u
1 2 tan2 u
12a
2 tan u
b (tan u)
1 2 tan2 u
2 tan u
tan u 2 tan3 u
1
2
1 2 tan u
1 2 tan2 u
2 tan2 u
12
1 2 tan2 u
2 tan u
tan u 2 tan3 u
1
1 2 tan2 u
1 2 tan2 u
1 2 tan2 u
2 tan2 u
2
2
1 2 tan u
1 2 tan2 u
2 tan u 1 tan u 2 tan3 u
1 2 tan2 u
1 2 tan2 u 2 2 tan2 u
1 2 tan2 u
3 tan u 2 tan3 u
1 2 tan2 u
1 2 3 tan2 u
1 2 tan2 u
1 2 tan2 u
3 tan u 2 tan3 u
3
1 2 tan2 u
1 2 3 tan2 u
3
3 tan u 2 tan u
5
1 2 3 tan2 u
5
13. a) Since sin2 x 5 89, and since the angle x is in
2 !2
the second quadrant, sin x 5 # 89 5 !8
!9 5 3 .
Since sin x 5 2 !2
3 , the leg opposite the angle x in a
right triangle has a length of 2 !2, while the
hypotenuse of the right triangle has a length of 3.
For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 A2 !2 B 2 5 32
x2 1 8 5 9
2
x 18285928
x2 5 1
x 5 21, in quadrant II
Advanced Functions Solutions Manual
Since cos x 5
adjacent leg
,
hypotenuse
and since the angle x is in
the second quadrant, cos x 5 2 13.
Since sin 2u 5 2 sin u cos u,
2 !2
1
4!2
sin 2x 5 2 sin x cos x 5 (2)a
b a2 b 5 2
3
3
9
b) Since sin2 x 5 89, and since the angle x is in the
2 !2
second quadrant, sin x 5 # 89 5 !8
!9 5 3 .
Since sin x 5 2 !2
3 , the leg opposite the angle x in
a right triangle has a length of 2 !2, while the
hypotenuse of the right triangle has a length of 3.
For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 (2 !2)2 5 32
x2 1 8 5 9
2
x 18285928
x2 5 1
x 5 21, in quadrant II
Since cos x 5
adjacent leg
,
hypotenuse
and since the angle x is
in the second quadrant, cos x 5 2 13. Since
cos 2u 5 cos2 u 2 sin2 u,
cos 2x 5 cos2 x 2 sin2 x
2 !2 2
1 2
b
5 a2 b 2 a
3
3
8
7
1
5 2 52
9
9
9
(The formulas cos 2u 5 2 cos2 u 2 1 and cos 2u
5 1 2 2 sin2 u could also have been used.)
c) Since sin2 x 5 89, and since the angle x is in the
second quadrant, sin x 5 # 89 5
!8
!9
5 2 !2
3 .
Since sin x 5 2 !2
3 , the leg opposite the angle x in a
right triangle has a length of 2 !2, while the
hypotenuse of the right triangle has a length of 3.
For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 (2 !2)2 5 32
x2 1 8 5 9
2
x 18285928
x2 5 1
x 5 21, in quadrant II
Since cos x 5
adjacent leg
,
hypotenuse
and since the angle x is
in the second quadrant, cos x 5 2 13. Since
u
2
cos 2u 5 2 cos2 u 2 1, cos u 5 2 cos2 2 1, and
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Page 20
u
2
since cos x 5 2 13, 2 13 5 2 cos2 2 1. The value of
u
2
cos can now be determined as follows:
1
u
2 5 2 cos2 2 1
3
2
u
1
2 1 1 5 2 cos2 2 1 1 1
3
2
u
2
5 2 cos2
3
2
2
u
3
2 cos2 2
5
2
2
u
2
1
cos2 5 3
2
3
2
u
2
1
cos2 5 5
2
6
3
cos
angle a can be found with the formula x 2 1 y 2 5 r 2.
x
Once x is determined, cos a can be written as r , and
since the terminal arm of angle a lies in the second
quadrant, cos a is negative. With sin a and cos a
known, sin 2a can be found with the formula
sin 2u 5 2 sin u cos u, Therefore, an appropriate
flow chart would be as follows:
y
Write sin a in terms of r .
Solve for x using the Pythagorean
theorem, x 2 1 y 2 5 r 2.
Choose the negative value of x since
aP c , pd , and determine cos a.
p
2
u
1
!1
1
!3
5
5
5
5
2 Ä3
!3
!3
3
x
Write cos a in terms of r .
d) Since sin2 x 5 89, and since the angle x is in the
2 !2
second quadrant, sin x 5 # 89 5 !8
!9 5 3 . Since
sin x 5 2 !2
3 , the leg opposite the angle x in a right
triangle has a length of 2 !2, while the hypotenuse
of the right triangle has a length of 3. For this
reason, the other leg of the right triangle can be
calculated as follows:
x 2 1 (2!2)2 5 32
x2 1 8 5 9
x2 1 8 2 8 5 9 2 8
x2 5 1
x 5 21, in quadrant II
Since cos x 5
adjacent leg
,
hypotenuse
and since the angle x is
in the second quadrant, cos x 5 2 13. Since
sin 3u 5 3 cos2 u sin u 2 sin3 u,
sin 3x 5 3 cos2 x sin x 2 sin3 x
1 2 2!2
2 !2 3
5 (3)a2 b a
b2a
b
3
3
3
1 2!2
16!2
5 (3)a b a
b2
9
3
27
5
6 !2
16 !2
2
27
27
52
15. a) Since sin 2u 5 2 sin u cos u,
sin 2x 5 2 sin x cos x. For this reason,
sin 2x
2 sin x cos x
5
5 sin x cos x.
2
2
Therefore, the
graph of f(x) 5 sin x cos x is the same as that of
sin 2x
.
2
sin 2x
The graph of f(x) 5
can be
2
obtained by vertically compressing f(x) 5 sin x by
a factor of 12 and horizontally compressing it by a
factor of 12. The graph is shown below:
f(x) 5
0.5
–p
–p
2
0
y
x
p
2
p
–0.5
10!2
27
y
14. If sin a is written as r , where y is the side
opposite angle a in a right triangle, and r is the
radius of the right triangle, the side x adjacent to
7-20
Use the formula sin 2a 5 2 sin a cos a
to evaluate sin 2a.
b) Since cos 2u 5 2 cos2 u 2 1,
cos 2x 5 2 cos2 x 2 1. For this reason,
cos 2x 1 1 5 2 cos2 x 2 1 1 1 5 2 cos2 x.
Chapter 7: Trigonometric Identities and Equations
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Page 21
Therefore, the graph of f(x) 5 2 cos2 x is the same
as that of f(x) 5 cos 2x 1 1. The graph of
f(x) 5 cos 2x 1 1 can be obtained by horizontally
compressing f(x) 5 cos x by a factor of 12 and
vertically translating it 1 unit up. The graph is
shown below:
y
2
1
–p
–p
2
x
0
p
2
p
–1
2 tan u
2 tan x
, tan 2x 5
.
1 2 tan2 u
1 2 tan2 x
2 tan x
tan 2x
1 2 tan2 x
5
For this reason,
2
2
2 tan x
1
tan x
5
3 5
. Therefore, the graph of
1 2 tan2 x
2
1 2 tan2 x
tan x
tan 2x
f(x) 5
.
is the same as that of f(x) 5
1 2 tan2 x
2
tan 2x
The graph of f(x) 5
can be obtained by
2
c) Since tan 2u 5
vertically compressing f(x) 5 tan x by
a factor of 12 and horizontally compressing it by a
factor of 12. The graph is shown below:
4
y
2
–p
–p
2
0
–2
x
p
2
p
–4
16. a) To eliminate A from the equations x 5 tan 2A
and y 5 tan A to find an equation that relates x to y,
first take the tan21 of both sides of both equations
and solve for A as follows:
x 5 tan 2A
tan21 x 5 tan21 (tan 2A)
Advanced Functions Solutions Manual
tan21 x 5 2A
tan21 x
A5
2
y 5 tan A
tan21 y 5 tan21 (tan A)
A 5 tan21 y
tan21 x
Since A 5
and A 5 tan21 y,
2
tan21 x
5 tan21 y
2
b) To eliminate A from the equations x 5 cos 2A
and y 5 cos A to find an equation that relates x to y,
first take the cos21 of both sides of both equations
and solve for A as follows:
x 5 cos 2A
cos21 x 5 cos21 (cos 2A)
cos21 x 5 2A
cos21 x
A5
2
y 5 cos A
cos21 y 5 cos21 (cos A)
A 5 cos21 y
cos21 x
Since A 5
and A 5 cos21 y,
2
cos21 x
5 cos21 y
2
c) To eliminate A from the equations x 5 cos 2A
and y 5 csc A to find an equation that relates x to y,
first take the cos21 of both sides of the first equation
and the csc21 of both sides of the second equation
and solve for A as follows:
x 5 cos 2A
cos21 x 5 cos21 (cos 2A)
cos21 x 5 2A
cos21 x
A5
2
A 5 csc A
csc21 y 5 csc21 (csc A)
A 5 csc21 y
cos21 x
Since A 5
and A 5 cos21 y,
2
1
cos21 x
cos21 x
5 csc21 y, or
5 sin21 a b
y
2
2
d) To eliminate A from the equations x 5 sin 2A
and y 5 sec 4A to find an equation that relates x
to y, first take the sin21 of both sides of the first
7-21
08-035_07_AFSM_C07_001-060.qxd
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7:23 PM
Page 22
equation and the sec21 of both sides of the second
equation and solve for A as follows:
x 5 sin 2A
sin21 x 5 sin21 (sin 2A)
sin21 x 5 2A
sin21 x
A5
2
y 5 sec 4A
sec21 y 5 sec21 (sec 4A)
sec21 y 5 4A
sec21 y
A5
4
sec21 y
sin21 x
Since A 5
and A 5
,
2
4
cos21 ( 1y)
sin21 x
sec21 y
sin21 x
5
, or
5
2
4
2
4
17. a) Since cos 2u 5 1 2 2 sin2 u, the equation
cos 2x 5 sin x can be rewritten 1 2 2 sin2 x 5 sin x.
This equation can be solved as follows:
1 2 2 sin2 x 5 sin x
2
1 2 2 sin x 1 2 sin2 x 2 1 5 sin x 1 2 sin2 x 2 1
2 sin2 x 1 sin x 2 1 5 0
(2 sin x 2 1)(sin x 1 1) 5 0
2 sin x 2 1 5 0
2 sin x 2 1 1 1 5 0 1 1
2 sin x 5 1
1
2 sin x
5
2
2
1
sin x 5
2
5p
p
x 5 or
6
6
or sin x 1 1 5 0
sin x 1 1 2 1 5 0 2 1
sin x 5 21
3p
x5
2
p 5p
3p
, or
.
Therefore, x 5 ,
6 6
2
b) Since sin 2u 5 2 sin u cos u, and since
cos 2u 5 2 cos2 u 2 1, the equation
sin 2x 2 1 5 cos 2x can be rewritten
2 sin x cos x 2 1 5 2 cos2 x 2 1. This equation
can be solved as follows:
2 sin x cos x 2 1 5 2 cos2 x 2 1
2 sin x cos x 2 1 1 1 5 2 cos2 x 2 1 1 1
2 sin x cos x 5 2 cos2 x
2 sin x cos x 2 2 sin x cos x
5 2 cos2 x 2 2 sin x cos x
7-22
2 cos2 x 2 2 sin x cos x 5 0
(2 cos x)(cos x 2 sin x) 5 0
2 cos x 5 0
0
2 cos x
5
2
2
cos x 5 0
p
3p
x 5 or
2
2
or cos x 2 sin x 5 0
cos x 2 sin x 1 sin x 5 0 1 sin x
cos x 5 sin x
p
5p
x 5 or
4
4
p p 5p
3p
, or
.
Therefore, x 5 , ,
4 2 4
2
18. First write sin u and cos u in terms of tan u.
sin u
3 cos u
sin u 5
cos u
tan u
5
sec u
tan u
5
"1 1 tan2 u
sin u
, so
tan u 5
cos u
sin u
cos u 5
tan u
5
5
tan u
!1 1 tan2 u
tan u
1
"1 1 tan2 u
a) sin 2u 5 2 sin u cos u
tan u
1
5 2a
ba
b
2
"1 1 tan u "1 1 tan2 u
2 tan u
5
1 1 tan2 u
b) cos 2u 5 cos 2 u 2 sin2 u
2
2
1
tan u
5a
b
2
a
b
"1 1 tan2 u
"1 1 tan2 u
2
1 2 tan u
5
1 1 tan2 u
c) Use the results from parts a) and b)
2 tan u
sin 2u
1 1 tan2 u
5
1 2 tan2 u
1 1 cos 2u
11
2
1 1 tan u
Chapter 7: Trigonometric Identities and Equations
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Page 23
2 tan u
1 1 tan2 u
5 1 1 tan2 u 1 1 2 tan2 u
1 1 tan2 u
2 tan u
1 1 tan2 u
5
2
1 1 tan2 u
1 1 tan2 u
2 tan u
3
5
2
1 1 tan u
2
5 tan u
d) Use the results from parts a) and b)
1 2 tan2 u
12
1 2 cos 2u
1 1 tan2 u
5
2
tan u
sin 2u
1 1 tan2 u
1 1 tan2 u 2 1 1 tan2 u
1 1 tan2 u
5
2 tan u
1 1 tan2 u
2 tan2 u
1 1 tan2 u
5
2 tan u
1 1 tan2 u
1 1 tan2 u
2 tan2 u
3
2
1 1 tan u
2 tan u
5 tan u
5
d) Since 2cos u 5 cos (p 1 u),
2cos
2p
2p
5 cos ap 1 b
5
5
5 cos a
5p
2p
7p
1 b 5 cos
5
5
5
e) Since 2sin u 5 sin (p 1 u),
2sin
9p
9p
5 sin ap 1 b
7
7
5 sin a
7p
9p
16p
1 b 5 sin
7
7
7
Since sin u 5 sin (u 2 2p),
16p
16p
sin
5 sin a
2 2pb
7
7
16p
14p
5 sin a
2
b
7
7
2p
5 sin
7
9p
2p
5 sin
.
Therefore, 2sin
7
7
f) Since tan (p 1 u) 5 tan u, tan u 5 tan (p 1 u).
3p
3p
5 tan ap 1
b
Therefore, tan
4
4
3p
7p
4p
1
b 5 tan
5 tan a
4
4
4
2. Since cos u 5 sin Q u 1 R , the equation
2
p
Mid-Chapter Review, p. 411
1. a) Since cos (2p 2 u) 5 cos u,
cos u 5 cos (2p 2 u). Therefore,
p
p
cos
5 cos a2p 2 b
16
16
p
32p
2 b
5 cos a
16
16
31p
5 cos
16
b) Since sin (p 2 u) 5 sin u, sin u 5 sin (p 2 u).
Therefore, sin
7p
7p
5 sin ap 2 b
9
9
5 sin a
9p
7p
2p
2 b 5 sin
9
9
9
c) Since tan (p 1 u) 5 tan u, tan u 5 tan (p 1 u).
Therefore, tan
9p
9p
5 tan ap 1 b
10
10
5 tan a
10p
9p
19p
1 b 5 tan
10
10
10
Advanced Functions Solutions Manual
y 5 26 cos Q x 1 R 1 4 can be rewritten
2
p
p
y 5 26 sin Q x 1 1 R 1 4
p
2
2
5 26 sin (x 1 p) 1 4. Since a horizontal
translation of p to the left or right is equivalent
to a reflection in the x-axis, the equation
y 5 26 sin (x 1 p) 1 4 can be rewritten
y 5 6 sin x 1 4. Therefore, the equation
p
y 5 26 cos Q x 1 R 1 4 can be rewritten
2
y 5 6 sin x 1 4.
3. a) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
5p
5p
5p
b 5 cos x cos
2 sin x sin
cos ax 1
3
3
3
1
!3
5 (cos x)a b 2 (sin x)a2
b
2
2
1
!3
5 cos x 1
sin x
2
2
7-23
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Page 24
b) Since sin (a 1 b) sin a cos b 1 cos a sin b,
5p
5p
5p
sin ax 1
b 5 (sin x)acos
b 1 (cos x)asin
b
6
6
6
!3
1
5 (sin x)a2
b 1 (cos x)a b
2
2
1
!3
5 cos x 2
sin x
2
2
tan a 1 tan b
,
c) Since tan (a 1 b) 5
1 2 tan a tan b
5p
5p
4
tan ax 1
b5
5p
4
1 2 tan x tan
4
tan x 1 tan
tan x 1 1
5
1 2 (tan x)(1)
1 1 tan x
5
1 2 tan x
d) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
4p
4p
4p
cos ax 1
b 5 cos x cos
2 sin x sin
3
3
3
!3
1
b
5 (cos x)a2 b 2 (sin x)a2
2
2
!3
1
5
sin x 2 cos x
2
2
4. a) Since sin (a 2 b) 5 sin a cos b 2 cos a cos b,
11p
11p
11p
sin ax 2
b 5 sin x cos
2 cos x sin
6
6
6
1
!3
b 2 (cos x)a2 b
5 (sin x)a
2
2
1
!3
sin x 1 cos x
5
2
2
tan a 2 tan b
,
b) Since tan (a 2 b) 5
1 1 tan a tan b
p
tan x 2 tan 3
p
tan ax 2 b 5
p
3
1 1 tan x tan
3
5
tan x 2 !3
tan x 2 !3
5
1 1 (tan x)( !3)
1 1 !3 tan x
c) Since cos (a 2 b) 5 cos a cos b 1 sin a sin b,
7p
7p
7p
cos ax 2 b 5 (cos x)acos
b 1 (sin x)asin
b
4
4
4
!2
!2
5 (cos x)a
b 1 (sin x)a2
b
2
2
!2
!2
5
cos x 2
sin x
2
2
7-24
d) Since sin (a 2 b) 2 sin a cos b 2 cos a sin b,
2p
2p
2p
sin ax 2
b 5 (sin x)acos
b 2 (cos x) asin
b
3
3
3
!3
1
b
5 (sin x)a2 b 2 (cos x)a
2
2
1
!3
5 2 sin x 2
cos x
2
2
tan a 2 tan b
,
5. a) Since tan (a 2 b) 5
1 1 tan a tan b
8p
5p
2 tan
8p
5p
9
9
8p
5p 5 tan a 9 2 9 b
1 1 tan tan
9
9
tan
p
3p
5 tan 5 !3
9
3
b) Since sin (a 2 b) 5 sin a cos b 2 cos a sin b,
p
299p
p
299p
cos
2 cos
sin
sin
298
298
298
298
298p
299p
p
5 sin a
2
b 5 sin
298
298
298
5 sin p 5 0
c) Since sin (a 2 b) 5 sin a cos b 2 cos a sin b,
sin 50° cos 20° 2 cos 50° sin 20°
1
5 sin (50° 2 20°) 5 sin 30° 5
2
d) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
p
3p
3p
p
3p
p
sin
cos 1 cos
sin 5 sin a
1 b
8
8
8
8
8
8
4p
p
5 sin
5 sin 5 1
8
2
tan a 1 tan b
,
6. a) Since tan (a 1 b) 5
1 2 tan a tan b
tan x 1 tan x
2 tan x
5
2
1 2 tan x
1 2 (tan x)(tan x)
5 tan (x 1 x) 5 tan 2x
b) Since sin (a 1 b) 5 sin a cos b 1 cos a sin b,
x
x
4x
x
4x
4x
sin cos
1 cos sin
5 sin a 1 b
5
5
5
5
5
5
5x
5 sin
5 sin x
5
p
p
p
c) cos a 2 xb 5 cos cos x 1 sin sin x
2
2
2
5 (0) cos x 1 (1) sin x 5 sin x
p
p
p
d) sin a 1 xb 5 sin cos x 1 cos sin x
2
2
2
5 (1) cos x 1 (0) sin x 5 cos x
5 tan
Chapter 7: Trigonometric Identities and Equations
08-035_07_AFSM_C07_001-060.qxd
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Page 25
p
p
p
1 xb 1 cos a 1 xb 5 2 cos a 1 xb
4
4
4
p
p
5 2 c cos cos x 2 sin sin xd
4
4
!2
!2
5 2ca
b cos x 2 a
b sin xd
2
2
e) cos a
5 !2(cos x 2 sin x)
tan x 2 tan Q R
p
tan x 2 1
4
f) tan ax 2 b 5
5
p
4
1 1 tan x
1 1 tan x tan Q R
p
4
7. a 5 !3 and b 5 23, so
R 5 ! ( !3 )2 1 (23)2 5 !12 5 2 !3
1
!3
5
cos a 5
2 !3
2
23
2 !3
sin a 5
5
2 !3
2
Since cos a is positive and sin a is negative, a is in
p
the fourth quadrant. a 5 2
3
p
So, !3 cos x 2 3 sin x 5 2 !3 cos ax 1 b.
3
8. a) Since cos 2u 5 2 cos2 u 2 1,
2p
2p
2 cos2
2 1 5 cos a(2)a bb
3
3
1
4p
52
5 cos
3
2
b) Since sin 2u 5 2 sin u cos u,
11p
11p
11p
2 sin
cos
5 sin a(2)a
bb
12
12
12
11p
22p
1
5 sin
5 sin
52
12
6
2
2
2
c) Since cos 2u 5 cos u 2 sin u,
7p
7p
7p
cos2
2 sin2
5 cos a(2)a bb
8
8
8
14p
7p
"2
5 cos
5 cos
5
8
8
2
2
d) Since cos 2u 5 1 2 2 sin u,
p
p
1 2 2 sin2 5 cos a(2)a bb 5 cos p 5 21
2
2
10
2
9. a) Since cos x 5 11, cos x 5 6# 10
11
!110
and
since
angle
x
is
in the
5
6
,
5 6 !10
11
!11
third quadrant, cos x is negative. For this reason,
!110
cos x 5 2 !110
11 . Since cos x 5 2 11 , the leg
Advanced Functions Solutions Manual
adjacent to the angle x in a right triangle has a length
of !110, while the hypotenuse of the right triangle
has a length of 11. For this reason, the other leg of
the right triangle can be calculated as follows:
( !110)
2
1 y 2 5 112
110 1 y 2 5 121
110 1 y 2 2 110 5 121 2 110
y 2 5 11
y 5 2 !11, in quadrant II
Since sin x 5
opposite leg
,
hypotenuse
and since angle x is in
the third quadrant, sin x 5 2 !11
11 .
10
!10
cos
x
5
6
,
b) Since cos2 x 5 10
#11 5 6 !11
11
5 6 !110
11 , and since angle x is in the third quadrant,
cos x is negative. For this reason, cos x 5 2 !110
11 .
10
!10
c) Since cos2 x 5 10
11 , cos x 5 6#11 5 6 !11
5 6 !110
11 , and since angle x is in the third quadrant,
cos x is negative. For this reason, cos x 5 2 !110
11 .
!110
Since cos x 5 2 11 , the leg adjacent to the angle
x in a right triangle has a length of !110, while the
hypotenuse of the right triangle has a length of 11.
For this reason, the other leg of the right triangle
can be calculated as follows:
( !110)2 1 y 2 5 112
110 1 y 2 5 121
110 1 y 2 2 110 5 121 2 110
y 2 5 11
y 5 2 !11, in quadrant II
Since sin x 5
opposite leg
,
hypotenuse
and since angle x is in
the third quadrant, sin x 5 2 !11
11 . Since
sin 2u 5 2 sin u cos u, sin 2x 5 2 sin x cos x
!110
2 !10
!11
b a2
b5
.
5 (2)a2
11
11
11
10
!10
d) Since cos2 x 5 10
11 , cos x 5 6#11 5 6 !11
5 6 !110
11 , and since angle x is in the third quadrant,
cos x is negative. For this reason, cos x 5 2 !110
11 .
!110
Since cos x 5 2 11 , the leg adjacent to the angle x
in a right triangle has a length of !110, while the
hypotenuse of the right triangle has a length of 11.
For this reason, the other leg of the right triangle can
be calculated as follows:
( !110)2 1 y 2 5 112
110 1 y 2 5 121
110 1 y 2 2 110 5 121 2 110
7-25
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Page 26
y 2 5 11
y 5 2 !11, in quadrant II
Since sin x 5
opposite leg
,
hypotenuse
and since angle x
is in the third quadrant, sin x 5 2 !11
11 . Since
2
2
cos 2u 5 cos u 2 sin u, cos 2x 5 cos2 x 2 sin2 x
!11 2
10
1
9
5 10
11 2 (2 11 ) 5 11 2 11 5 11 . (The formulas
cos 2u 5 2 cos2 u 2 1 and cos 2u 5 1 2 2 sin2 u
could also have been used.)
10. Since sin x 5 35, the leg opposite the angle x in
a right triangle has a length of 3, while the
hypotenuse of the right triangle has a length of 5.
For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 32 5 52
x 2 1 9 5 25
x 2 1 9 2 9 5 25 2 9
x 2 5 16
x 5 4, in quadrant I
Since cos x 5
adjacent leg
, cos x 5 45.
hypotenuse
Therefore,
since sin 2u 5 2 sin u cos u,
3 4
24
sin 2x 5 2 sin x cos x 5 (2)a b a b 5 .
5 5
25
Also, since cos 2u 5 cos2 u 2 sin2 u,
cos 2x 5 cos2 x 2 sin2 x 5 ( 45)2 2 ( 35)2
9
7
2
5 16
25 2 25 5 25 . (The formulas cos 2u 5 2 cos u 2 1
2
and cos 2u 5 1 2 2 sin u could also have been used.)
11. Since sin x 5 135 , the leg opposite the angle x
in a right triangle has a length of 5, while the
hypotenuse of the right triangle has a length of 13.
For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 52 5 132
x 2 1 25 5 169
2
x 1 25 2 25 5 169 2 25
x 2 5 144
x 5 12, in quadrant I
Since cos x 5
adjacent leg
, cos x 5
hypotenuse
Since tan x 5
opposite leg
, tan x 5 34.
adjacent leg
(The reason
the sign is positive is because angle x is in the third
quadrant.) Since tan 2u 5
tan 2x 5
5
2 tan u
,
1 2 tan2 u
3
(2)( 34)
2 tan x
2
5
5
1 2 tan2 x
1 2 169
1 2 ( 34) 2
3
2
16
16
2 169
5
3
2
7
16
5
3
16
24
3
5
2
7
7
7.4 Proving Trigonometric Identities,
pp. 417–418
p
1. Although sin x 5 cos x is true for x 5 , it is not
4
true for all values of x, and therefore, it is not an
identity. A counterexample is a value of x for which
sin x 5 cos x is not true. Many counterexamples
exist, so answers may vary. One counterexample is
p
6
x 5 , since sin
p
5 12,
6
and since cos
p
5
6
!3
2 .
2. a) The graphs of f(x) 5 sin x and
g(x) 5 tan x cos x are as follows: f(x) 5 sin x:
g(x) 5 tan x cos x:
12
13 .
Therefore, since sin 2u 5 2 sin u cos u,
5
12
120
sin 2x 5 2 sin x cos x 5 (2)a b a b 5
.
13 13
169
12. Since cos x 5 2 45, the leg adjacent to the angle
x in a right triangle has a length of 4, while the
hypotenuse of the right triangle has a length of 5.
7-26
For this reason, the other leg of the right triangle
can be calculated as follows:
42 1 y 2 5 52
16 1 y 2 5 25
16 1 y 2 2 16 5 25 2 16
y2 5 9
y 5 3, in quadrant III
b) Since the graphs of f(x) 5 sin x and
g(x) 5 tan x cos x are the same, sin x 5 tan x cos x.
c) To prove that the identity sin x 5 tan x cos x is
true, tan x cos x can be simplified as follows:
sin x
sin x cos x
tan x cos x 5 a
b (cos x) 5
5 sin x
cos x
cos x
Chapter 7: Trigonometric Identities and Equations
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Page 27
d) The identity is not true when cos x 5 0 because
The graph of y 5 1 1 2 sin x cos x is as follows:
sin x
,
cos x
when cos x 5 0, tan x, or
is undefined. This
is because 0 cannot be in the denominator of a fraction.
3. a) The graph of y 5 sin x cot x is as follows:
2
1
y
1
–2p
–p
x
0
p
2p
–2p
–p
8
The graph of y 5 cos x is as follows:
y
1
–p
x
p
2p
–2p
–p
Since the graphs are the same, the answer is C.
b) The graph of y 5 1 2 2 sin2 x is as follows:
–p
2p
–2p
y
x
0
p
2p
–1
Since the graphs are the same, the answer is D.
c) The graph of y 5 (sin x 1 cos x)2 is as follows:
2
y
1
–2p
–p
0
p
2p
y
x
p
The graph of y 5 2 cos2 x 2 1 is as follows:
–p
x
4
–1
–2p
y
0
8
y
0
1
2p
The graph of y 5 sin2 x 1 cos2 x 1 tan2 x is as
follows:
–1
–2p
p
4
0
1
x
0
Since the graphs are the same, the answer is B.
d) The graph of y 5 sec2 x is as follows:
–1
–2p
y
x
p
2p
Advanced Functions Solutions Manual
–p
0
x
p
2p
Since the graphs are the same, the answer is A.
4. a) The identity sin x cot x 5 cos x can be proven
as follows:
cos x
sin x cos x
sin x cot x 5 (sin x)a
b5
5 cos x
sin x
sin x
b) The identity 1 2 2 sin2 x 5 2 cos2 x 2 1 can be
proven as follows:
1 2 2 sin2 x 5 2 cos2 x 2 1
2
1 2 2 sin x 2 2 cos2 x 1 1 5 0
2 2 2 sin2 x 2 2 cos2 x 5 0
2 2 2(sin2 x 1 cos2 x) 5 0
2 2 2(1) 5 0
22250
050
c) The identity (sin x 1 cos x)2 5 1 1 2 sin x cos x
can be proven as follows:
(sin x 1 cos x)2 5 sin2 x 1 2 sin x cos x 1 cos2 x
5 (sin2 x 1 cos2 x) 1 2 sin x cos x
5 1 1 2 sin x cos x
7-27
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Page 28
d) The identity sec2 x 5 sin2 x 1 cos2 x 1 tan2 x
can be proven as follows:
sin2 x 1 cos2 x 1 tan2 x
5 (sin2 x 1 cos2 x) 1 tan2 x
5 1 1 tan2 x
cos2 x
sin2 x
5
1
2
cos x
cos2 x
2
sin x 1 cos2 x
5
cos2 x
1
5
cos2 x
5 sec2 x
1
5. a) The equation cos x 5
is not true for all
p
One counterexample is x 5 , since
3
p
2p
1
cos a2a bb 5 cos a b 5 2 , and since
3
3
2
!3 2
p
1 1 2 sin2 a b 5 1 1 (2)a
b
3
2
3
4
6
10
5
5 1 1 (2)a b 5 1 5
5
4
4
4
4
2
6. Answers may vary. For example, the graph of
the function y 5
1
cos x
values of x, and therefore, it is not an identity.
A counterexample is a value of x for which
1
cos x 5
is not true. Many counterexamples
cos x
exist, so answers may vary. One counterexample
1
p
p
is x 5 , since cos 5 !3
p
2 , and since
6
6
cos
6
1
2
5 "3 5 !3
5 2 !3
3 .
2
b) The equation 1 2 tan2 x 5 sec2 x is not true for
all values of x, and therefore, it is not an identity.
A counterexample is a value of x for which
1 2 tan2 x 5 sec2 x is not true. Many counterexamples
exist, so answers may vary. One counterexample
p
is
since 1 2 tan2 Q R 5 1 2 (1)2
4
p
5 1 2 1 5 0, and since sec2 Q R 5 ( !2)2 5 2.
4
p
x5 ,
4
c) The equation
sin (x 1 y) 5 cos x cos y 1 sin x sin y is not true for
all values of x and y, and therefore, it is not an identity.
A counterexample is values for x and y for which
sin (x 1 y) 5 cos x cos y 1 sin x sin y is not true.
Many counterexamples exist, so answers may vary.
p
One counterexample is x 5 and y 5 p, since
2
sin Q 1 p R 5 sin Q R 5 21, and since
2
2
p
p
cos Q R cos p 1 sin Q R sin p
2
2
p
3p
5 (0)(21) 1 (1)(0) 5 0 1 0 5 0.
d) The equation cos 2x 5 1 1 2 sin2 x is not true
for all values of x, and therefore, it is not an
identity. A counterexample is a value of x for
which cos 2x 5 1 1 2 sin2 x is not true. Many
counterexamples exist, so answers may vary.
–2p–3p –p – p 0
2
2
–1
1 2 tan2 x
1 1 tan2 x
is as follows:
y
x
p p 3p 2p
2
2
The graph is the same as that of the function
y 5 cos 2x, so an appropriate conjecture is that
cos 2x is another expression that is equivalent to
1 2 tan2 x
.
1 1 tan2 x
7. The identity
1 2 tan2 x
5 cos 2x
1 1 tan2 x
can be proven as
follows:
cos2 x
sin2 x
cos2 x 2 sin2 x
2 2
2
1 2 tan2 x
cos2 x
5 cos x 2 cos x 5
2
1 1 tan x
sec x
sec2 x
2
2
1
cos x 2 sin x
3
5
2
cos x
sec2 x
2
2
cos x 2 sin x
5
3 cos2 x 5 cos2 x 2 sin2 x
cos2 x
5 cos 2x
8. The identity
1 1 tan x
1 2 tan x
5
1 1 cot x
cot x 2 1
as follows:
1 1 tan x
LS 5
1 1 cot x
1 1 tan x
5
1
11
RS 5
5
can be proven
1 2 tan x
cot x 2 1
1 2 tan x
tan x
1
21
tan x
1 1 tan x
5 tan x 1 1
1 2 tan x
5 1 2 tan x
tan x
tan x
5 tan x
5 tan x
Since the right side and the left side are equal,
1 1 tan x
1 2 tan x
5
1 1 cot x
cot x 2 1
7-28
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Page 29
cos2 u 2 sin2 u
9. a) The identity 2
5 1 2 tan u can
cos u 1 sin u cos u
be proven as follows:
cos2 u 2 sin2 u
cos2 u 1 sin u cos u
(cos u 2 sin u)(cos u 1 sin u)
5
(cos u)(cos u 1 sin u)
cos u 2 sin u
sin u
cos u
5
2
5 1 2 tan u
5
cos u
cos u
cos u
b) The identity tan2 x 2 sin2 x 5 sin2 x tan2 x can
be proven as follows:
LS 5 tan2 x 2 sin2 x
sin2 x
5
2 sin2 x
cos2 x
1
5 sin2 xa 2 2 1b
cos x
2
5 sin x(sec2 x 2 1)
5 sin2 x tan2 x
5 RS
So tan2 x 2 sin2 x 5 sin2 x tan2 x.
Since csc2 x 5 1 1 cot2 x is a known identity,
tan2 x 2 sin2 x must equal sin2 x tan2 x.
c) The identity
tan2 x 2 cos2 x 5
1
2 1 2 cos2 x
cos2 x
can be proven
as follows:
1
2 1 2 cos2 x;
cos2 x
1
5
21
cos2 x
2 cos2 x 1 cos2 x;
1
5
2 1;
cos2 x
1
cos2 x
5
2
;
2
cos x
cos2 x
1 2 cos2 x
5
;
cos2 x
sin2 x
5
;
cos2 x
5 tan2 x
tan2 x 2 cos2 x 5
tan2 x 2 cos2 x 1 cos2 x
tan2 x
tan2 x
tan2 x
tan2 x
tan2 x
d) The identity
1
1
2
1
5 2
1 1 cos u
1 2 cos u
sin u
can be
proven as follows:
1
1
1 2 cos u
1
5
1 1 cos u
1 2 cos u
(1 1 cos u)(1 2 cos u)
1 1 cos u
1
(1 2 cos u)(1 1 cos u)
Advanced Functions Solutions Manual
1 1 cos u
1 2 cos u
1
2
1 2 cos u
1 2 cos2 u
2
2
1 2 cos u 1 1 1 cos u
5
5
5
2
2
1 2 cos u
1 2 cos u
sin2 u
3
2
10. a) The identity cos x tan x 5 sin x tan x can
be proven as follows:
cos x tan3 x 5 sin x tan2 x
cos x tan3 x
sin x tan2 x
5
2
tan x
tan2 x
cos x tan x 5 sin x
sin x
(cos x)a
b 5 sin x
cos x
sin x 5 sin x
b) The identity sin2 u 1 cos4 u 5 cos2 u 1 sin4 u can
be proven as follows:
sin2 u 1 cos4 u 5 cos2 u 1 sin4 u
sin2 u 1 cos4 u 2 sin4 u 5 cos2 u 1 sin4 u 2 sin4 u
sin2 u 1 cos4 u 2 sin4 u 5 cos2 u
sin2 u 1 cos4 u 2 sin4 u 2 sin2 u 5 cos2 u 2 sin2 u
cos4 u 2 sin4 u 5 cos2 u 2 sin2 u
(cos2 u 1 sin2 u)(cos2 u 2 sin2 u) 5 cos2 u 2 sin2 u
cos2 u 1 sin2 u 5 1
151
c) The identity
tan2 x 1 1
1
1
(sin x 1 cos x)a
b5
1
cos x
tan x
sin x
can be proven as follows:
5
tan2 x 1 1
1
1
b5
1
cos x
tan x
sin x
2
cos x
sec x
sin x
1
(sin x 1 cos x)a
b5
tan x
cos x sin x
sin x cos x
1
1
sin x 1 cos x
(sin x 1 cos x)a 2 b a
b5
cos x tan x
cos x sin x
1
cos x
sin x 1 cos x
(sin x 1 cos x)a 2 b a
b5
cos x sin x
cos x sin x
1
sin x 1 cos x
(sin x 1 cos x)a
b5
cos x sin x
cos x sin x
sin x 1 cos x
sin x 1 cos x
5
cos x sin x
cos x sin x
(sin x 1 cos x)a
d) The identity tan2 b 1 cos2 b 1 sin2 b 5
1
cos2 b
can be proven as follows:
tan2 b 1 cos2 b 1 sin2 b 5
tan2 b 1 1 5
1
cos2 b
1
cos2 b
7-29
08-035_07_AFSM_C07_001-060.qxd
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Page 30
tan2 b 1 1 5 sec2 b
Since tan2 b 1 1 5 sec2 b is a known identity,
tan2 b 1 cos2 b 1 sin2 b must equal
1
.
cos2 b
e) The identity
p
p
sin Q 1 x R 1 sin Q 2 x R 5 !2 cos x
4
4
can be proven as follows:
p
p
sin a 1 xb 1 sin a 2 xb 5 !2 cos x;
4
4
p
p
p
sin cos x 1 cos sin x 1 sin cos x
4
4
4
p
2 cos sin x 5 !2 cos x;
4
p
2 sin cos x 5 !2 cos x;
4
!2
(2)a
b (cos x) 5 !2 cos x;
2
!2 cos x 5 !2 cos x
f) The identity sin Q 2 x R cot Q 1 x R 5 2sin x
2
2
can be proven as follows:
p
p
sin a 2 xb cot a 1 xb 5 2sin x;
2
2
p
p
p
cos ( 2 1 x)
p
¢ 5 2sin x;
sin a 2 xb °
p
2
sin ( 2 1 x)
asin
p
p
cos x 2 cos sin xb
2
2
p
p
cos x 2 sin sin x
2
2
3 °
¢ 5 2sin x;
p
p
sin cos x 1 cos sin x
2
2
cos
((1)(cos x) 2 (0)(sin x))
(0)(cos x) 2 (1)(sin x)
a
b 5 2sin x;
(1)(cos x) 1 (0)(sin x)
0 2 sin x
(cos x 2 0)a
b 5 2sin x;
cos x 1 0
sin x
(cos x)a2
b 5 2sin x;
cos x
2sin x 5 2sin x
11. a) The identity
cos 2x 1 1
5 cot x
sin 2x
as follows:
cos 2x 1 1
5 cot x
sin 2x
2 cos2 x 2 1 1 1
5 cot x
2 sin x cos x
7-30
can be proven
2 cos2 x
5 cot x
2 sin x cos x
cos x
5 cot x
sin x
cot x 5 cot x
b) The identity
sin 2x
5 cot x
1 2 cos 2x
can be proven as
follows:
sin 2x
5 cot x
1 2 cos 2x
2 sin x cos x
5 cot x
1 2 (1 2 2 sin2 x)
2 sin x cos x
5 cot x
1 2 1 1 2 sin2 x
2 sin x cos x
5 cot x
2 sin2 x
cos x
5 cot x
sin x
cot x 5 cot x
c) The identity (sin x 1 cos x)2 5 1 1 sin 2x can
be proven as follows:
(sin x 1 cos x)2 5 1 1 sin 2x;
sin2 x 1 sin x cos x 1 sin x cos x 1 cos2 x
5 1 1 2 sin x cos x;
sin2 x 1 2 sin x cos x 1 cos2 x 5 1 1 2 sin x cos x;
(cos2 x 1 sin2 x) 1 2 sin x cos x
5 1 1 2 sin x cos x;
1 1 2 sin x cos x 5 1 1 2 sin x cos x
d) The identity cos4 u 2 sin4 u 5 cos 2u can be
proven as follows:
cos4 u 2 sin4 u 5 cos 2u
(cos2 u 1 sin2 u)(cos2 u 2 sin2 u) 5 cos2 u 2 sin2 u
(1)(cos2 u 2 sin2 u) 5 cos2 u 2 sin2 u
cos2 u 2 sin2 u 5 cos2 u 2 sin2 u
e) The identity cot u 2 tan u 5 2 cot 2u can be
proven as follows:
cot u 2 tan u 5 2 cot 2u
sin u
cos 2u
cos u
2
52
sin u
cos u
sin 2u
2
2
sin u
cos 2u
cos u
2
5 (2)a
b
sin u cos u
cos u sin u
2 cos u sin u
cos 2u
cos2 u 2 sin2 u
5
sin u cos u
cos u sin u
cos 2u
cos 2u
5
cos u sin u
cos u sin u
Chapter 7: Trigonometric Identities and Equations
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Page 31
f) The identity cot u 1 tan u 5 2 csc 2u
can be proven as follows:
cot u 1 tan u 5 2 csc 2u
sin u
1
cos u
1
52
sin u
cos u
sin 2u
cos2 u
sin2 u
1
1
5 (2)a
b
sin u cos u
cos u sin u
2 cos u sin u
1
cos2 u 1 sin2 u
5
sin u cos u
cos u sin u
1
1
5
cos u sin u
cos u sin u
g) The identity
1 1 tan x
p
5 tan Q x 1 R
1 2 tan x
4
can be
proven as follows:
1 1 tan x
p
5 tan ax 1 b
1 2 tan x
4
p
tan
x
1
tan
1 1 tan x
4
5
p
1 2 tan x
1 2 tan x tan
4
1 1 tan x
tan x 1 1
5
1 2 tan x
1 2 (tan x)(1)
1 1 tan x
1 1 tan x
5
1 2 tan x
1 2 tan x
h) The identity csc 2x 1 cot 2x 5 cot x can be
proven as follows:
csc 2x 1 cot 2x 5 cot x;
1
1
1
5 cot x;
sin 2x
tan 2x
1
1
1 2 tan x 5 cot x;
2 sin x cos x
2 cos2 x
;
2 sin x cos x
1
cos2 x 2 sin2 x
2 cos2 x
1
5
;
2 sin x cos x
2 sin x cos x
2 sin x cos x
1 1 cos2 x 2 sin2 x
2 cos2 x
5
;
2 sin x cos x
2 sin x cos x
1 1 cos2 x 2 sin2 x
2 cos2 x
2
2 sin x cos x
2 sin x cos x
2 cos2 x
2 cos2 x
5
2
;
2 sin x cos x
2 sin x cos x
1 1 cos2 x 2 sin2 x 2 2 cos2 x
5 0;
2 sin x cos x
1 2 sin2 x 2 cos2 x
5 0;
2 sin x cos x
1 2 (sin2 x 1 cos2 x)
5 0;
2 sin x cos x
121
5 0;
2 sin x cos x
0
5 0;
2 sin x cos x
050
2 tan x
i) The identity
5
sin
2x
2
5
1 1 tan x
can be proven as follows:
2 tan x
5 sin 2x
1 1 tan2 x
2 tan x
5 sin 2x
sec2 x
2 tan x
5 sin 2x
1
cos2 x
1 2 tan2 x
2
1
1 2 tan x
1
5 cot x;
2 sin x cos x
2 tan x
1 2 tan2 x
1
1
5 cot x;
sin x
2 sin x cos x
2 cos x
1
(cos x)(1 2 tan2 x)
cos x
1
5
;
2 sin x cos x
2 sin x
sin x
1
(cos x)(1 2 tan2 x)(cos x)
1
2 sin x cos x
2 sin x cos x
(cos x)(2 cos x)
5
;
(sin x)(2 cos x)
1
(cos2 x)(1 2 tan2 x)
1
2 sin x cos x
2 sin x cos x
2 cos2 x
5
;
2 sin x cos x
1
cos2 x 2 (tan2 x)(cos2 x)
1
2 sin x cos x
2 sin x cos x
Advanced Functions Solutions Manual
(2 tan x)(cos2 x) 5 sin 2x
2 sin x
a
b (cos2 x) 5 sin 2x
cos x
sin 2x 5 2 sin x cos x
Since sin 2x 5 2 sin x cos x is a known identity,
2 tan x
1 2 tan2 x
must equal sin 2x.
csc t
j) The identity sec 2t 5
csc t 2 2 sin t
can be proven as follows:
csc t
sec 2t 5
csc t 2 2 sin t
1
1
5 1 sin t
cos 2t
2 2 sin t
sin t
1
1
sin t
5 1
2 sin2 t
cos 2t
2
sin t
sin t
7-31
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Page 32
The graph is the same as that of the function
y 5 tan x, so an expression equivalent to
1
1
sin t
5 1 2 2 sin2 t
cos 2t
sin t
1
sin t
1
5
3
cos 2t
sin t
1 2 2 sin2 t
1
1
5
cos 2t
1 2 2 sin2 t
1
1
5
cos 2t
cos 2t
k) The identity csc 2u 5 12 sec u csc u
can be proven as follows:
1
csc 2u 5 sec u csc u
2
1
1
1
1
5 a ba
ba
b
sin 2u
2 cos u sin u
1
1
5
sin 2u
2 cos u sin u
1
1
5
2 sin u cos u
2 sin u cos u
l) The identity sec t 5
sin 2t
cos 2t
2
sin t
cos t
can be proven as follows:
1
2 sin t cos t
2 cos2 t 2 1
5
2
cos t
sin t
cos t
2
sin t
2 sin t cos t
(sin t)(2 cos2 t 2 1)
5
2
cos t sin t
sin t cos t
cos t sin t
sin t
2 sin t cos2 t
2 cos2 t sin t 2 sin t
5
2
cos t sin t
sin t cos t
cos t sin t
2
sin t
2 sin t cos t 2 (2 cos2 t sin t 2 sin t)
5
cos t sin t
sin t cos t
2
sin t
2 sin t cos t 2 2 sin t cos2 t 1 sin t
5
cos t sin t
sin t cos t
sin t
sin t
5
cos t sin t
cos t sin t
12. Answers may vary. For example, the graph of
the function y 5
4
3
2
1
–2p
7-32
0
–p
–1
–2
–3
–4
sin x 1 sin 2x
1 1 cos x 1 cos 2x
y
sin x 1 sin 2x
1 1 cos x 1 cos 2x
13. The identity
is tan x.
sin x 1 sin 2x
5 tan x
1 1 cos x 1 cos 2x
can be
proven as follows:
sin x 1 sin 2x
5 tan x
1 1 cos x 1 cos 2x
sin x 1 2 sin x cos x
5 tan x
1 1 cos x 1 cos 2x
(sin x)(1 1 2 cos x)
5 tan x
1 1 cos x 1 cos 2x
(sin x)(1 1 2 cos x)
5 tan x
1 1 cos x 1 2 cos2 x 2 1
(sin x)(1 1 2 cos x)
5 tan x
cos x 1 2 cos2 x
(sin x)(1 1 2 cos x)
5 tan x
(cos x)(1 1 2 cos x)
sin x
5 tan x
cos x
tan x 5 tan x
14. A trigonometric identity is a statement of the
equivalence of two trigonometric expressions. To
prove it, both sides of the equation must be shown
to be equivalent through graphing or simplifying/
rewriting. Therefore, the chart can be completed as
follows:
Definition
A statement of the
equivalence of two
trigonometric
expressions
Methods of Proof
Both sides of the
equation must be
shown to be equivalent
through graphing or
simplifying/rewriting.
Trigonometric
Identities
is as follows:
Examples
Non-Examples
cos 2x 1 sin2 x 5 cos2 x cos 2x 2 2 sin2 x 5 1
cos 2x 1 1 5 2 cos2 x
cot2 x 1 csc2 x 5 1
x
p
2p
15. She can determine whether the equation
2 sin x cos x 5 cos 2x is an identity by trying to
simplify and/or rewrite the left side of the equation
so that it is equivalent to the right side of the
Chapter 7: Trigonometric Identities and Equations
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equation. Alternatively, she can graph the functions
y 5 2 sin x cos x and y 5 cos 2x and see if the
graphs are the same. If they’re the same, it’s an
identity, but if they’re not the same, it’s not an
identity. By doing this she can determine it’s not
an identity, but she can make it an identity by
changing the equation to 2 sin x cos x 5 sin 2x.
16. a) To write the expression 2 cos2 x 1 4 sin x cos x
in the form a sin 2x 1 b cos 2x 1 c, rewrite the
expression as follows:
2 cos2 x 1 4 sin x cos x
5 2 cos2 x 1 (2)(2 sin x cos x)
5 (2 cos2 x 2 1) 1 (2)(2 sin x cos x) 1 1
5 cos 2x 1 2 sin 2x 1 1
5 2 sin 2x 1 cos 2x 1 1
Since the expression can be rewritten as
2 sin 2x 1 cos 2x 1 1, the values of a, b, and c are
a 5 2, b 5 1, and c 5 1.
b) To write the expression 22 sin x cos x 2 4 sin2 x
in the form a sin 2x 1 b cos 2x 1 c, rewrite the
expression as follows:
22 sin x cos x 2 4 sin2 x
5 22 sin x cos x 1 (2 2 4 sin2 x) 2 2
5 22 sin x cos x 1 (2)(1 2 2 sin2 x) 2 2
5 2sin 2x 1 2 cos 2x 2 2
Since the expression can be rewritten as
2sin 2x 1 2 cos 2x 2 2, the values of a, b, and c
are a 5 21, b 5 2, and c 5 22.
17. To write the expression 8 cos4 x in the form
a cos 4x 1 b cos 2x 1 c, it’s first necessary to develop
a formula for cos 4x. Since cos 2x 5 2 cos2 x 2 1,
cos 4x 5 2 cos2 2x 2 1. The formula
cos 2x 5 2 cos2 x 2 1 can now be used again to
further develop the formula for cos 4x as follows:
cos 4x 5 2 cos2 2x 2 1
cos 4x 5 (2)(2 cos2 x 2 1)2 2 1
cos 4x 5 (2)(4 cos4 x 2 2 cos2 x
2 2 cos2 x 1 1) 2 1
cos 4x 5 (2)(4 cos4 x 2 4 cos2 x 1 1) 2 1
cos 4x 5 8 cos4 x 2 8 cos2 x 1 2 2 1
cos 4x 5 8 cos4 x 2 8 cos2 x 1 1
Since cos 4x 5 8 cos4 x 2 8 cos2 x 1 1, the value of
a must be 1. Now, to turn the expression
8 cos4 x 2 8 cos2 x 1 1 into the expression 8 cos4 x,
the terms 28 cos2 x and 1 must be eliminated.
First, to eliminate 28 cos2 x, 4 cos 2x, or
(4)(2 cos2 x 2 1), can be added to the expression
8 cos4 x 2 8 cos2 x 1 1 as follows:
8 cos4 x 2 8 cos2 x 1 1 1 (4)(2 cos2 x 2 1)
8 cos4 x 2 8 cos2 x 1 1 1 8 cos2 x 2 4
8 cos4 x 2 3
Since adding 4 cos 2x to the expression eliminated
the term 28 cos2 x, the value of b must be 4. Now,
to turn the expression 8 cos4 x 2 3 into the expression
8 cos4 x, 3 must be added to it as follows:
8 cos4 x 2 3 1 3
8 cos4 x
Therefore, the value of c must be 3, and the
expression must be cos 4x 1 4 cos 2x 1 3.
a 5 1, b 5 4, c 5 3
7.5 Solving Linear Trigonometric
Equations, pp. 426–428
1. a) From the graph of y 5 sin u, the equation
p
sin u 5 1 is true when u 5 .
2
b) From the graph of y 5 sin u, the equation
3p
sin u 5 21 is true when u 5 .
2
c) From the graph of y 5 sin u, the equation
p
5p
sin u 5 0.5 is true when u 5 or .
6
7p
11p
or
sin u 5 20.5 is true when u 5
.
6
6
e) From the graph of y 5 sin u, the equation
sin u 5 0 is true when u 5 0, p, or 2p.
f) From the graph of y 5 sin u, the equation
sin u 5
!3
2
p
2p
is true when u 5 or .
3
3
2. a) From the graph of y 5 cos u, the equation
cos u 5 1 is true when u 5 0 or 2p.
b) From the graph of y 5 cos u, the equation
cos u 5 21 is true when u 5 p.
c) From the graph of y 5 cos u, the equation
cos u 5 0.5 is true when u 5
p
3
or
5p
.
3
d) From the graph of y 5 cos u, the equation
2p
4p
or .
cos u 5 20.5 is true when u 5
3
3
e) From the graph of y 5 cos u, the equation
cos u 5 0 is true when u 5
p
2
or
3p
.
2
f) From the graph of y 5 cos u, the equation
cos u 5
!3
2
is true when u 5
3. a) Given sin x 5
Advanced Functions Solutions Manual
6
d) From the graph of y 5 sin u, the equation
!3
2
p
6
or
11p
.
6
and 0 # x # 2p,
7-33
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Page 34
2 solutions must be possible, since sin x 5
in 2 of the 4 quadrants.
!3
2
c) Since cos
b) Given sin x 5
and 0 # x # 2p, the
solutions for x must occur in the 1st and 2nd
quadrants, since the sine function is positive in
these quadrants.
c) Given sin x 5
acute angle for the equation must be
sin
p
5
3
!3
2 .
d) Given sin x 5
!3
2
sin
!3
2
and
p
x5
3
2p
sin
5
3
or
!3
2 .
2p
,
3
since
4
4
b) Since sin
p
5
4
1
!2
and sin
3p
5
4
solutions to the equation sin u 5
7-34
p
4
1
!2 ,
1
!2
or
since
5p
.
3
5p
.
4
f) Since tan
5p
.
4
the solutions
are u 5
p
4
or
11p
5
6
4p
5 2 !3
2
3
and sin
3p
1
5 2 !2
4
and cos
!3
2 ,
!3
2
the
are u 5
p
6
5p
5 2 !3
2 ,
3
the
5p
1
5 2 !2
,
4
the
4p
3
1
solutions to the equation cos u 5 2 !2
are u 5
or
or
and cos
solutions to the equation sin u 5 2 !3
2 are u 5
e) Since cos
4. a) Given cos x 5 20.8667 and 0° # x # 360°,
two solutions must be possible, since
cos x 5 20.8667 in two of the four quadrants.
b) Given cos x 5 20.8667 and 0° # x # 360°, the
solutions for x must occur in the second and third
quadrants, since the cosine function is negative in
these quadrants.
c) Given cos x 5 20.8667 and 0° # x # 360°, the
related acute angle for the equation must be x 5 30°,
since cos 30° 5 20.866.
d) Given cos x 5 20.8667 and 0° # x # 360°, the
solutions to the equation must be x 5 150°
or 210°, since cos 150° 5 20.866 and
cos 210° 5 20.866.
5. a) Given tan u 5 2.7553 and 0 # u # 2p, two
solutions must be possible, since tan u 5 2.7553
in two of the four quadrants.
b) Given tan u 5 2.7553 and 0 # u # 2p, the
solutions for u must occur in the first and third
quadrants, since the tangent function is positive in
these quadrants.
c) Given tan u 5 2.7553 and 0 # u # 2p, the
related acute angle for the equation must be
u 5 1.22, since tan 1.22 5 2.7553.
d) Given tan u 5 2.7553 and 0 # u # 2p, the
solutions to the equation must be u 5 1.22 or 4.36,
since tan 1.22 5 2.7553 and tan 4.36 5 2.7553.
p
5p
5 1, the solutions
6. a) Since tan 5 1 and tan
to the equation tan u 5 1 are u 5
11p
.
6
d) Since sin
and 0 # x # 2p, the solutions
to the equation must be
p
5
3
or
and 0 # x # 2p, the related
p
x5 ,
3
!3
2
solutions to the equation cos u 5
!3
2
!3
2
p
5
6
p
5 !3
3
and tan
4p
5 !3,
3
the
solutions to the equation tan u 5 !3 are u 5
4p
.
3
3p
4
p
3
or
7. a) The equation 2 sin u 5 21 can be rewritten as
follows:
2 sin u 5 21
1
2 sin u
52
2
2
1
sin u 5 2
2
Given sin u 5 2 12 and 0° # u # 360°, the solutions
to the equation must be u 5 210° or 330°, since
sin 210° 5 2 12 and sin 330° 5 2 12.
b) The equation 3 cos u 5 22 can be rewritten as
follows:
3 cos u 5 22
2
3 cos u
52
3
3
2
cos u 5 2
3
Given cos u 5 2 23 and 0° # u # 360°, the solutions
to the equation must be u 5 131.8° or 228.2°.
c) The equation 2 tan u 5 3 can be rewritten as
follows:
2 tan u 5 3
3
2 tan u
5
2
2
3
tan u 5
2
Given tan u 5 32 and 0° # u # 360°, the solutions to
the equation must be u 5 56.3° or 236.3°.
3p
.
4
Chapter 7: Trigonometric Identities and Equations
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Page 35
d) The equation 23 sin u 2 1 5 1 can be rewritten
as follows:
23 sin u 2 1 5 1
23 sin u 2 1 1 1 5 1 1 1
23 sin u 5 2
23 sin u
2
5
23
23
2
sin u 5 2
3
Given sin u 5 2 23 and 0° # u # 360°, the solutions
to the equation must be u 5 221.8° or 318.2°.
e) The equation 25 cos u 1 3 5 2 can be rewritten
as follows:
25 cos u 1 3 5 2
25 cos u 1 3 2 3 5 2 2 3
25 cos u 5 21
25 cos u
21
5
25
25
1
cos u 5
5
1
Given cos u 5 5 and 0° # u # 360°, the solutions to
the equation must be u 5 78.5° or 281.5°.
f) The equation 8 2 tan u 5 10 can be rewritten as
follows:
8 2 tan u 5 10
8 2 tan u 1 tan u 5 10 1 tan u
8 5 10 1 tan u
8 2 10 5 10 1 tan u 2 10
tan u 5 22
Given tan u 5 22 and 0° # u # 360°, the solutions
to the equation must be u 5 116.6° or 296.6°.
8. a) The equation 3 sin x 5 sin x 1 1 can be
rewritten as follows:
3 sin x 5 sin x 1 1
3 sin x 2 sin x 5 sin x 1 1 2 sin x
2 sin x 5 1
2 sin x
1
5
2
2
1
sin x 5
2
Given sin x 5 12 and 0 # x # 2p, the solutions to
the equation must be x 5 0.52 or 2.62.
b) The equation 5 cos x 2 !3 5 3 cos x can be
rewritten as follows:
5 cos x 2 !3 5 3 cos x
5 cos x 2 !3 2 3 cos x 5 3 cos x 2 3 cos x
2 cos x 2 !3 5 0
2 cos x 2 !3 1 !3 5 0 1 !3
Advanced Functions Solutions Manual
2 cos x 5 !3
2 cos x
!3
5
2
2
!3
cos x 5
2
!3
Given cos x 5 2 and 0 # x # 2p, the solutions
to the equation must be x 5 0.52 or 5.76.
c) The equation cos x 2 1 5 2cos x can be
rewritten as follows:
cos x 2 1 5 2cos x
cos x 2 1 1 cos x 5 2cos x 1 cos x
2 cos x 2 1 5 0
2 cos x 2 1 1 1 5 0 1 1
2 cos x 5 1
2 cos x
1
5
2
2
1
cos x 5
2
Given cos x 5 12 and 0 # x # 2p, the solutions to
the equation must be x 5 1.05 or 5.24.
d) The equation 5 sin x 1 1 5 3 sin x can be
rewritten as follows:
5 sin x 1 1 5 3 sin x
5 sin x 1 1 2 3 sin x 5 3 sin x 2 3 sin x
2 sin x 1 1 5 0
2 sin x 1 1 2 1 5 0 2 1
2 sin x 5 21
2 sin x
1
52
2
2
1
sin x 5 2
2
Given sin x 5 2 12 and 0 # x # 2p, the solutions to
the equation must be x 5 3.67 or 5.76.
9. a) The equation 2 2 2 cot x 5 0 can be rewritten
as follows:
2 2 2 cot x 5 0
2 2 2 cot x 1 2 cot x 5 0 1 2 cot x
2 5 2 cot x
2
2 cot x
5
2
2
cot x 5 1
Given cot x 5 1 and 0 # x # 2p, the solutions to
the equation must be x 5 0.79 or 3.93.
b) The equation csc x 2 2 5 0 can be rewritten as
follows:
csc x 2 2 5 0
csc x 2 2 1 2 5 0 1 2
csc x 5 2
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Given csc x 5 2 and 0 # x # 2p, the solutions to
the equation must be x 5 0.52 or 2.62.
c) The equation 7 sec x 5 7 can be rewritten as follows:
7 sec x 5 7
7
7 sec x
5
7
7
sec x 5 1
Given sec x 5 1 and 0 # x # 2p, the solutions to
the equation must be x 5 0 or 6.28.
d) The equation 2 csc x 1 17 5 15 1 csc x can be
rewritten as follows:
2 csc x 1 17 5 15 1 csc x
2 csc x 1 17 2 csc x 5 15 1 csc x 2 csc x
csc x 1 17 5 15
csc x 1 17 2 17 5 15 2 17
csc x 5 22
Given csc x 5 22 and 0 # x # 2p, the solutions to
the equation must be x 5 3.67 or 5.76.
e) The equation 2 sec x 1 1 5 6 can be rewritten as
follows:
2 sec x 1 1 5 6
2 sec x 1 1 2 1 5 6 2 1
2 sec x 5 5
5
2 sec x
5
2
2
5
sec x 5
2
Given sec x 5 52 and 0 # x # 2p, the solutions to
the equation must be x 5 1.16 or 5.12.
f) The equation 8 1 4 cot x 5 10 can be rewritten
as follows:
8 1 4 cot x 5 10
8 1 4 cot x 2 8 5 10 2 8
4 cot x 5 2
2
4 cot x
5
4
4
1
cot x 5
2
Given cot x 5 12 and 0 # x # 2p, the solutions to
the equation must be x 5 1.11 or 4.25.
1
10. a) Given sin 2x 5 !2
and 0 # x # 2p, 2x must
equal 0.79 1 2kp or 2.36 1 2kp. Therefore, the
solutions to the equation must be x 5 0.79
2 5 0.39,
2.36
5
1.18,
3.53,
or
4.32.
2
b) Given sin 4x 5 12 and 0 # x # 2p, 4x must equal
0.52 1 2kp or 2.62 1 2kp. Therefore, the solutions to
2.62
the equation must be x 5 0.52
4 5 0.13, 4 5 0.65, 1.70,
2.23, 3.27, 3.80, 4.84, and 5.37.
c) Given sin 3x 5 2 !3
2 and 0 # x # 2p, 3x must
7-36
equal 4.19 1 2kp or 5.24 1 2kp. Therefore, the
solutions to the equation must be x 5 4.19
3 5 1.40,
5.24
3 5 1.75, 3.49, 3.84, 5.59, or 5.93.
1
d) Given cos 4x 5 2 !2
and 0 # x # 2p, 4x must
equal 2.36 1 2kp or 3.93 1 2kp. Therefore, the
solutions to the equation must be x 5 2.36
4 5 0.59,
3.93
5
0.985,
2.16,
2.55,
3.73,
4.12,
5.30,
or 5.697.
4
e) Given cos 2x 5 2 12 and 0 # x # 2p, 2x must
equal 2.09 1 2kp or 4.19 1 2kp. Therefore, the
solutions to the equation must be x 5 2.09
2 5 1.05,
4.19
5
2.09,
4.19,
or
5.24.
2
f) Given cos 5 !3
must equal
2 and 0 # x # 2p,
2
2
0.52 or 5.76. Therefore, the solution to the equation
must be x 5 (2)(0.52) 5 1.05. 5.76 is not a solution
to the equation since (2)(5.76) 5 11.52 is greater
than 2p.
11. When graphed, the function modelling the city’s
daily high temperature is as follows:
x
x
40
30
Temperature (˚C)
08-035_07_AFSM_C07_001-060.qxd
20
10
0
–10
–20
73
146
219
292
365
Day of the year
First it’s necessary to find the first day when the
temperature is approximately 32°C. This can be
done as follows:
2p
d 1 10
32 5 228 cos
365
2p
d 1 10 2 10
32 2 10 5 228 cos
365
2p
d
22 5 228 cos
365
228
2p
22
5
cos
d
228
228
365
2p
20.7857 5 cos
d
365
2p
cos21 (20.7857) 5 cos21 acos
db
365
2p
2.4746 5
d
365
Chapter 7: Trigonometric Identities and Equations
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Page 37
2p
365
365
5
d3
2p
365
2p
d 5 143.76
Since the first day when the temperature is
approximately 32 °C is day 144, and since the
period of the function is 365 days, the other day
when the temperature is approximately 32 °C is
365 2 144 5 221. Therefore, the temperature is
approximately 32 °C or above from about day 144 to
about day 221, and those are the days of the year
when the air conditioners are running at the City Hall.
12. When graphed, the function modelling the
height of the nail above the surface of the water is
as follows:
8
2.4746 3
Height
6
t 2 1 1 1 5 0.8596 1 1
t 5 1.86
Since the first time when the nail is at the surface of the
water is 1.86 s, and since the period of the function is
8 s, the next time the nail is at the surface of the water
is 6 2 1.86 5 4.14 s. Therefore, the nail is below the
water when 1.86 s , t , 4.14 s. Since the cycle
repeats itself two more times in the first 24 s that the
wheel is rotating, the nail is also below the water when
9.86 s , t , 12.14 and when 17.86 s , t , 20.14 s.
13. To solve sin Q x 1 R 5 !2 cos x for
4
p
0 # x # 2p, graph the functions y 5 sin Q x 1 R
4
and y 5 !2 cos x on the same coordinate grid as
follows:
y
1
4
x
0
2
0
–2
–4
p
4
8
12
16
20 24
p
2
–1
3p
2
p
2p
Since the graphs intersect when x 5
Time (s)
First it’s necessary to find the first time when the
nail is at the surface of the water. This can be done
as follows:
p
0 5 24 sin (t 2 1) 1 2.5
4
p
0 2 2.5 5 24 sin (t 2 1) 1 2.5 2 2.5
4
p
22.5 5 24 sin (t 2 1)
4
24
p
22.5
5
sin (t 2 1)
24
24
4
p
0.625 5 sin (t 2 1)
4
p
sin21 (0.625) 5 sin21 asin (t 2 1)b
4
p
0.6751 5 (t 2 1)
4
4
p
4
0.6751 3 5 (t 2 1) 3
p
p
4
t 2 1 5 0.8596
Advanced Functions Solutions Manual
x5
5p
,
4
p
4
and when
the solution to the equation is x 5
p
4
or
5p
.
4
1
14. Given sin 2u 5 2 !2
and 0 # x # 2p, 2u must
equal
5p
1 2pk
4
or
7p
1 2pk.
4
Therefore, the
solutions to the equation must be
u5
5p
1 2pk
4
2
5
5p
1 pk or
8
7p
1 2pk
4
7p
5
1 pk
2
8
5p 7p 13p
15p
, and
. When
So the solution are , ,
8 8
8
8
the solutions are plotted on the graph of the
function y 5 sin 2u for 0 # u # 2p, the graph
appears as follows:
y
1.5
2
13p
1.0
8 ,– 2
0.5
x
0
3p
p
p
–0.5
2
2
–1.0
2
15p
5p 2
7p 2
–1.5 8 , – 2
8 ,– 2
8,– 2
7-37
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15. The value of f(x) 5 sin x is the same at x
and p 2 x. In other words, it is the same at x
and half the period minus x. Since the period of
p
f(x) 5 25 sin (x 1 20) 2 55 is 100, if the
50
function were not horizontally translated, its value at x
would be the same as at 50 2 x. The function is
horizontally translated 20 units to the left, however,
so it goes through half its period from x 5 220 to
x 5 30. At x 5 3, the function is 23 units away
from the left end of the range, so it will have the
same value at x 5 30 2 23 or x 5 7, which is
23 units away from the right end of the range.
16. To solve a trigonometric equation algebraically,
first isolate the trigonometric function on one side of
the equation. For example, the trigonometric equation
5 cos x 2 3 5 2 would become 5 cos x 5 5, which
would then become cos x 5 1. Next, apply the
inverse of the trigonometric function to both sides of
the equation. For example, the trigonometric equation
cos x 5 1 would become x 5 cos21 1. Finally,
simplify the equation. For example, x 5 cos21 1
would become x 5 0 1 2np, where nPI.
To solve a trigonometric equation graphically, first
isolate the trigonometric function on one side of the
equation. For example, the trigonometric equation
5 cos x 2 3 5 2 would become 5 cos x 5 5, which
would then become cos x 5 1. Next, graph both
sides of the equation. For example, the functions
f(x) 5 cos x and f(x) 5 1 would both be graphed.
Finally, find the points where the two graphs
intersect. For example, f(x) 5 cos x and f(x) 5 1
would intersect at x 5 0 1 2np, where nPI.
Similarity: Both trigonometric functions are first
isolated on one side of the equation.
Differences: The inverse of a trigonometric function is
not applied in the graphical method, and the points of
intersection are not obtained in the algebraic method.
17. To solve the trigonometric equation
2 sin x cos x 1 sin x 5 0, first factor sin x from the
left side of the equation as follows:
2 sin x cos x 1 sin x 5 0
(sin x)(2 cos x 1 1) 5 0
Since (sin x)(2 cos x 1 1) 5 0, either sin x 5 0
or 2 cos x 1 1 5 0 (or both). If sin x 5 0, one
solution to the equation is x 5 0 1 np, where nPI.
If 2 cos x 1 1 5 0, x can be found by first rewriting
the equation as follows:
2 cos x 1 1 5 0
2 cos x 1 1 2 1 5 0 2 1
2 cos x 5 21
7-38
2 cos x
1
52
2
2
1
cos x 5 2
2
The solutions to the equation cos x 5 2 12 occur at
2p
4p
x5
1 2np or
1 2np, where nPI, since
3
3
the period of the cosine function is 2p. Therefore, the
solutions to the equation 2 sin x cos x 1 sin x 5 0
are x 5 0 1 np,
2p
1 2np,
3
or
4p
1 2np,
3
where
nPI.
18. a) Since sin 2u 5 2 sin u cos u, the equation
sin 2x 2 2 cos2 x 5 0 can be rewritten as follows:
sin 2x 2 2 cos2 x 5 0
2 sin x cos x 2 2 cos2 x 5 0
(2 cos x)(sin x 2 cos x) 5 0
Since (2 cos x)(sin x 2 cos x) 5 0, either
2 cos x 5 0 or sin x 2 cos x 5 0 (or both). If
2 cos x 5 0, x can be found by first rewriting the
equation as follows:
2 cos x 5 0
2 cos x
0
5
2
2
cos x 5 0
The solutions to the equation cos x 5 0 occur at
p
3p
x 5 or . If sin x 2 cos x 5 0, x can be found by
2
2
first rewriting the equation as follows:
sin x 2 cos x 5 0
sin x 2 cos x 1 cos x 5 0 1 cos x
sin x 5 cos x
The solutions to the equation sin x 5 cos x occur
p
5p
at x 5 or . Therefore, the solutions to the
4
4
p p 5p
equation sin 2x 2 2 cos2 x 5 0 are x 5 , , ,
4 2 4
3p
or .
2
b) Since cos 2u 5 1 2 2 sin2 u, the equation
3 sin x 1 cos 2x 5 2 can be rewritten as follows:
3 sin x 1 cos 2x 5 2
3 sin x 1 1 2 2 sin2 x 5 2
3 sin x 1 1 2 2 sin2 x 2 2 5 2 2 2
3 sin x 2 1 2 2 sin2 x 5 0
22 sin2 x 1 3 sin x 2 1 5 0
(21)(22 sin2 x 1 3 sin x 2 1) 5 (21)(0)
2 sin2 x 2 3 sin x 1 1 5 0
(2 sin x 2 1)(sin x 2 1) 5 0
Since (2 sin x 2 1)(sin x 2 1) 5 0, either
2 sin x 2 1 5 0 or sin x 2 1 5 0 (or both).
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If sin x 2 1 5 0, x can be found by first rewriting
the equation as follows:
2 sin x 2 1 5 0
2 sin x 2 1 1 1 5 0 1 1
2 sin x 5 1
2 sin x
1
5
2
2
1
sin x 5
2
The solutions to the equation sin x 5 12 occur at
x5
p
6
or
5p
.
6
If sin x 2 1 5 0, x can be found by
first rewriting the equation as follows:
sin x 2 1 5 0
sin x 2 1 1 1 5 0 1 1
sin x 5 1
The solution to the equation sin x 5 1 occurs at
p
x 5 . Therefore, the solutions to the equation
2
p p
6 2
3 sin x 1 cos 2x 5 2 are x 5 , , or
5p
.
6
7.6 Solving Quadratic Trigonometric
Equations, pp. 435–437
1. a) The expression sin2 u 2 sin u can be factored
as follows:
sin2 u 2 sin u
(sin u)(sin u 2 1)
b) The expression cos2 u 2 2 cos u 1 1 can be
factored as follows:
cos2 u 2 2 cos u 1 1
(cos u 2 1)(cos u 2 1)
c) The expression 3 sin2 u 2 sin u 2 2 can be
factored as follows:
(3 sin u 1 2)(sin u 2 1)
d) The expression 4 cos2 u 2 1 can be factored as
follows:
4 cos2 u 2 1
(2 cos u 2 1)(2 cos u 1 1)
e) The expression 24 sin2 x 2 2 sin x 2 2 can be
factored as follows:
24 sin2 x 2 2 sin x 2 2
(6 sin x 2 2)(4 sin x 1 1)
f) The expression 49 tan2 x 2 64 can be factored as
follows:
49 tan2 x 2 64
(7 tan x 1 8)(7 tan x 2 8)
2. a) The equation y 2 5 13 can be solved as follows:
1
y2 5
3
1
"y 2 5 6
Å3
1
y56
Å3
!3
y56
3
Likewise, the equation tan2 x 5 13 can be solved
as follows:
1
tan2 x 5
3
1
"tan2 x 5 6
Å3
1
tan x 5 6
Å3
!3
tan x 5 6
3
The solutions to the equation tan x 5 6 !3
3 occur
p 5p 7p
, ,
6 6 6
at x 5 ,
or
11p
.
6
b) The equation y 2 1 y 5 0 can be solved as
follows:
y2 1 y 5 0
(y)(y 1 1) 5 0
Since (y)(y 1 1) 5 0, either y 5 0 or y 1 1 5 0
(or both). If y 1 1 5 0, y can be solved for as
follows:
y1150
y11215021
y 5 21
Therefore, the solutions to the equation y 2 1 y 5 0
are y 5 0 or y 5 21. Likewise, the equation
sin2 x 1 sin x 5 0 can be solved as follows:
sin2 x 1 sin x 5 0
(sin x)(sin x 1 1) 5 0
Since (sin x)(sin x 1 1) 5 0, either sin x 5 0 or
sin x 1 1 5 0 (or both). The solutions to the
equation sin x 5 0 occur at x 5 0, p, or 2p. If
sin x 1 1 5 0, x can be solved for as follows:
sin x 1 1 5 0
sin x 1 1 2 1 5 0 2 1
sin x 5 21
The solution to the equation sin x 5 21 occurs
3p
at x 5 , so the solutions to the equation
2
2
sin x 1 sin x 5 0 are x 5 0, p,
Advanced Functions Solutions Manual
3p
,
2
or 2p.
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c) The equation y 2 2yz 5 0 can be solved as
follows:
y 2 2yz 5 0
(y)(1 2 2z) 5 0
Since (y)(1 2 2z) 5 0, either y 5 0 or 1 2 2z 5 0
(or both). If 1 2 2z 5 0, z can be solved for as
follows:
1 2 2z 5 0
1 2 2z 1 2z 5 0 1 2z
1 5 2z
1
2z
5
2
2
1
z5
2
Likewise, the equation cos x 2 2 cos x sin x 5 0
can be solved as follows:
cos x 2 2 cos x sin x 5 0
(cos x)(1 2 2 sin x) 5 0
Since (cos x)(1 2 2 sin x) 5 0, either cos x 5 0 or
1 2 2 sin x 5 0 (or both). The solutions to the
equation cos x 5 0 occur at x 5
p
2
or
3p
.
2
If
1 2 2 sin x 5 0, x can be solved for as follows:
1 2 2 sin x 5 0
1 2 2 sin x 1 2 sin x 5 0 1 2 sin x
1 5 2 sin x
1
2 sin x
5
2
2
1
sin x 5
2
The solutions to the equation sin x 5 12 occur at
x5
p
6
or
5p
,
6
so the solutions to the equation
p p 5p
,
6 2 6
cos x 2 2 cos x sin x 5 0 are x 5 , ,
or
3p
.
2
d) The equation yz 5 y can be solved as follows:
yz 5 y
yz 2 y 5 y 2 y
yz 2 y 5 0
(y)(z 2 1) 5 0
Since (y)(z 2 1) 5 0, either y 5 0 or z 2 1 5 0
(or both). If z 2 1 5 0, z can be solved for as
follows:
z2150
z21115011
z51
Therefore, the solutions to the equation yz 5 y
are y 5 0 or z 5 1. Likewise, the equation
tan x sec x 5 tan x can be solved as follows:
tan x sec x 5 tan x
tan x sec x 2 tan x 5 tan x 2 tan x
7-40
tan x sec x 2 tan x 5 0
(tan x)(sec x 2 1) 5 0
Since (tan x)(sec x 2 1) 5 0, either tan x 5 0 or
sec x 2 1 5 0 (or both). The solutions to the equation
tan x 5 0 occur at x 5 0, p, or 2p. If sec x 2 1 5 0,
x can be solved for as follows:
sec x 2 1 5 0
sec x 2 1 1 1 5 0 1 1
sec x 5 1
The solutions to the equation sec x 5 1 occur at
x 5 0 or 2p, so the solutions to the equation
(tan x)(sec x 2 1) 5 0 are x 5 0, p, or 2p.
3. a) The equation 6y 2 2 y 2 1 5 0 can be solved
as follows:
6y 2 2 y 2 1 5 0
(2y 2 1)(3y 1 1) 5 0
Since (2y 2 1)(3y 1 1) 5 0, either 2y 2 1 5 0 or
3y 1 1 5 0 (or both). If 2y 2 1 5 0, y can be
solved for as follows:
2y 2 1 5 0
2y 2 1 1 1 5 0 1 1
2y 5 1
2y
1
5
2
2
1
y5
2
If 3y 1 1 5 0, y can be solved for as follows:
3y 1 1 5 0
3y 1 1 2 1 5 0 2 1
3y 5 21
3y
1
52
3
3
1
y52
3
Therefore, the solutions to the equation
6y 2 2 y 2 1 5 0 are y 5 12 or 2 13.
b) The equation 6 cos2 x 2 cos x 2 1 5 0 for
0 # x # 2p can be solved as follows:
6 cos2 x 2 cos x 2 1 5 0
(2 cos x 2 1)(3 cos x 1 1) 5 0
Since (2 cos x 2 1)(3 cos x 1 1) 5 0, either
2 cos x 2 1 5 0 or 3 cos x 1 1 5 0 (or both). If
2 cos x 2 1 5 0, x can be solved for as follows:
2 cos x 2 1 5 0
2 cos x 2 1 1 1 5 0 1 1
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2 cos x 5 1
2 cos x
1
5
2
2
1
cos x 5
2
The solutions to the equation cos x 5 12 occur at
x 5 1.05 or 5.24. If 3 cos x 1 1 5 0, x can be
solved for as follows:
3 cos x 1 1 5 0
3 cos x 1 1 2 1 5 0 2 1
3 cos x 5 21
3 cos x
1
52
3
3
1
cos x 5 2
3
The solutions to the equation cos x 5 2 13 occur at
x 5 1.91 or 4.37. Therefore, the solutions to the
equation 6 cos2 x 2 cos x 2 1 5 0 are x 5 1.05,
1.91, 4.37, or 5.24.
4. a) The equation sin2 u 5 1 can be solved as
follows:
sin2 u 5 1
"sin2 u 5 6 !1
sin u 5 61
The solutions to the equation sin u 5 61 occur at
u 5 90° or 270°.
b) The equation cos2 u 5 1 can be solved as follows:
cos2 u 5 1
"cos2 u 5 6 !1
cos u 5 61
The solutions to the equation cos u 5 61 occur at
u 5 0°, 180°, or 360°.
c) The equation tan2 u 5 1 can be solved as follows:
tan2 u 5 1
"tan2 u 5 6 !1
tan u 5 61
The solutions to the equation tan u 5 61 occur at
u 5 45°, 135°, 225°, or 315°.
d) The equation 4 cos2 u 5 1 can be solved as
follows:
4 cos2 u 5 1
4 cos2 u
1
5
4
4
1
cos2 u 5
4
1
"cos2 u 5 6
Å4
Advanced Functions Solutions Manual
1
2
The solutions to the equation cos u 5 612 occur at
u 5 60°, 120°, 240°, or 300°.
e) The equation 3 tan2 u 5 1 can be solved as
follows:
3 tan2 u 5 1
3 tan2 u
1
5
3
3
1
tan2 u 5
3
cos u 5 6
"tan2 u 5 6
tan u 5 6
1
Å3
!3
3
The solutions to the equation tan u 5 6 !3
3 occur
at u 5 30°, 150°, 210°, or 330°.
f) The equation 2 sin2 u 5 1 can be solved as follows:
2 sin2 u 5 1
2 sin2 u
1
5
2
2
1
sin2 u 5
2
1
"sin2 u 5 6
Å2
!2
sin u 5 6
2
The solutions to the equation sin u 5 6 !2
2 occur at
u 5 45°, 135°, 225°, or 315°.
5. a) Since sin x cos x 5 0, either sin x 5 0 or
cos x 5 0 (or both). If sin x 5 0, the solutions
for x occur at x 5 0°, 180°, or 360°. If cos x 5 0,
the solutions for x occur at x 5 90° or 270°.
Therefore, the solutions to the equation occur at
x 5 0°, 90°, 180°, 270°, or 360°.
b) Since sin x(cos x 2 1) 5 0, either sin x 5 0
or cos x 2 1 5 0 (or both). If sin x 5 0, the
solutions for x occur at x 5 0°, 180°, or 360°. If
cos x 2 1 5 0, x can be solved for as follows:
cos x 2 1 5 0
cos x 2 1 1 1 5 0 1 1
cos x 5 1
The solutions to the equation cos x 5 1 occur at
x 5 0° or 360°. Therefore, the solutions to the
equation sin x(cos x 2 1) 5 0 occur at x 5 0°,
180°, or 360°.
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c) Since (sin x 1 1) cos x 5 0, either sin x 1 1 5 0
or cos x 5 0 (or both). If sin x 1 1 5 0, x can be
solved for as follows:
sin x 1 1 5 0
sin x 1 1 2 1 5 0 2 1
sin x 5 21
The solution to the equation sin x 5 21 occurs at
x 5 270°. If cos x 5 0, the solutions for x occur at
x 5 90° or 270°. Therefore, the solutions to the
equation (sin x 1 1) cos x 5 0 occur at x 5 90°
or 270°.
d) Since cos x (2 sin x 2 !3) 5 0, either cos x 5 0
or 2 sin x 2 !3 5 0 (or both). If cos x 5 0, the
solutions for x occur at x 5 90° or 270°. If
2 sin x 2 !3 5 0, x can be solved for as follows:
2 sin x 2 !3 5 0
2 sin x 2 !3 1 !3 5 0 1 !3
2 sin x 5 !3
!3
2 sin x
5
2
2
!3
sin x 5
2
The solutions to the equation sin x 5 !3
2 occur at
x 5 60° and 120°. Therefore, the solutions to the
equation cos x(2 sin x 2 !3) 5 0 occur at x 5 60°,
90°, 120°, or 270°.
e) Since ( !2 sin x 2 1)( !2 sin x 1 1) 5 0, either
!2 sin x 2 1 5 0 or !2 sin x 1 1 5 0 (or both). If
!2 sin x 2 1 5 0, x can be solved for as follows:
!2 sin x 2 1 5 0
!2 sin x 2 1 1 1 5 0 1 1
!2 sin x 5 1
!2 sin x
1
5
!2
!2
1
sin x 5
!2
!2
sin x 5
2
The solutions to the equation sin x 5 !2
2 occur at
x 5 45° or 135°. If !2 sin x 1 1 5 0, x can be
solved for as follows:
!2 sin x 1 1 5 0
!2 sin x 1 1 2 1 5 0 2 1
!2 sin x 5 21
7-42
!2 sin x
1
52
!2
!2
sin x 5 2
sin x 5 2
1
!2
!2
2
The solutions to the equation sin x 5 2 "2
2 occur
at x 5 225° or 315°. Therefore, the solutions to the
equation ("2 sin x 2 1)("2 sin x 1 1) 5 0 occur
at x 5 45°, 135°, 225°, or 315°.
f) Since (sin x 2 1)(cos x 1 1) 5 0, either
sin x 2 1 5 0 or cos x 1 1 5 0 (or both). If
sin x 2 1 5 0, x can be solved for as follows:
sin x 2 1 5 0
sin x 2 1 1 1 5 0 1 1
sin x 5 1
The solution to the equation sin x 5 1 occurs at
x 5 90°. If cos x 1 1 5 0, x can be solved for as
follows:
cos x 1 1 5 0
cos x 1 1 2 1 5 0 2 1
cos x 5 21
The solution to the equation cos x 5 21 occurs at
x 5 180°. Therefore, the solutions to the equation
(sin x 2 1)(cos x 1 1) 5 0 occur at x 5 90°
or 180°.
6. a) Since (2 sin x 2 1) cos x 5 0, either
2 sin x 2 1 5 0 or cos x 5 0. If 2 sin x 2 1 5 0,
x can be solved for as follows:
2 sin x 2 1 5 0
2 sin x 2 1 1 1 5 0 1 1
2 sin x 5 1
1
2 sin x
5
2
2
1
sin x 5
2
The solutions to the equation sin x 5 12 occur at
x5
p
6
or
5p
. Also,
6
the solutions to the equation
cos x 5 0 occur at x 5
p
2
or
3p
.
2
Therefore, the
solutions to the equation (2 sin x 2 1) cos x 5 0
p p 5p
,
6 2 6
occur at x 5 , ,
or
3p
.
2
b) Since (sin x 1 1)2 5 0, sin x 1 1 5 0, so x can
be solved for as follows:
sin x 1 1 5 0
sin x 1 1 2 1 5 0 2 1
sin x 5 21
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The solution to the equation sin x 5 21 occurs at
3p
x 5 , so the solution to the equation
2
(sin x 1 1)2 5 0 occurs at x 5
3p
.
2
c) Since (2 cos x 1 !3) sin x 5 0, either
2 cos x 1 !3 5 0 or sin x 5 0 (or both). If
2 cos x 1 !3 5 0, x can be solved for as follows:
2 cos x 1 !3 5 0
2 cos x 1 !3 2 !3 5 0 2 !3
2 cos x 5 2 !3
"3
2 cos x
52
2
2
!3
cos x 5 2
2
The solutions to the equation cos x 5 2 !3
2 occur
at x 5
5p
6
or
7p
. Also,
6
the solutions to the equation
sin x 5 0 occur at x 5 0, p, or 2p. Therefore, the
solutions to the equation (2 cos x 1 !3) sin x 5 0
occur at x 5 0,
5p
7p
, p, ,
6
6
or 2p.
d) Since (2 cos x 2 1)(2 sin x 1 !3) 5 0, either
2 cos x 2 1 5 0 or 2 sin x 1 !3 5 0. If
2 cos x 2 1 5 0, x can be solved for as follows:
2 cos x 2 1 5 0
2 cos x 2 1 1 1 5 0 1 1
2 cos x 5 1
1
2 cos x
5
2
2
1
cos x 5
2
The solutions to the equation cos x 5 12 occur at
x 5 or . If 2 sin x 1 !3 5 0, x can be solved
3
3
for as follows:
2 sin x 1 !3 5 0
p
5p
2 sin x 1 !3 2 !3 5 0 2 !3
2 sin x 5 2 !3
2 sin x
!3
52
2
2
!3
2
The solutions to the equation sin x 5 2 !3
2 occur
sin x 5 2
at x 5
4p
3
or
5p
.
3
Therefore, the solutions to the
e) Since ( !2 cos x 2 1)( !2 cos x 1 1) 5 0, either
!2 cos x 2 1 5 0 or !2 cos x 1 1 5 0 (or both).
If !2 cos x 2 1 5 0, x can be solved for as follows:
!2 cos x 2 1 5 0
!2 cos x 2 1 1 1 5 0 1 1
!2 cos x 5 1
!2 cos x
1
5
!2
!2
1
cos x 5
!2
!2
cos x 5
2
The solutions to the equation cos x 5 !2
2 occur at
or . If !2 cos x 1 1 5 0, x can be solved
4
for as follows:
!2 cos x 1 1 5 0
!2 cos x 1 1 2 1 5 0 2 1
!2 cos x 5 21
!2 cos x
1
52
!2
!2
1
cos x 5 2
!2
!2
cos x 5 2
2
The solutions to the equation cos x 5 2 !2
2 occur
x5
p
4
at x 5
7p
3p
4
or
5p
.
4
Therefore, the solutions to the
equation ( !2 cos x 2 1)( !2 cos x 1 1) 5 0 occur
p 3p 5p
, ,
4 4 4
at x 5 ,
or
7p
.
4
f) Since (sin x 1 1)(cos x 2 1) 5 0, either
sin x 1 1 5 0 or cos x 2 1 5 0 (or both). If
sin x 1 1 5 0, x can be solved for as follows:
sin x 1 1 5 0
sin x 1 1 2 1 5 0 2 1
sin x 5 21
The solution to the equation sin x 5 21 occurs at
3p
x 5 . If cos x 2 1 5 0, x can be solved for as
2
follows:
cos x 2 1 5 0
cos x 2 1 1 1 5 0 1 1
cos x 5 1
equation (2 cos x 2 1)(2 sin x 1 !3) 5 0 occur at
p 4p
,
3 3
x5 ,
or
5p
.
3
Advanced Functions Solutions Manual
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Page 44
The solutions to the equation cos x 5 1 occur at
x 5 0 or 2p. Therefore, the solutions to the equation
(sin x 1 1)(cos x 2 1) 5 0 occur at
3p
x 5 0, ,
2
7. a) Since 2 cos2 u 1 cos u 2 1 5 0,
(2 cos u 2 1)(cos u 1 1) 5 0. For this reason,
either 2 cos u 2 15 0 or cos u 1 1 5 0 (or both).
If 2 cos u 2 1 5 0, u can be solved for as follows:
2 cos u 2 1 5 0
2 cos u 2 1 1 1 5 0 1 1
2 cos u 5 1
2 cos u
1
5
2
2
1
cos u 5
2
The solutions to the equation cos u 5 12 occur at
p
3
5p
or . If cos u 1 1 5 0, u can be solved for
3
as follows:
cos u 1 1 5 0
cos u 1 1 2 1 5 0 2 1
cos u 5 21
The solution to the equation cos u 5 21 occurs at
u 5 p. Therefore, the solutions to the equation
p
5p
2 cos2 u 1 cos u 2 1 5 0 occur at u 5 , p, or .
3
3
2
b) The equation 2 sin u 5 1 2 sin u can be rewritten
and factored as follows:
2 sin2 u 5 1 2 sin u
2
2 sin u 2 1 1 sin u 5 1 2 sin u 2 1 1 sin u
2 sin2 u 1 sin u 2 1 5 0
(2 sin u 2 1)(sin u 1 1) 5 0
Since (2 sin u 2 1)(sin u 1 1) 5 0, either
2 sin u 2 1 5 0 or sin u 1 1 5 0 (or both). If
2 sin u 2 1 5 0, u can be solved for as follows:
2 sin u 2 1 5 0
2 sin u 2 1 1 1 5 0 1 1
2 sin u 5 1
2 sin u
1
5
2
2
1
sin u 5
2
The solutions to the equation sin u 5 12 occur at
u5
p
6
or
5p
.
6
If sin u 1 1 5 0, u can be solved for
as follows:
sin u 1 1 5 0
sin u 1 1 2 1 5 0 2 1
sin u 5 21
7-44
x5
3p
.
2
Therefore, the solutions to the equation
p 5p
,
6 6
2 sin2 u 5 1 2 sin u occur at u 5 ,
or 2p.
u5
The solution to the equation sin u 5 21 occurs at
or
3p
.
2
c) The equation cos2 u 5 2 1 cos u can be rewritten
and factored as follows:
cos2 u 5 2 1 cos u
2
cos u 2 2 2 cos u 5 2 1 cos u 2 2 2 cos u
cos2 u 2 cos u 2 2 5 0
(cos u 2 2)(cos u 1 1) 5 0
Since (cos u 2 2)(cos u 1 1) 5 0, either
cos u 2 2 5 0 or cos u 1 1 5 0. If cos u 2 2 5 0,
the equation can be rewritten as follows:
cos u 2 2 5 0
cos u 2 2 1 2 5 0 1 2
cos u 5 2
Since cos u can never equal 2, the factor cos u 2 2
can be ignored. If cos u 1 1 5 0, u can be solved
for as follows:
cos u 1 1 5 0
cos u 1 1 2 1 5 0 2 1
cos u 5 21
The solution to the equation cos u 5 21 occurs at
u 5 p. Therefore, the solution to the equation
cos2 u 5 2 1 cos u occurs at u 5 p.
d) Since 2 sin2 u 1 5 sin u 2 3 5 0,
(2 sin u 2 1)(sin u 1 3) 5 0. For this reason, either
2 sin u 2 1 5 0 or sin u 1 3 5 0. If 2 sin u 2 1 5 0,
u can be solved for as follows:
2 sin u 2 1 5 0
2 sin u 2 1 1 1 5 0 1 1
2 sin u 5 1
2 sin u
1
5
2
2
1
sin u 5
2
The solutions to the equation sin u 5 12 occur at
u5
p
6
or
5p
.
6
If sin u 1 3 5 0, the equation can be
rewritten as follows:
sin u 1 3 5 0
sin u 1 3 2 3 5 0 2 3
sin u 5 23
Since sin u can never equal 23, the factor sin u 1 3
can be ignored. Therefore, the solutions to the
p
equation 2 sin2 u 1 5 sin u 2 3 5 0 occur at u 5
or
5p
.
6
6
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Page 45
e) The equation 3 tan2 u 2 2 tan u 5 1 can be
rewritten and factored as follows:
3 tan2 u 2 2 tan u 5 1
3 tan2 u 2 2 tan u 2 1 5 1 2 1
3 tan2 u 2 2 tan u 2 1 5 0
(3 tan u 1 1)(tan u 2 1) 5 0
Since (3 tan u 1 1)(tan u 2 1) 5 0, either
3 tan u1 1 5 0 or tan u 2 1 5 0 (or both). If
3 tan u 1 1 5 0, u can be solved for as follows:
3 tan u 1 1 5 0
3 tan u 1 1 2 1 5 0 2 1
3 tan u 5 21
1
3 tan u
52
3
3
1
tan u 5 2
3
The solutions to the equation tan u 5 2 13 occur at
u 5 2.81 or 5.96. If tan u 2 1 5 0, u can be solved
for as follows:
tan u 2 1 5 0
tan u 2 1 1 1 5 0 1 1
tan u 5 1
The solutions to the equation tan u 5 1 occur at
u5
p
4
or
5p
. Therefore,
4
the solutions to the equation
p
5p
3 tan2 u 2 2 tan u 5 1 occur at u 5 , 2.82, ,
4
4
or 5.96.
f) Since 12 sin2 u 1 sin u 2 6 5 0,
(4 sin u 1 3) (3 sin u 2 2) 5 0. For this reason,
either 4 sin u 1 3 5 0 or 3 sin u 2 2 5 0 (or both).
If 4 sin u 1 3 5 0, u can be solved for as follows:
4 sin u 1 3 5 0
4 sin u 1 3 2 3 5 0 2 3
4 sin u 5 23
3
4 sin u
52
4
4
3
sin u 5 2
4
The solutions to the equation sin u 5 2 34 occur at
u 5 3.99 or 5.44. If 3 sin u 2 2 5 0, u can be solved
for as follows:
3 sin u 2 2 5 0
3 sin u 2 2 1 2 5 0 1 2
3 sin u 5 2
2
3 sin u
5
2
3
2
sin u 5
3
Advanced Functions Solutions Manual
The solutions to the equation sin u 5 23 occur at
u 5 0.73 or 2.41. Therefore, the solutions to the
equation 12 sin2 u 1 sin u 2 6 5 0 occur at
u 5 0.73, 2.41, 3.99, or 5.44.
8. a) Since sec x csc x 2 2 csc x 5 0,
(csc x) (sec x 2 2) 5 0. For this reason, either
csc x 5 0 or sec x 2 2 5 0. Since csc x can never
equal 0, the factor csc x can be ignored. If
sec x 2 2 5 0, x can be solved for as follows:
sec x 2 2 5 0
sec x 2 2 1 2 5 0 1 2
sec x 5 2
The solutions to the equation sec x 5 2 occur at
x5
p
3
or
5p
. Therefore,
3
the solutions to the equation
sec x csc x 2 2 csc x 5 0 occur at x 5
p
3
or
5p
.
3
b) Since 3 sec2 x 2 4 5 0,
( !3 sec x 2 2) (!3 sec x 1 2) 5 0. For this reason,
either !3 sec x 2 2 5 0 or !3 sec x 1 2 5 0
(or both). If !3 sec x 2 2 5 0, x can be solved for
as follows:
!3 sec x 2 2 5 0
!3 sec x 2 2 1 2 5 0 1 2
!3 sec x 5 2
!3 sec x
2
5
!3
!3
2
sec x 5
!3
2 !3
sec x 5
3
The solutions to the equation sec x 5 2"3
3 occur at
x5
p
6
or
11p
.
6
If !3 sec x 1 2 5 0, x can be solved
for as follows:
!3 sec x 1 2 5 0
!3 sec x 1 2 2 2 5 0 2 2
!3 sec x 5 22
!3 sec x
2
52
!3
!3
2
sec x 5 2
!3
2!3
sec x 5 2
3
The solutions to the equation sec x 5 2 2"3
3 occur
7-45
08-035_07_AFSM_C07_001-060.qxd
at x 5
5p
6
or
7p
.
6
7/18/08
12:55 PM
Page 46
csc x 1 !2 5 0. The solutions to the equation
Therefore, the solutions to the
2
equation 3 sec x 2 4 5 0 occur at
11p
p 5p 7p
x5 , , ,
6 6 6
or
.
6
c) Since 2 sin x sec x 2 2 !3 sin x 5 0,
(sin x) (2 sec x 2 2 !3) 5 0. For this reason,
either sin x 5 0 or 2 sec x 2 2 !3 5 0 (or both).
The solutions to the equation sin x 5 0 occur at
x 5 0, p, or 2p. If 2 sec x 2 2!3 5 0, x can be
solved for as follows:
2 sec x 2 2 !3 5 0
2 sec x 2 2 !3 1 2 !3 5 0 1 2 !3
2 sec x 5 2 !3
2 sec x
2 !3
5
2
2
sec x 5 !3
The solutions to the equation sec x 5 !3 occur at
x 5 0.96 or 5.33. Therefore, the solutions to the
equation 2 sin x sec x 2 2!3 sin x 5 0 occur at
x 5 0, 0.96, p, 5.33, or 2p.
d) To solve the equation 2 cot x 1 sec2 x 5 0, graph
the function y 5 2 cot x 1 sec2 x as follows:
10
8
6
4
2
0
–2
–4
y
x
p
2
p
3p
2
Since the graph intersects the x-axis at x 5
3p
4
and
7p
, the solutions to the equation 2 cot x 1 sec2 x 5 0
4
3p
7p
occur at x 5
or .
4
4
e) The equation cot x csc2 x 5 2 cot x can be
rewritten and factored as follows:
cot x csc2 x 5 2 cot x
2
cot x csc x 2 2 cot x 5 2 cot x 2 2 cot x
cot x csc2 x 2 2 cot x 5 0
(cot x)(csc x 2 !2)(csc x 1 !2) 5 0
Since (cot x)(csc x 2 !2)(csc x 1 !2) 5 0,
either cot x 5 0, csc x 2 !2 5 0, or
7-46
cot x 5 0 occur at x 5
p
2
or
3p
.
2
If csc x 2 !2 5 0,
x can be solved for as follows:
csc x 2 !2 5 0
csc x 2 !2 1 !2 5 0 1 !2
csc x 5 !2
The solutions to the equation csc x 5 !2 occur at
x5
p
4
or
3p
.
4
If csc x 1 !2 5 0, x can be solved for
as follows:
csc x 1 !2 5 0
csc x 1 !2 2 !2 5 0 2 !2
csc x 5 2 !2
The solutions to the equation csc x 5 2 !2 occur at
7p
. Therefore, the solutions to the equation
4
p p 3p 5p 3p
cot x csc2 x 5 2 cot x occur at x 5 , , , ,
4 2 4 4 2
7p
or .
4
x5
5p
4
or
f) Since 3 tan3 x 2 tan x 5 0,
(tan x)( !3 tan x 2 1)( !3 tan x 1 1) 5 0. For
this reason, either tan x 5 0, !3 tan x 2 1 5 0, or
!3 tan x 1 1 5 0. The solutions to the equation
tan x 5 0 occur at x 5 0, p, and 2p. If
!3 tan x 2 1 5 0, x can be solved for as follows:
!3 tan x 2 1 5 0
!3 tan x 2 1 1 1 5 0 1 1
!3 tan x 5 1
!3 tan x
1
5
!3
!3
1
tan x 5
!3
!3
tan x 5
3
The solutions to the equation tan x 5 "3
3 occur at
x5
p
6
or
7p
.
6
If !3 tan x 1 1 5 0, x can be solved
for as follows:
!3 tan x 1 1 5 0
!3 tan x 1 1 2 1 5 0 2 1
!3 tan x 5 21
!3 tan x
1
52
!3
!3
Chapter 7: Trigonometric Identities and Equations
08-035_07_AFSM_C07_001-060.qxd
7/23/08
tan x 5 2
tan x 5 2
2:09 PM
Page 47
1
!3
!3
3
The solutions to the equation tan x 5 "3
3 occur
at x 5
5p
6
or
11p
.
6
Therefore, the solutions to the
p
6
3
equation 3 tan x 2 tan x 5 0 occur at x 5 0, ,
5p
7p 11p
, p, ,
,
6
6
6
or 2p.
9. a) Since cos 2u 5 2 cos2 u 2 1, the equation
5 cos 2x 2 cos x 1 3 5 0 can be rewritten and
factored as follows:
5 cos 2x 2 cos x 1 3 5 0
(5)(2 cos2 x 2 1) 2 cos x 1 3 5 0
10 cos2 x 2 5 2 cos x 1 3 5 0
10 cos2 x 2 cos x 2 2 5 0
(5 cos x 1 2)(2 cos x 2 1) 5 0
For this reason, either 5 cos x 1 2 5 0 or
2 cos x 2 1 5 0 (or both). If 5 cos x 1 2 5 0, x can
be solved for as follows:
5 cos x 1 2 5 0
5 cos x 1 2 2 2 5 0 2 2
5 cos x 5 22
2
5 cos x
52
5
5
2
cos x 5 2
5
The solutions to the equation cos x 5 2 25 occur at
x 5 1.98 or 4.30.
If 2 cos x 2 1 5 0, x can be solved for as follows:
2 cos x 2 1 5 0
2 cos x 2 1 1 1 5 0 1 1
2 cos x 5 1
1
2 cos x
5
2
2
1
cos x 5
2
The solutions to the equation cos x 5 12 occur at
x5
p
3
or
5p
. Therefore,
3
the solutions to the equation
p
3
5 cos 2x 2 cos x 1 3 5 0 occur at x 5 , 1.98,
4.30, or
5p
.
3
b) Since cos 2u 5 2 cos2 u 2 1, the equation
10 cos 2x 2 8 cos x 1 1 5 0 can be rewritten and
factored as follows:
10 cos 2x 2 8 cos x 1 1 5 0
(10)(2 cos2 x 2 1) 2 8 cos x 1 1 5 0
Advanced Functions Solutions Manual
20 cos2 x 2 10 2 8 cos x 1 1 5 0
20 cos2 x 2 8 cos x 2 9 5 0
(10 cos x 2 9)(2 cos x 1 1) 5 0
For this reason, either 10 cos x 2 9 5 0 or
2 cos x 1 1 5 0 (or both). If 10 cos x 2 9 5 0,
x can be solved for as follows:
10 cos x 2 9 5 0
10 cos x 2 9 1 9 5 0 1 9
10 cos x 5 9
9
10 cos x
5
10
10
9
cos x 5
10
The solutions to the equation cos x 5 109 occur at
x 5 0.45 or 5.83. If 2 cos x 1 1 5 0, x can be solved
for as follows:
2 cos x 1 1 5 0
2 cos x 1 1 2 1 5 0 2 1
2 cos x 5 21
1
2 cos x
52
2
2
1
cos x 5 2
2
The solutions to the equation cos x 5 2 12 occur at
4p
. Therefore, the solutions to the equation
3
2p
10 cos 2x 2 8 cos x 1 1 5 0 occur at x 5 0.45, ,
3
4p
x5
3
2p
3
or
, or 5.83.
c) Since cos 2u 5 1 2 2 sin2 u, the equation
4 cos 2x 1 10 sin x 2 7 5 0 can be rewritten and
factored as follows:
4 cos 2x 1 10 sin x 2 7 5 0
(4)(1 2 2 sin2 x) 1 10 sin x 2 7 5 0
4 2 8 sin2 x 1 10 sin x 2 7 5 0
28 sin2 x 1 10 sin x 2 3 5 0
(21)(28 sin2 x 1 10 sin x 2 3) 5 (21)(0)
8 sin2 x 2 10 sin x 1 3 5 0
(4 sin x 2 3)(2 sin x 2 1) 5 0
For this reason, either 4 sin x 2 3 5 0 or
2 sin x 2 1 5 0 (or both). If 4 sin x 2 3 5 0, x can
be solved for as follows:
4 sin x 2 3 5 0
4 sin x 2 3 1 3 5 0 1 3
4 sin x 5 3
3
4 sin x
5
4
4
3
sin x 5
4
7-47
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7/18/08
12:55 PM
Page 48
The solutions to the equation sin x 5 34 occur at
x 5 0.85 or 2.29. If 2 sin x 2 1 5 0, x can be solved
for as follows:
2 sin x 2 1 5 0
2 sin x 2 1 1 1 5 0 1 1
2 sin x 5 1
2 sin x
1
5
2
2
1
sin x 5
2
The solutions to the equation sin x 5 12 occur at
x5
p
6
or
5p
. Therefore, the solutions to the equation
6
p
4 cos 2x 1 10 sin x 2 7 5 0 occur at x 5 , 0.85,
6
5p
, or 2.29.
6
d) Since cos 2u 5 1 2 2 sin2 u, the equation
22 cos 2x 5 2 sin x can be rewritten and factored
as follows:
22 cos 2x 5 2 sin x
(22)(1 2 2 sin2 x) 5 2 sin x
22 1 4 sin2 x 5 2 sin x
22 1 4 sin2 x 2 2 sin x 5 2 sin x 2 2 sin x
4 sin2 x 2 2 sin x 2 2 5 0
(2)(2 sin x 1 1)(sin x 2 1) 5 0
For this reason, either 2 sin x 1 1 5 0 or
sin x 2 1 5 0 (or both). If 2 sin x 1 1 5 0, x can
be solved for as follows:
2 sin x 1 1 5 0
2 sin x 1 1 2 1 5 0 2 1
2 sin x 5 21
2 sin x
1
52
2
2
1
sin x 5 2
2
The solutions to the equation sin x 5 2 12 occur at
x5
7p
6
or
11p
.
6
If sin x 2 1 5 0, x can be solved
for as follows:
sin x 2 1 5 0
sin x 2 1 1 1 5 0 1 1
sin x 5 1
The solution to the equation sin x 5 1 occurs at
p
x 5 . Therefore, the solutions to the equation
2
p 7p
,
2 6
22 cos 2x 5 2 sin x occur at x 5 ,
or
11p
.
6
10. To solve the equation 8 sin2 x 2 8 sin x 1 1 5 0,
first substitute u for sin x. The equation then becomes
8u2 2 8u 1 1. Next, use the quadratic formula to
solve for u as follows:
7-48
2b 6 "b 2 2 4ac
2a
2 (28) 6 "(28)2 2 4(8)(1)
5
2(8)
8 6 !64 2 32
5
16
8 6 !32
5
16
8 6 4 !2
5
16
2 6 !2
5
4
5 0.1464 or 0.8536
Since u 5 0.1464 or 0.8536, sin x 5 0.1464 or
0.8536. If sin x 5 0.1464, x 5 0.15 or 2.99. If
sin x 5 0.8536, x 5 1.02 or 2.12. Therefore, the
solutions to the equation 8 sin2 x 2 8 sin x 1 1 5 0
occur at x 5 0.15, 1.02, 2.12, or 2.99.
11. Since the solutions of the quadratic trigonometric
p p 7p
equation cot2 x 2 b cot x 1 c 5 0 are , , , and
u5
5p
,
4
6 4
6
the cotangent of these solutions must be found.
The cotangent of both
cotangent of both
p
4
p
6
and
and
5p
4
7p
6
is !3, while the
is 1. For this reason,
cot x 5 !3 and cot x 5 1, so cot x 2 !3 5 0
and cot x 2 1 5 0. If the factors cot x 2 !3
and cot x 2 1 are multiplied together as follows,
the quadratic trigonometric equation
cot2 x 2 b cot x 1 c 5 0 is formed:
(cot x 2 !3)(cot x 2 1) 5 0
2
cot x 2 !3 cot x 2 cot x 1 (21)(2 !3) 5 0
cot2 x 2 (1 1 !3) cot x 1 !3 5 0
Therefore, b 5 1 1 !3 and c 5 !3.
12. It’s clear from the graph that the quadratic
trigonometric equation sin2 x 2 c 5 0 has solutions
p 3p 5p
, ,
4 4 4
at x 5 ,
and
7p
,
4
so the sine of these
solutions must be found. The sine of both
is
!2
2 ,
while the sine of both
5p
4
and
7p
4
p
4
and
is 2 !2
2 .
3p
4
!2
For this reason, sin x 5 !2
2 and sin x 5 2 2 , so
!2
sin x 2 !2
2 5 0 and sin x 1 2 5 0. If the factors
!2
sin x 2 !2
2 and sin x 1 2 are multiplied together
Chapter 7: Trigonometric Identities and Equations
08-035_07_AFSM_C07_001-060.qxd
7/18/08
12:55 PM
Page 49
as follows, the quadratic trigonometric equation
sin2 x 2 c 5 0 is formed:
!2
!2
b asin x 1
b50
asin x 2
2
2
!2
!2
sin2 x 2
sin x 1
sin x
2
2
!2 !2
1 a2
ba
b50
2
2
2
sin2 x 2 5 0
4
1
sin2 x 2 5 0
2
Therefore, c 5 12.
13. To solve the problem, first the zeros of the function
h(d) 5 4 cos2 d 2 1 must be found as follows:
h(d) 5 4 cos2 d 2 1
0 5 4 cos2 d 2 1
(2 cos d 2 1)(2 cos d 1 1) 5 0
Since (2 cos d 2 1)(2 cos d 1 1) 5 0, either
2 cos d 2 1 5 0 or 2 cos d 1 1 5 0 (or both). If
2 cos d 2 1 5 0, d can be solved for as follows:
2 cos d 2 1 5 0
2 cos d 2 1 1 1 5 0 1 1
2 cos d 5 1
2 cos d
1
5
2
2
1
cos d 5
2
The solutions to the equation cos d 5 12 occur at
d5
p
3
or
5p
.
3
If 2 cos d 1 1 5 0, d can be solved
for as follows:
2 cos d 1 1 5 0
2 cos d 1 1 2 1 5 0 2 1
2 cos d 5 21
2 cos d
1
52
2
2
1
cos d 5 2
2
The solutions to the equation cos d 5 2 12 occur at
d5
2p
3
or
4p
.
3
Therefore, the zeros of the function
p 2p 4p
, ,
3 3 3
h(d) 5 4 cos2 d 2 1 are at d 5 ,
or
5p
.
3
With the zeros known, the 2p stretch of rolling hills
can be broken down into the following regions:
Advanced Functions Solutions Manual
p p
2p 2p
4p 4p
5p
,d, ;
,d, ;
,d, ;
3 3
3 3
3 3
3
5p
p
d . . First, the region d , can be tested by
3
3
p
finding h Q R as follows:
6
d, ;
p
p
ha b 5 4 cos2 a b 2 1
6
6
!3 2
5 4a
b 21
2
3
5 4a b 2 1 5 3 2 1 5 2
4
p
Since h Q R is positive, the height of the rolling
6
hills above sea level relative to Natasha’s home is
p
positive in the region d , . Next, the region
p
2p
,d,
3
3
3
can be tested by finding h Q R as
2
p
follows:
p
p
ha b 5 4 cos2 a b 2 1
2
2
5 4(0)2 2 1 5 4(0) 2 1
5 0 2 1 5 21
Since h Q R is negative, the height of the rolling
2
hills above sea level relative to Natasha’s home is
2p
p
negative in the region , d , . Next, the region
p
3
3
2p
4p
,d,
3
3
can be tested by finding h(p) as
in the region
2p
4p
,d, .
3
3
follows:
h(1) 5 4 cos2 (p) 2 1
5 4(21)2 2 1 5 4(1) 2 1
542153
Since h (p) is positive, the height of the rolling hills
above sea level relative to Natasha’s home is positive
4p
5p
,d,
3
3
Next, the region
can be tested by finding h Q
3p
R
2
as
follows:
3p
3p
ha b 5 4 cos2 a b 2 1
2
2
5 4(0)2 2 1
5 4(0) 2 1 5 0 2 1 5 21
Since h Q
3p
R
2
is negative, the height of the rolling
negative in the region
4p
5p
,d, .
3
3
Finally, the
7-49
08-035_07_AFSM_C07_001-060.qxd
region d .
5p
3
7/23/08
7:27 PM
Page 50
can be tested by finding h Q
11p
R
6
as follows:
11p
11p
b 5 4 cos2 a
b21
ha
6
6
!3 2
5 4a
b 21
2
3
5 4a b 2 1 5 3 2 1 5 2
4
Since h Q R is positive, the height of the rolling
6
hills above sea level relative to Natasha’s home is
5p
positive in the region d . . Therefore, the
3
intervals at which the height of the rolling hills
above sea level relative to Natasha’s home is
p
2p
negative are km , d , km and
11p
1
3 cos x
52
3
3
1
cos x 5 2
3
The solutions to the equation cos x 5 2 13 occur at
x 5 1.91 or 4.37. Therefore, the solutions to the
equation 6 sin2 x 5 17 cos x 1 11 occur at x 5 1.91
or 4.37.
15. a) Since sin2 u 1 cos2 u 5 1, or
sin2 u 5 1 2 cos2 u, the equation
sin2 x 2 !2 cos x 5 cos2 x 1 !2 cos x 1 2
can be rewritten and factored as follows:
sin2 x 2 !2 cos x 5 cos2 x 1 !2 cos x 1 2;
1 2 cos2 x 2 !2 cos x 5 cos2 x 1 !2 cos x 1 2;
1 2 cos2 x 2 !2 cos x 2 cos2 x 2 !2 cos x 2 2
3
3
5 cos2 x 1 !2 cos x 1 2 2 cos2 x 2 !2 cos x 2 2;
4p
5p
km , d , km.
1 2 cos2 x 2 !2 cos x 2 cos2 x
3
3
2 !2 cos x 2 2 5 0;
14. Since sin2 u 1 cos2 u 5 1, or sin2 u 5 1 2 cos2 u,
2
the equation 6 sin2 x 5 17 cos x 1 11 can be
22 cos x 2 2 !2 cos x 2 1 5 0;
rewritten and factored as follows:
1
a2 b (22 cos2 x 2 2 !2 cos x 2 1) 5 0;
6 sin2 x 5 17 cos x 1 11;
2
6 sin2 x 2 17 cos x 2 11 5 17 cos x 1 11
1
2 17 cos x 2 11;
cos2 x 1 !2 cos x 1 5 0;
2
6 sin2 x 2 17 cos x 2 11 5 0;
!2 2
6(1 2 cos2 x) 2 17 cos x 2 11 5 0;
acos
x
1
b 50
2
6 2 6 cos2 x 2 17 cos x 2 11 5 0;
2
!2
26 cos2 x 2 17 cos x 2 5 5 0;
Since (cos x 1 !2
2 ) 5 0, cos x 1 2 5 0. For
2
(21)(26 cos x 2 17 cos x 2 5) 5 (21)(0);
this reason, x can be solved for as follows:
6 cos2 x 1 17 cos x 1 5 5 0;
!2
cos x 1
50
(2 cos x 1 5)(3 cos x 1 1) 5 0
2
For this reason, either 2 cos x 1 5 5 0 or
3 cos x 1 1 5 0. If 2 cos x 1 5 5 0, x can be solved cos x 1 !2 2 !2 5 0 2 !2
2
2
2
for as follows:
2 cos x 1 5 5 0
!2
cos x 5 2
2 cos x 1 5 2 5 5 0 2 5
2
2 cos x 5 25
The solutions to the equation cos x 5 2 !2
2 occur
5
2 cos x
3p
5p
52
at x 5
or , so the solutions to the equation
2
2
4
4
5
2
sin x 2 !2 cos x 5 cos2x 1 !2 cos x 1 2 occur at
cos x 5 2
2
3p
5p
or .
x5
Since cos x can never equal 2 52, the factor 2 cos x 1 5
4
4
can be ignored. If 3 cos x 1 1 5 0, x can be solved for
b)
Since
the
period of the cosine function is 2p, a
as follows:
general
solution
for the equation in part (a) is
3 cos x 1 1 5 0
3p
5p
3 cos x 1 1 2 1 5 0 2 1
x5
1 2np or
1 2np, where nPI.
4
4
3 cos x 5 21
16. It is possible to have different numbers of solutions
for quadratic trigonometric equations because, when
factored, a quadratic trigonometric equation can be
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08-035_07_AFSM_C07_001-060.qxd
7/18/08
12:56 PM
Page 51
one expression multiplied by another expression or
it can be a single expression squared. For example,
the equation cos2 x 1 32 cos x 1 12 becomes
(cos x 1 1)(cos x 1 12 ) when factored, and it
has the solutions
2p
, p,
3
and
4p
3
and
4p
Q3R
because
cos x 5 2 12 for two different values of x. The
expression cos x 1 1, however, produces only one
solution in the interval 0 # x # 2p(p), because
cos x 5 21 for only one value of x.
17. To determine all the values of a such that
f(x) 5 tan (x 1 a), first graph the functions
f(x) 5
tan x
cot x
2
1 2 tan x
1 2 cot x
and g(x) 5 tan x
on the same coordinate grid as follows:
8
6
4
2
0
–2
–4
–6
–8
y
p
3p
2
x
0
p
2
–2
tan x
cot x
2
.
1 2 tan x
1 2 cot x
p
2
at x 5 0.72, , p,
p
4
the graph of the function g(x) 5 tan (x 1 )
would be the same as the graph of the function
true for
18. To solve the equation 2 cos 3x 1 cos 2x 1 1 5 0,
graph the function f(x) 5 2 cos 3x 1 cos 2x 1 1 on
a coordinate grid as follows:
Advanced Functions Solutions Manual
2p
3p
,
2
and 5.56. Therefore, the
solutions to the equation 2 cos 3x 1 cos 2x 1 1 5 0
p
2
3p
, or 5.56.
2
2 tan u
, the equation
tan 2u 5
1 2 tan2 u
occur at x 5 0.72, , p,
19. Since
3 tan2 2x 5 1 can be rewritten as
3 tan2 2x
2 tan x 2
3a
b
1 2 tan2 x
4 tan2 x
3a
b
1 2 tan2 x 2 tan2 x 1 tan4 x
12 tan2 x
1 2 2 tan2 x 1 tan4 x
1 2 2 tan2 x 1 tan4 x
1 2 2 tan2 x 1 tan4 x 2 12 tan2 x
5
5
For this reason,
tan x
cot x
f(x) 5
2
. The same is
1 2 tan x
1 2 cot x
5p
. Therefore, the values of a such that
4
p
5p
f(x) 5 tan (x 1 a) are a 5 and .
4
4
3p
2
p
It’s apparent from the graph that the x-intercepts are
tan2 x 5
It’s apparent from the graphs that if the graph of the
p
function g(x) 5 tan x were translated units to the
4
left, it would be the same as the graph of the function
f(x) 5
2
follows:
5 1;
5 1;
5 1;
5 1;
5 12 tan2 x;
5 12 tan2 x
2 12 tan2 x;
4
2
tan x 2 14 tan x 1 1 5 0
At this point the quadratic formula can be used to
solve for tan2 x as follows:
x
p
2
y
in the interval
0 # x # 2p. In comparison, the equation
cos2 x 1 2 cos x 1 1 5 0 becomes (cos x 1 1)2
when factored, and it has only one solution, p, in the
interval 0 # x # 2p. Also, different expressions
produce different numbers of solutions. For example,
the expression cos x 1 12 produces two solutions in
2p
the interval 0 # x # 2p Q R
3
4
5
5
2b 6 "b 2 2 4ac
2a
2 (214) 6 "(214)2 2 4(1)(1)
2(1)
14 6 !196 2 4
2
14 6 !192
2
14 6 8 !3
2
5 7 6 4!3
5 0.0718 or 13.9282
Since tan2 x 5 0.0718 or 13.9282, tan x 5 60.2679
or 63.7321. Therefore, four solutions for x are
x 5 15°, 75°, 285°, or 345°. Also, since the value of
tan x repeats itself every 180°, four more solutions
for x are x 5 105°, 165°, 195°, or 255°.
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Page 52
20. To solve the equation !2 sin u 5 !3 2 cos u,
first square both sides of the equation as follows:
!2 sin u 5 !3 2 cos u
( !2 sin u)2 5 ( !3 2 cos u)2
2 sin2 u 5 3 2 !3 cos u 2 !3 cos u 1 cos2 u
2 sin2 u 5 3 2 2!3 cos u 1 cos2 u
Since sin2 u 1 cos2 u 5 1, or sin2 u 5 1 2 cos2 u,
the equation 2 sin2 u 5 3 2 2!3 cos u 1 cos2 u can
be rewritten as follows:
2 sin2 u 5 3 2 2!3 cos u 1 cos2 u;
2(1 2 cos2 u) 5 3 2 2!3 cos u 1 cos2 u;
2 2 2 cos2 u 5 3 2 2!3 cos u 1 cos2 u;
2 2 2 cos2 u 2 2 1 2 cos2 u
5 3 2 2 !3 cos u 1 cos2 u 2 2 1 2 cos2 u;
3 cos2 u 2 2 !3 cos u 1 1 5 0
At this point the quadratic formula can be used to
solve for cos u as follows:
2b 6 "b 2 2 4ac
cos u 5
2a
2 (22 !3) 6 "(22!3)2 2 4(3)(1)
5
2(3)
2 !3 6 !12 2 12
5
6
2 !3 6 !0
5
6
2!3 6 0
2 !3
!3
5
5
5
6
6
3
!3
, u 5 0.96
Since cos u 5
3
Chapter Review, p. 440
1. a) Answers may vary. For example: Since
sin (p 2 u) 5 sin u, sin u 5 sin (p 2 u).
3p
3p
b
5 sin ap 2
Therefore, sin
10
10
10p
3p
7p
5 sin a
2
b 5 sin
10
10
10
b) Answers may vary. For example: Since
cos (2p 2 u) 5 cos u, cos u 5 cos (2p 2 u).
Therefore,
6p
6p
cos
b
5 cos a2p 2
7
7
6p
8p
14p
2
b 5 cos
5 cos a
7
7
7
7-52
c) Answers may vary. For example: Since
2sin u 5 sin (p 1 u),
13p
13p
b
5 sin ap 1
7
7
20p
7p
13p
5 sin a
1
b 5 sin
7
7
7
Since sin u 5 sin (u 2 2p),
2sin
20p
20p
2 2pb
5 sin a
7
7
20p
14p
6p
5 sin a
2
b 5 sin
7
7
7
13p
6p
Therefore, 2sin
5 sin
.
7
7
d) Answers may vary. For example: Since
2cos u 5 cos (p 1 u),
sin
8p
8p
b
5 cos ap 1
7
7
7p
8p
15p
5 cos a
1
b 5 cos
7
7
7
Since cos u 5 cos (u 2 2p),
15p
15p
cos
2 2pb
5 cos a
7
7
15p
14p
p
5 cos a
2
b 5 cos
7
7
7
8p
p
Therefore, 2cos
5 cos .
7
7
p
2. Since sin u 5 cos au 2 b, the equation
2
p
y 5 25 sin ax 2 b 2 8 can be rewritten
2
p
p
y 5 25 cos ax 2 2 b 2 8
2
2
5 25 cos (x 2 p) 2 8. Since a horizontal
translation of p to the left or right is equivalent
to a reflection in the x-axis, the equation
y 5 25 cos (x 2 p) 2 8 can be rewritten
y 5 5 cos x 2 8. Therefore, the equation
2cos
y 5 25 sin x Q 2 2 R 2 8 can be rewritten
y 5 5 cos x 2 8.
3. a) Since sin (a 2 b) 5 sin a cos b 2 cos a sin b,
p
sin ax 2
4p
4p
4p
b 5 sin x cos
2 cos x sin
3
3
3
1
!3
5 (sin x)a2 b 2 (cos x)a2
b
2
2
!3
1
5
cos x 2 sin x
2
2
Chapter 7: Trigonometric Identities and Equations
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Page 53
b) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
3p
3p
3p
cos ax 1
b 5 cos x cos
2 sin x sin
4
4
4
!2
!2
5 (cos x)a2
b 2 (sin x)a
b
2
2
!2
!2
52
cos x 2
sin x
2
2
tan a 1 tan b
c) Since tan (a 1 b) 5
,
1 2 tan a tan b
p
tan x 1 tan
p
3
tan ax 1 b 5
p
3
1 2 tan x tan
3
tan x 1 !3
1 2 (tan x)( !3)
tan x 1 !3
5
1 2 !3 tan x
d) Since cos (a 2 b) 5 cos a cos b 1 sin a sin b,
5
cos ax 2
5p
b
4
5p
5p
b 1 (sin x) asin
b
4
4
!2
!2
5(cos x)a2
b 1 (sin x)a2
b
2
2
!2
!2
cos x 2
sin x
52
2
2
tan a 1 tan b
4. a) Since tan (a 1 b) 5
,
1 2 tan a tan b
5(cos x)acos
p
7p
tan 1 tan
7p
p
12
4
p
7p 5 tan a 12 1 4 b
1 2 tan tan
12
4
p
21p
22p
!3
11p
1
b 5 tan
52
5 tan
12
12
12
6
3
b) Since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
19p
p
19p
p
2 sin sin
cos cos
9
18
9
18
p
19p
2p
19p
5 cos a 1
b 5 cos a
1
b
9
18
18
18
7p
!3
21p
5 cos
52
5 cos
18
6
2
5 tan a
5. a) Since sin 2u 5 2 sin u cos u,
p
p
p
2p
5 sin a(2)a bb sin
2 sin
cos
12
12
12
12
p
1
5 sin 5
6
2
Advanced Functions Solutions Manual
b) Since cos 2u 5 cos2 u 2 sin2 u,
p
p
p
5 cos a(2)a bb
2 sin2
cos2
12
12
12
2p
!3
p
5 cos
5 cos 5
12
6
2
2
c) Since cos 2u 5 1 2 2 sin u,
3p
3p
5 cos a(2)a bb
1 2 2 sin2
8
8
6p
!2
3p
5 cos
52
5 cos
8
4
2
d) Since tan 2u 5
2 tan
p
6
2 tan u
,
1 2 tan2 u
5 !3
p 5 tan 2a b 5 tan
6
3
1 2 tan
6
p
p
2
6. a) Since sin x 5 35, the leg opposite to the angle
x in a right triangle has a length of 3, while the
hypotenuse of the right triangle has a length of 5.
For this reason, the other leg of the right triangle
can be calculated as follows:
32 1 y2 5 52
9 1 y2 5 25
9 1 y2 2 9 5 25 2 9
y2 5 16
y 5 4, in quadrant I
Since cos x 5
4
adjacent leg
, cos x 5 .
hypotenuse
5
Therefore, since sin 2x 5 2 sin x cos x,
3 4
24
sin 2x 5 (2)a b a b 5 .
5 5
25
Also, since cos 2x 5 cos2 x 2 sin2 x,
4 2
3 2 16
9
7
cos 2x 5 a b 2 a b 5
2
5 .
5
5
25
25
25
Since tan x 5
opposite leg
, tan x 5 34.
adjacent leg
Since tan 2x 5
3
2
(2) Q 34 R
2 tan x
tan
2x
5
,
1 2 tan2 x
1 2 Q 34 R 2
3
2
3
2
7
16
3
16
24
3
5
12
2
2
7
7
7
b) Since cot x 5 2 24, the leg opposite the angle
x in a right triangle has a length of 24, while the leg
adjacent to the angle x has a length of 7. For this
reason, the hypotenuse of the right triangle can be
calculated as follows:
72 1 242 5 c2
49 1 576 5 c2
5
9
16
5
16
16
9
16
5
5
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Page 54
625 5 c2
c 5 25
Since tan 2x 5
opposite leg
, sin x 5 24
25 , and since
hypotenuse
adjacent leg
cos x 5
, cos x 5 2 257 . (The reason
hypotenuse
Since sin x 5
the sign is negative is because angle x is in the
second quadrant.) Therefore, since
24
7
sin 2x 5 2 sin x cos x, sin 2x 5 (2)a b a2 b
25
25
336
. Also, since cos 2x 5 cos2 x 2 sin2 x,
52
625
7 2
24 2
49
576
527
cos 2x 5 a2 b 2 a b 5
2
52
.
25
25
625
625
625
24
7
Since cot x 5 2 , tan x 5 2
24
7.
2 tan x
,
Since tan 2x 5
1 2 tan2 x
tan 2x 5
5
(2) Q 2 247 R
1 2 Q 2 247 R 2
5
2 487
5
1 2 576
49
49
49
2 487
2 576
49
2 487
48
49
336
3
5
527 5 2
2 49
7
527
527
c) Since cos x 5 12
13 , the leg adjacent to the angle x
in a right triangle has a length of 12, while the
hypotenuse of the right triangle has a length of 13.
For this reason, the other leg of the right triangle
can be calculated as follows:
x2 1 122 5 132
x2 1 144 5 169
2
x 1 144 2 144 5 169 2 144
x2 5 25
x55
opposite leg
, sin x 5 2 135 . (Sine is
Since sin x 5
tan 2x 5
5
5
12
120
ba b 5 2
. Also, since
13 13
169
cos 2x 5 cos2 x 2 sin2 x,
12 2
5 2
cos 2x 5 a b 2 a2 b
13
13
144
25
119
5
2
5
169
169
169
sin 2x 5 (2)a2
Finally, since tan x 5
opposite leg
, tan x 5 2 125 .
adjacent leg
(The reason the sign is negative is because angle x
is in the fourth quadrant.)
7-54
(2) Q 2 125 R
1 2 Q 2 125 R
2 56
25 5
1 2 144
2
144
144
2 56
2 56
25 5 119
2 144
144
144
120
5
52
52 3
6
119
119
7. a) Since sin 2 u 5 2 sin u cos u, and since
cos 2u 5 1 2 2 sin2 u, tan 2x 5
2 sin x cos x
sin 2x
2 sin x cos x
5
.
cos 2x
1 2 2 sin2 x
Therefore, tan 2x 5
is a trigonometric
1 2 2 sin2 x
identity.
b) Since 1 1 tan2 x 5 sec2 x, sec2 x 2 tan2 x 5 1.
Therefore, since sec2 x 2 tan2 x 5 1 is a
trigonometric identity, sec2 x 2 tan2 x 5 cos x must
be a trigonometric equation, because cos x does not
always equal 1.
c) Since 1 1 cot2 x 5 csc2 x, csc2 x 2 cot2 x 5 1.
Therefore, since sin2 x 1 cos2 x 5 1,
csc2 x 2 cot2 x5 sin2 x 1 cos2 x is a trigonometric
identity.
d) Since tan2 x 5 1, tan x 5 6 1. Therefore, since
tan x does not always equal 21 or 1, tan2 x 5 1
must be a trigonometric equation.
1 2 sin2 x
5 1 2 cos2 x
8. The trigonometric identity
cot2 x
can be proven as follows:
1 2 sin2 x
5 1 2 cos2 x
cot2 x
cos2 x
5 1 2 cos2 x
cot2 x
cos2 x
2
cos2 x 5 1 2 cos x
hypotenuse
negative because x is in the fourth quadrant.)
Therefore, since sin 2x 5 2 sin x cos x,
2 tan x
,
1 2 tan2 x
sin2 x
2
(cos x)(sin2 x)
5 1 2 cos2 x
cos2 x
sin2 x 5 1 2 cos2 x
1 2 cos2 x 5 1 2 cos2 x
9. The trigonometric identity
2 sec2 x 2 2 tan2 x
5 sin 2x sec x
csc x
can be proven as
follows:
2 sec2 x 2 2 tan2 x
5 sin 2x sec x
csc x 2
2
2(sec x 2 tan x)
5 sin 2x sec x
csc x
Chapter 7: Trigonometric Identities and Equations
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Page 55
10 sec x
20
52
10
10
sec x 5 22
The solutions to the equation sec x 5 22 occur at
2(1)
5 sin 2x sec x
csc x
2
5 sin 2x sec x
csc x
2 sin x 5 sin 2x sec x
2 sin x cos x
5 sin 2x sec x
cos x
sin 2x
5 sin 2x sec x
cos x
sin 2x sec x 5 sin 2x sec x
10. a) The trigonometric equation
x5
2
1 10 5 6
sin x
can be solved as follows:
2
1 10 5 6
sin x
2
1 10 2 10 5 6 2 10
sin x
2
5 24
sin x
24 sin x 5 2
24 sin x
2
5
24
24
1
sin x 5 2
2
The solutions to the equation sin x 5 2 12 occur at
x5
7p
6
or
11p
.
6
b) The trigonometric equation 2
5 cot x
1
2
7
3
5 2 16
can be solved as follows:
5 cot x
7
1
2
1 52
2
3
6
15 cot x
14
1
2
1
52
6
6
6
215 cot x 1 14 5 21
215 cot x 1 14 2 14 5 21 2 14
215 cot x 5 215
215 cot x
215
5
215
215
cot x 5 1
The solutions to the equation cot x 5 1 occur at
p
5p
x 5 or .
4
Advanced Functions Solutions Manual
or
4p
.
3
11. a) The equation y 2 2 4 5 0 can be solved as
follows:
y2 2 4 5 0
(y 2 2)(y 1 2) 5 0
Since (y 2 2)(y 1 2) 5 0, either y 2 2 5 0 or
y 1 2 5 0 (or both). If y 2 2 5 0, y can be solved
for as follows:
y2250
y22125012
y52
If y 1 2 5 0, y can be solved for as follows:
y1250
y12225022
y 5 22
Therefore, the solutions to the equation y 2 2 4 5 0
are y 5 2 or y 5 22.
b) The equation csc2 x 2 4 5 0 can be solved as
follows:
csc2 x 2 4 5 0
(csc x 2 2)(csc x 1 2) 5 0
Since (csc x 2 2)(csc x 1 2) 5 0, either
csc x 2 2 5 0 or csc x 1 2 5 0 (or both). If
csc x 2 2 5 0, x can be solved for as follows:
csc x 2 2 5 0
csc x 2 2 1 2 5 0 1 2
csc x 5 2
The solutions to the equation csc x 5 2 occur at
x5
p
6
or
5p
.
6
If csc x 1 2 5 0, x can be solved for
as follows:
csc x 1 2 5 0
csc x 1 2 2 2 5 0 2 2
csc x 5 22
The solutions to the equation csc x 5 22 occur at
x5
7p
6
or
11p
.
6
Therefore, the solutions to the
p 5p 7p
, ,
6 6 6
equation csc2 x 2 4 5 0 occur at x 5 ,
4
c) The trigonometric equation
3 1 10 sec x 2 1 5 218 can be solved as follows:
3 1 10 sec x 2 1 5 218
2 1 10 sec x 5 218
2 1 10 sec x 2 2 5 218 2 2
10 sec x 5 220
2p
3
or
11p
.
6
12. a) The equation 2 sin2 x 2 sin x 2 1 5 0 can
be solved as follows:
2 sin2 x 2 sin x 2 1 5 0
(2 sin x 1 1)(sin x 2 1) 5 0
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Page 56
Since (2 sin x 1 1)(sin x 2 1) 5 0, either
2 sin x 1 1 5 0 or sin x 2 1 5 0 (or both). If
2 sin x 1 1 5 0, x can be solved for as follows:
2 sin x 1 1 5 0
2 sin x 1 1 2 1 5 0 2 1
2 sin x 5 21
2 sin x
1
52
2
2
1
sin x 5 2
2
The solutions to the equation sin x 5 2 12 occur at
x5
7p
6
or
11p
.
6
If sin x 2 1 5 0, x can be solved for
as follows:
sin x 2 1 5 0
sin x 2 1 1 1 5 0 1 1
sin x 5 1
The solution to the equation sin x 5 1 occurs at
p
2
x 5 . Therefore, the solutions to the equation
p 7p
,
2 6
2 sin2 x 2 sin x 2 1 5 0 occur at x 5 ,
or
11p
.
6
b) The equation tan2 x sin x 2
sin x
50
3
can be
solved as follows:
sin x
50
3
1
(sin x)atan2 x 2 b 5 0
3
!3
!3
(sin x)atan x 2
b atan x 1
b50
3
3
!3
Since (sin x)( tan x 2 !3
3 )( tan x 1 3 ) 5 0, either
!3
sin x 5 0, tan x 2 3 5 0, or tan x 1 !3
3 5 0. The
solutions to the equation sin x 5 0 occur at x 5 0,
!3
p, or 2p. If tan x 2
5 0, x can be solved for as
3
follows:
!3
tan x 2
50
3
!3
!3
!3
tan x 2
1
501
3
3
3
!3
tan x 5
3
tan2 x sin x 2
The solutions to the equation tan x 5
p
x5
6
or
7p
.
6
If tan x 1
!3
3
occur at
5 0, x can be solved for
as follows:
!3
50
3
!3
!3
!3
tan x 1
2
502
3
3
3
!3
tan x 5 2
3
The solutions to the equation tan x 5 2 !3
3 occur
tan x 1
5p
6
at x 5
11p
.
6
or
Therefore, the solutions to the
equation tan2 x sin x 2
5p
7p 11p
, p, ,
,
6
6
6
sin x
50
3
p
6
occur at x 5 0, ,
or 2p.
c) The equation
1 2 !2
!2
cos2 x 1 a
b cos x 2
50
2
4
can be solved as follows:
1 2 !2
!2
cos2 x 1 a
b cos x 2
50
2
4
1
!2
b acos x 1 b 5 0
acos x 2
2
2
1
cos
x
1
5
0, either
Since ( cos x 2 !2
)(
)
2
2
!2
1
cos x 2 2 5 0 or cos x 1 2 5 0 (or both). If
cos x 2 !2
2 5 0, x can be solved for as follows:
!2
cos x 2
50
2
!2
!2
!2
cos x 2
1
501
2
2
2
!2
cos x 5
2
The solutions to the equation cos x 5 !2
2 occur at
x5
p
4
or
7p
.
4
If cos x 1 12 5 0, x can be solved for
as follows:
1
cos x 1 5 0
2
1
1
1
cos x 1 2 5 0 2
2
2
2
1
cos x 5 2
2
The solutions to the equation cos x 5 2 12 occur at
x5
2p
3
or
4p
.
3
Therefore, the solutions to the
equation cos x 1 ( 1 22 !2) cos x 2
2
occur at
7-56
!3
3
p 2p 4p
x5 , , ,
4 3 3
or
!2
4
50
7p
.
4
Chapter 7: Trigonometric Identities and Equations
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Page 57
d) The equation 25 tan2 x 2 70 tan x 5 249 can be
solved as follows:
25 tan2 x 2 70 tan x 5 249
25 tan2 x 2 70 tan x 1 49 5 249 1 49
25 tan2 x 2 70 tan x 1 49 5 0
(5 tan x 2 7)2 5 0
Since (5 tan x 2 7)2 5 0, 5 tan x 2 7 5 0. For this
reason, x can be solved for as follows:
5 tan x 2 7 5 0
5 tan x 2 7 1 7 5 0 1 7
5 tan x 5 7
5 tan x
7
5
5
5
7
tan x 5
5
The solutions to the equation tan x 5 75 occur at
x 5 0.95 and 4.09. Therefore, the solutions to the
equation 25 tan2 x 2 70 tan x 5 249 occur at
x 5 0.95 or 4.09.
13. Since 1 1 tan2 x 5 sec2 x, the equation
1
5 2cos x
1 1 tan2 x
can be rewritten and factored as
follows:
1
5 2cos x
1 1 tan2 x
1
5 2cos x
sec2 x
cos2 x 5 2cos x
2
cos x 1 cos x 5 2cos x 1 cos x
cos2 x 1 cos x 5 0
(cos x)(cos x 1 1) 5 0
Since (cos x)(cos x 1 1) 5 0, either cos x 5 0 or
cos x 1 1 5 0 (or both). The solutions to the
equation cos x 5 0 occur at x 5
p
2
or
3p
.
2
If
cos x 1 1 5 0, x can be solved for as follows:
cos x 1 1 5 0
cos x 1 1 2 1 5 0 2 1
cos x 5 21
The solution to the equation cos x 5 21 occurs at
x 5 p. Therefore, the solutions to the equation
1
5 2cos x
1 1 tan2 x
occur at
p
x 5 , p,
2
Chapter Self-Test, p. 441
1. The identity
x
1 2 2 sin2 x
x
1 2 sin cos 5 cos x
cos x 1 sin x
2
2
can be proven as follows:
Advanced Functions Solutions Manual
or
3p
.
2
1 2 2 sin2 x
x
x
1 2 sin cos 5 cos x
cos x 1 sin x
2
2
1 2 2 sin2 x
1 sin x 5 cos x
cos x 1 sin x
1 2 2 sin2 x
1 sin x 2 sin x 5 cos x 2 sin x
cos x 1 sin x
1 2 2 sin2 x
5 cos x 2 sin x
cos x 1 sin x
1 2 2 sin2 x 5 (cos x 2 sin x)
3 (cos x 1 sin x)
cos 2x 5 (cos x 2 sin x)
3 (cos x 1 sin x)
cos 2x 5 cos2 x 2 sin2 x
cos 2x 5 cos 2x
2. Since cos 2u 5 1 2 2 sin2 u, the equation
cos 2x 1 2 sin2 x 2 3 5 22 can be rewritten and
solved as follows:
cos 2x 1 2 sin2 x 2 3 5 22
1 2 2 sin2 x 1 2 sin2 x 2 3 5 22
22 5 22
Since 22 always equals 22, the equation
cos 2x 1 2 sin2 x 2 3 5 22 is an identity and is
true for all real numbers x, where 0 # x # 2p.
3. a) The solutions to the equation cos x 5 !3
2
occur at x 5
p
6
or
11p
.
6
b) The solutions to the equation tan x 5 2 !3
occur at x 5
2p
3
or
5p
.
3
c) The solutions to the equation sin x 5 2
at
5p
x5
4
or
!2
2
occur
7p
.
4
4. Since the quadratic trigonometric equation
p
a cos2 x 1 b cos x 2 1 5 0 has the solutions , p,
and
5p
,
3
3
the left side of the equation must have
factors of cos x 2 12 and cos x 1 1. This is because
the cosine of
p
3
and
5p
3
is 12, and the cosine of p
is 21. For this reason, the quadratic trigonometric
equation can be found as follows:
1
acos x 2 b (cos x 1 1) 5 0
2
1
1
cos2 x 2 cos x 1 cos x 2 5 0
2
2
1
1
cos2 x 1 cos x 2 5 0
2
2
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1
1
cos x 2 b 5 (2)(0)
2
2
2 cos2 x 1 cos x 2 1 5 0
Therefore, a 5 2 and b 5 1.
5. Since the depth of the ocean in metres can be
also at 3 metres at t 5 7 1 12 5 19 h or at
t 5 11 1 12 5 23 h.
modelled by the function d(t) 5 4 1 2 sin Q t R ,
The cosine of p is 21, and the cosine of
(2)acos2 x 1
p
6
p
when the depth is 3 metres, 3 5 4 1 2 sin Q t R .
6
p
The equation 3 5 4 1 2 sin Q t R can be solved as
6
follows:
p
3 5 4 1 2 sin a tb
6
p
3 2 4 5 4 1 2 sin a tb 2 4
6
p
21 5 2 sin a tb
6
2 sin a tb
p
6
1
2 5
2
2
1
p
sin a tb 5 2
6
2
p
1
sin21 asin a tbb 5 sin21 a2 b
6
2
p
7p 11p
t5
or
6
6
6
p
6
If t 5
7p
,
6
t can be solved for as follows:
p
7p
t5
6
6
6 7p
6 p
a b a tb 5 a b a b
p 6
p
6
t57
p
11p
, t can be solved for as follows:
If t 5
6
6
11p
p
t5
6
6
6 p
6 11p
a b a tb 5 a b a
b
p 6
p
6
t 5 11
Therefore, two times when the depth of the water is
3 metres are t 5 7 h or 11 h. Also, since the period
p
of the function d(t) 5 4 1 2 sin Q t R is
6
2p
6
p 5 (2p)Q p R 5 12 h, the depth of the water is
6
7-58
6. Nina can find the cosine of
11p
4
by using the
formula cos (x 1 y) 5 cos x cos y 2 sin x sin y.
7p
is
4
7p
Also, the sine of p is 0, and the sine of
is
4
11p
7p
5 cos Q p 1 R
2 !2
2 . Therefore, cos
4
4
!2
2 .
!2
!2
b 2 a0 3 2
b
2
2
!2
!2
52
2052
2
2
7. The equation 3 sin x 1 2 5 1.5 can be solved as
follows:
3 sin x 1 2 5 1.5
3 sin x 1 2 2 2 5 1.5 2 2
3 sin x 5 20.5
0.5
3 sin x
52
3
3
sin x 5 20.1667
The solutions to the equation sin x 5 20.1667
occur at x 5 3.31 or 6.12.
8. Since tan a 5 0.75, the leg opposite the angle a
in a right triangle has a length of 3, while the leg
adjacent to angle a has a length of 4. For this
reason, the hypotenuse of the right triangle can be
calculated as follows:
32 1 42 5 z 2
9 1 16 5 z 2
25 5 z 2
z55
5 a21 3
adjacent leg
, cos a 5 45. Also, since
hypotenuse
opposite leg
sin a 5
, sin a 5 35. In addition, since
hypotenuse
Since cos a 5
tan b 5 2.4, the leg opposite the angle b in a right
triangle has a length of 12, while the leg adjacent to
angle b has a length of 5. For this reason, the
hypotenuse of the right triangle can be calculated as
follows:
122 1 52 5 z 2
144 1 25 5 z 2
169 5 z 2
z 5 13
adjacent leg
, cos b 5 135 . Also, since
hypotenuse
12
opposite leg
, sin b 5 . Therefore, since
sin b 5
hypotenuse
13
Since cos b 5
Chapter 7: Trigonometric Identities and Equations
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sin (a 2 b) 5 sin a cos b 2 cos a sin b,
sin (a 2 b) 5 sin a cos b 2 cos a sin b
5
4 12
15
48
33
3
2
52
5 a ba b 2 a ba b 5
5 13
5 13
65
65
65
Also, since cos (a 1 b) 5 cos a cos b 2 sin a sin b,
cos (a 1 b) 5 cos a cos b 2 sin a sin b
4
5
3 12
20
36
16
5 a ba b 2 a ba b 5
2
52
5 13
5 13
65
65
65
4
2
9. a) Since sin x 5 9, and since the angle x is in
!4
!9
the second quadrant, sin x 5 #49 5
5 23.
Since sin x 5 23, the leg opposite the angle x in a
right triangle has a length of 2, while the
hypotenuse of the right triangle has a length of 3.
For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 (2)2 5 32
x2 1 4 5 9
x2 1 4 2 4 5 9 2 4
x2 5 5
x 5 "5
Since
adjacent leg
,
cos x 5
hypotenuse
and since the angle
2 "5
3 .
x is in the second quadrant, cos x 5
sin 2u 5 2 sin u cos u,
sin 2x 5 2 sin x cos x
Since
!4
!9
5 23. Since
sin x 5 23, the leg opposite the angle x in a right
triangle has a length of 2, while the hypotenuse of
the right triangle has a length of 3. For this reason,
the other leg of the right triangle can be calculated
as follows:
x 2 1 (2)2 5 32
x2 1 4 5 9
2
x 14245924
x2 5 5
x 5 "5
Since cos x 5
adjacent leg
,
hypotenuse
2
positive. Since sin2 x 5 49, and since the angle x is in
2
the second quadrant, sin x 5 # 49 5 !4
!9 5 3 .
Since sin x 5 23, the leg opposite the angle x in a
right triangle has a length of 2, while the hypotenuse
of the right triangle has a length of 3. For this
reason, the other leg of the right triangle can be
calculated as follows:
x 2 1 (2)2 5 32
x2 1 4 5 9
x2 1 4 2 4 5 9 2 4
x2 5 5
x 5 "5
Since cos x 5
adjacent leg
,
hypotenuse
and since the angle
x is in the second quadrant, cos x 5 2 "5
3 . Since
2
2u
cos 2u 5 2 cos u 2 1, cos u 5 2 cos 2 1, and
2
!5
2
b
5 (2)a b a2
3
3
4 !5
.
52
9
b) Since sin2 x 5 49, and since the angle x is in the
second quadrant, sin x 5 # 49 5
2 2
"5 2
b 2a b
3
3
4
1
5
5 2 5
9
9
9
(The formulas cos 2u 5 2 cos2 u 2 1 and
cos 2u 5 1 2 2 sin2 u could also have been used.)
c) First note that because x is in the second quadrant,
x
is in the first quadrant, where the cosine is
5 a2
and since the angle
x is in the second quadrant, cos x 5 2 "5
3 . Since
cos 2u 5 cos2 u 2 sin2 u,
cos 2x 5 cos2 x 2 sin2 x
Advanced Functions Solutions Manual
"5
2 x
2 1. The value
since cos x 5 2 "5
3 , 2 3 5 2 cos
x
2
2
of cos can now be determined as follows:
"5
x
5 2 cos2 2 1
3
2
"5
x
2
1 1 5 2 cos2 2 1 1 1
3
2
3 2 "5
x
5 2 cos2
3
2
( 3 2 "5 )
2x
2 cos 2
3
5
2
2
x
(3 2 !5)
cos2 5
2
6
x
(3 2 !5)
cos 5 c
d
2 Å
6
2
d) Since sin2 x 5 49, and since the angle x is in
2
the second quadrant, sin x 5 # 49 5 !4
!9 5 3 .
Since sin x 5 23, the leg opposite the angle x in
a right triangle has a length of 2, while the
hypotenuse of the right triangle has a length of 3.
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For this reason, the other leg of the right triangle
can be calculated as follows:
x 2 1 (2)2 5 32
x2 1 4 5 9
2
x 14245924
x2 5 5
x 5 "5
Since cos x 5
adjacent leg
,
hypotenuse
and since the angle
x is in the second quadrant, cos x 5 2 !5
3 . Since
sin 3u 5 3 cos2 u sin u 2 sin3 u,
sin 3x 5 3 cos2 x sin x 2 sin3 x
!5 2 2
2 3
5(3)a2
b a b2 a b
3
3
3
5 2
8
5(3)a b a b 2
9 3
27
8
22
30
2
5
27
27
27
10. a) The equation 2 2 14 cos x 5 25 can be
solved as follows:
2 2 14 cos x 5 25
2 2 14 cos x 2 2 5 25 2 2
214 cos x 5 27
214 cos x
27
5
214
214
1
cos x 5
2
5
7-60
From the graph, the solutions to the equation
cos x 5 12 occur at x 5 2
5p
p p
,2 , ,
3
3 3
or
5p
.
3
b) The equation 9 2 22 cos x 2 1 5 19 can be
solved as follows:
9 2 22 cos x 2 1 5 19
8 2 22 cos x 5 19
8 2 22 cos x 2 8 5 19 2 8
222 cos x 5 11
222 cos x
11
5
222
222
1
cos x 5 2
2
From the graph, the solutions to the equation
cos x 5 2 12 occur at x 5 2
4p
2p 2p
,2 , ,
3
3 3
or
4p
.
3
c) The equation 2 1 7.5 cos x 5 25.5 can be solved
as follows:
2 1 7.5 cos x 5 25.5
2 1 7.5 cos x 2 2 5 25.5 2 2
7.5 cos x 5 27.5
7.5 cos x
27.5
5
7.5
7.5
cos x 5 21
From the graph, the solutions to the equation
cos x 5 21 occur at x 5 2p and p.
Chapter 7: Trigonometric Identities and Equations