Solutions for Trigonometry by Gelfand & Saul
Git commit f3d7863. Published January 7, 2023.
This work is licensed under a Creative Commons
“Attribution-ShareAlike 4.0 International” license.
Introduction
Trigonometry by Gelfand and Saul is often recommended as a precalculus text
for self-study. However, those who are learning without the help of a teacher
can struggle with the lack of solutions to exercises in the text. Partial sets of
solutions for Trigonometry have been published by John Beach1 and Fardeen
Ashraf2 . It is hoped that this document will eventually contain a complete set of
solutions. Contributions are welcome. These can take the form of pull requests
or issues submitted to the project’s GitHub repository3 .
Chapter 0: Trigonometry
Page 8
1. Statement I applies:
c2 = a2 + b2 = 102 + 242 = 100 + 576 = 676
√
c = 676 = 26
2. Statement I applies:
a2 + b2 = c2
a2 + 92 = 412
a2 + 81 = 1681
a2 = 1600
√
a = 1600 = 40
1 https://jbeach50.weebly.com/gelfand–saul-trig-solutions.html
2 https://archive.org/details/gelfand-trigonometry-solutions-manual
3 https://github.com/philip-healy/gelfand-trigonometry-solutions
1
3. 52 + 122 = 25 + 144 = 169 = 132 . By Statement II, a right triangle exists
with legs of length 5 and 12, and hypotenuse of length 13.
4. Statement I applies:
a2 + b2 = c2
a2 + 12 = 32
a2 + 1 = 9
a2 = 8
√
√ √
√
a= 8= 4 2=2 2
5. Statement I applies, where a = b:
a2 + a2 = c2
a2 + a2 = 12
2a2 = 1
1
a2 =
2
r
a=
√
1
1
1
=√ =√
2
2
2
6. From the diagram at the bottom of Page 11, we can see the shorter leg is
half the length of the hypotenuse. So in this instance the shorter leg has
length 1/2. We can use Statement 1 to find the length of the longer leg:
a2 + b2 = c2
2
1
a2 +
= 12
2
1
a2 + = 1
4
3
a2 =
4
r
a=
√
√
3
3
3
=√ =
4
2
4
7. For any point Y , we can draw a triangle with sides AY , BY and AB. Let
a be the length of side AY , b be the length of side BY and c be the length
of side AB. According to Statement II, the subset of these triangles where
a2 + b2 = c2 are right triangles with legs of length a and b and hypotenuse
c. Let X be the subset of Y that are vertices of these right triangles. This
set of points describes a circle with its centre at the midpoint of AB, and
radius AB/2.
8.
2
Page 9
1. 62 + 82 = 36 + 64 = 100 = 102 . By Statement II on Page 7 (converse of the
Pythagorean Theorem), this is a right triangle.
2. 10-24-26 (Exercise 1), 9-40-41 (Exercise 2), 5-12-13 (Exercise 3)
3. Using the Pythagorean Theorem:
c2 = a2 + b2 = 82 + 152 = 64 + 225 = 289
√
c = 289 = 17
4. The first column in the table increases by 3, the second increases by 4 and
the third increases by 5. Continuing to add rows yields triangles 12-16-20,
15-20-25 and 18-24-30.
5. Shortest side with length 10: 10-24-26. Shortest side with length 15: 15-3639.
6. Multiplying all sides by the common denominator (5), we get a similar triangle with sides 15/5 = 3, 20/5 = 4 and 5. We know that this is a right
triangle from the table in Question 4.
7. To find a similar triangle with shorter leg 1, divide all sides by 3, resulting in
sides 1-4/3-5/3. To find a similar triangle with longer leg 1, divide all sides
by 4, resulting in sides 3/4-1-5/4.
8. To find a similar triangle with hypotenuse 1, divide all sides by 13, resulting
in sides 5/13-12/13-1. To find a similar triangle with shorter leg 1, divide
all sides by 5, resulting in sides 1-12/5-13/5. To find a similar triangle with
longer leg 1, divide all sides by 12, resulting in sides 5/12-1-13/12.
9. To formula for the area of a triangle is 12 bh where b is the length of the base
and h is the height. For right triangles, finding the area is easy: one leg is the
base and the other leg is the height. For other triangles, finding the height
is more difficult: we need to find the length of the altitude drawn from the
base. The triangles with sides 5-12-13 and 9-12-15 are both right triangles:
see Exercise 3 on Page 8 and Exercise 4 on Page 9. The triangle with sides
13-14-15 is not a right triangle. We can confirm this using Statement I:
a2 + b2 = 132 + 142 = 365, c2 = 152 = 225, a2 + b2 ̸= c2 . However, if we join
the 5-12-13 and 9-12-15 triangles using their equal legs, the resulting triangle
has the dimensions we are looking for: 13-14-15. The base of this combined
triangle has length 5 + 9 = 14. We also know the length of the altitude from
the base of the combined triangle: 12. So, the area of the 13-14-15 triangle
is 12 · 14 · 12 = 84 units squared.
10. (a)
(b)
3
Page 11
1.
√1
2
(see the solution for Question 5 on page 8).
√
√
√
Challenge: √12 = 22 (multiplying above and below by 2). 2 is given to
9 decimal places in the diagram on the top of page 11: 1.4141213562373.
Dividing this decimal representation by 2 (using long division if necessary)
yields a figure of 0.707060678.
√
√ √
√
2. c2 = a2 + b2 = 32 + 32 = 9 + 9 = 18. c = 18 = 9 2 = 3 2.
3. The hypotenuse of a 30◦ right triangle is double the length of the shorter leg.
In this instance the hypotenuse is 10 units long. We can use the Pythagorean
Theorem to find the length of the longer leg:
a2 + b2 = c2
a2 + 52 = 102
a2 + 25 = 100
a2 = 75
√ √
√
√
a = 75 = 25 3 = 5 3
4. We can √
solve these by finding similar triangles to the √
30◦ right triangle with
◦
sides 1- 3-2, or the 45 right triangle with sides 1-1- 2.
√
(a) x = 3, y = 2
(b) x =
√1 ,
3
y=
√2
3
√
(c) x = 1/2, y = 3/2
√
(d) x = 4 3, y = 8
√
(e) x = y = 2 2
√
(f) x = 5, y = 5 2
Page 14 (Examples)
1. Why didn’t we need to compare 32 with 22 + 42 , or 22 with 32 + 42 ?
The obtuse angle will always be opposite the longest side.
2. This conclusion is incorrect. Why?
From the footnote at the beginning of Chapter 0: “Given three arbitrary
lengths. . . they form a triangle if and only if the sum of any two of them is
greater than the third.” In this case 1 + 2 = 3 which is equal to (not greater
than) the third side.
4
Page 14 (Exercise)
1. (a) c2 = 82 = 64. a2 + b2 = 62 + 72 = 36 + 49 = 85. c2 < a2 + b2 , so the
triangle is acute.
(b) c2 = 102 = 100. a2 + b2 = 62 + 82 = 36 + 64 = 100. c2 = a2 + b2 , so
the triangle is a right triangle.
(c) a and b are the same as in question b), but c is smaller, so the triangle
is acute.
(d) a and b are the same as in question b), but c is larger, so the triangle
is obtuse.
(e) c2 = 122 = 144. a2 + b2 = 52 + 122 = 25 + 144 = 169. c2 < a2 + b2 , so
the triangle is acute.
(f) c2 = 142 = 196. a2 + b2 = 169, as above. c2 > a2 + b2 , so the triangle
is obtuse.
(g) The sum of two sides must be larger than the third, but 12 + 5 = 17
in this case.
Chapter 1: Trigonometric Ratios in a Triangle
Page 23
1. (a) sin α = 5/13
(b) sin α = 4/5
(c) sin α = 5/13
√
√
(d) c = 62 + 82 = 100 = 10. sin α = 8/10.
(e) sin α = 3/5
(f) sin α = 12/13
(g) sin α = 3/5
√
√
√
(h) c = 72 + 32 = 58. sin α = 7/ 58.
2. (a) sin β = 12/13
(b) sin β = 3/5
(c) sin β = 12/13
(d) sin β = 6/10
(e) sin β = 4/5
(f) sin β = 5/13
(g) sin β = 4/5
√
(h) sin β = 3/ 58
5
√
3. The example 30-60-90 triangle given on page 11 has sides
1, 3, 2. Let β
√
represent the 60◦ angle. The opposite
√ leg b has length 3. The hypotenuse
c has length 2. So, sin β = b/c = 3/2 ≈ 1.732/2 = 0.866.
Crossing off the numbers listed:
0.9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Page 25
1. The Altitude-on-Hypotenuse Theorem tells us that when an altitude is drawn
to the hypotenuse of a right triangle, the two triangles formed are similar
to the given triangle and to each other. Therefore, the triangles with sides
a-b-c, a-p-d and d-b-q are similar, and the ratio for sin α appears in all of
them:
(a) b/c
(b) d/a
(c) q/b
2. (a) sin α = h/b
(b) Multiplying both sides of formula above by b: h = b sin α
(c) Substituting b sin α for h, the formula for the area of ABC can be
rewritten as: bc sin α/2.
(d) sin β = h/a. Rewriting this in terms of h: h = a sin β. Substituting
this for h in the area formula: ac sin β/2.
(e) Let h2 represent the altitude from A to BC. sinβ = h2 /c. Rewriting
in terms of h2 , we get h2 = c sin β.
3. (a) Expressing h in terms of sin α and b:
h
b
h = b sin α
sin α =
Expressing h in terms of sin β and a:
h
a
h = a sin β
sin β =
(b) Both expresssions are equal to h:
a sin β = h = b sin α
(c) Expressing h2 in terms of sin β and c:
h2
c
h2 = c sin β
sin β =
6
Expressing h2 in terms of sin γ and b:
h2
b
h2 = b sin γ
sin γ =
Both expressions are equal to h2 :
b sin α = h2 = c sin γ
(d)
i. We can rewrite the result from part (b) so that the expressions on
each side are fractions with sine denominators:
a sin β = b sin α
a sin β
b sin α
=
sin α sin β
sin α sin β
a
b
=
sin α
sin β
ii. We can rewrite the result from part (c) similarly:
c sin β = b sin γ
c sin β
b sin γ
=
sin β sin γ
sin β sin γ
b
c
=
sin γ
sin β
We can derive the Law of Sines by combining results i. and ii. using the
common expression b/ sin β:
a
b
c
=
=
sin α
sin β
sin γ
Page 26
1. (a) cos α = 12/13. cos β = 5/13.
(b) cos α = 3/5. cos β = 4/5.
(c) cos α = 12/13. cos β = 5/13.
(d) cos α = 6/10. cos β = 8/10.
(e) cos α = 4/5. cos β = 3/5.
(f) cos α = 5/13. cos β = 12/13.
(g) cos α = 4/5. cos β = 3/5.
√
√
(h) cos α = 3/ 58. cos β = 7/ 58.
√
√
√
2. (a) c = 82 + 62 = 64 + 36 = 100 = 10. cos α = 8/10. cos β = 6/10.
7
√
√
√
(b) c = 52 + 122 = 25 + 155 = 169 = 13. cos α = 12/13. cos β =
5/13.
√
(c) Scaling up the 1- 3-2 30◦ triangle gives us a value of 20 units for the
length of c. Next, we will use the Pythagorean Theorem to find the
length of the longer leg:
a2 + b2 = c2
102 + b2 = 202
b2 = 400 − 100 = 300
√
√
√
√
b = 300 = 100 3 = 10 3
We can now find cos α and cos β:
√
√
3
10 3
=
cos α =
20
2
1
10
=
cos β =
20
2
(d) The triangle is congruent to the one above, so the solution is the same.
(e) Consider
the 45◦ right√triangle with legs of length 1 and hypotenuse
√
2. cos α = cos β = 1/ 2.
√
√
√
(f) c = 32 + 42 = 9 + 16 = 25 = 5. cos α = 3/5. cos β = 4/5.
√
√
√
(g) b = x 3. cos α = x 3/2x = 3/2. cos β = x/2x = 1/2.
3. The Altitude-on-Hypotenuse Theorem tells us that when an altitude is drawn
to the hypotenuse of a right triangle, the two triangles formed are similar
to the given triangle and to each other. Therefore, the triangles with sides
a-b-c, a-p-d and d-b-q are similar, and the ratio for cos α appears in all of
them:
(a) a/c
(b) p/a
(c) d/b
Page 28
1. In this instance, α = 29◦ , β = 61◦ , and α + β = 90◦ . According to the
theorem above, if α + β = 90◦ , then sin α = cos β.
2. x = 90 − 35 = 55◦
3. If α + β = 90◦ , then β = 90◦ − α. According to the theorem above, sin α =
cos β. Substituting (90 − α) for β: sin α = cos (90 − α).
8
Page 29
First,
we need to find the length of the hypotenuse: c =
√
25 = 5.
2
1. sin2 α = 45 = 16
25
2
2
9
2. sin β = 35 = 25
2
9
(same as sin2 β)
3. cos2 α = 35 = 25
2
16
4. cos2 β = 45 = 25
(same as sin2 α)
5. sin2 α + cos2 α =
16
25
+
9
25
=
25
25
6. sin2 α + cos2 β =
16
25
+
16
25
=
32
25
7. cos2 α + sin2 β =
9
25
+
9
25
=
18
25
√
32 + 4 2 =
√
9 + 16 =
=1
Page 30
1. sin2 α + cos2 α =
2 2
4
3
16
9
25
+
=
+
=
=1
5
5
25 25
25
2. It’s not an error. According to the corollary of the Pythagoream Theorem,
this a right triangle: a2 + b2 = 32 + 42 = 9 + 16 = 25 = c2 .
2 2
3
4
9
16
25
2
2
3. sin β + cos β =
+
=
+
=
=1
5
5
25 25
25
4. cos2 α + sin2 α = 1
cos2 α = 1 − sin2 α = 1 −
r
144
12
cos α =
=
169
13
5
13
2
=1−
25
144
=
169
169
5. cos2 α + sin2 α = 1
2
5
25
24
cos2 α = 1 − sin2 α = 1 −
=1−
=
7
49
49
r
√
√ √
24
4 6
2 6
=
cos α =
= √
49
7
49
9
6. We will follow the proof at the bottom of Page 29:
a 2 b 2
sin2 α + sin2 β =
+
c
c
b2
a2
= 2 + 2
c
c
a 2 + b2
=
c2
a 2 + b2
= 2
a + b2
=1
7. Again, we will follow the proof at the bottom of Page 29:
2 b
a 2
2
2
cos α + cos β =
+
c
c
2
2
b
a
= 2+ 2
c
c
a2 + b2
=
c2
2
a + b2
= 2
a + b2
=1
Page 31
1.
2. cos 30◦ =
√
3
2
angle x
sin x
30◦
1
2
45◦
√1
2
60◦
√1
2
√
3
2
α
4
5
3
5
β
3
5
4
5
◦
√
3
2
1
2
= sin 60◦
2
1
3. sin 30 + cos 30 =
+
2
2
cos x
2
◦
√ !2
3
1 3
= + =1
2
4 4
10
4. We can observe from the table that sin x increases with the size of an acute
angle (sin 30◦ < sin 45◦ < sin 60◦ ), while cos x decreases with the size of an
acute angle. You can compare the fractions or convert to decimal make sure.
We know that sin α = 45 . We also know that α is an acute angle.
Is it larger or smaller than 30◦ ? Larger, 45 > 21 so sin α > sin 30◦ .
Than 45◦ ? Larger, 45 > √12 so sin α > sin 45◦ .
Than 60◦ ? Smaller,
4
5
√
<
3
2
so sin α < sin 60◦ .
Page 33 (First)
1. As the angle α get smaller, the ratio of the opposite side to the hypotenuse
approaches 0.
2. Recall from the theorem on page 28 that if α + β = 90◦ , then sin α = cos β
and cos α = sin β. So, if sin 90◦ = 1, then cos 0◦ = 1.
3. sin2 0◦ + cos2 0◦ = 02 + 12 = 0 + 1 = 1
4. sin2 90◦ + cos2 90◦ = 12 + 02 = 1 + 0 = 1
5. Our friend is mistaken; the sine of an angle can never be greater than 1.
Page 33 (Second)
1.
sin 0◦ + cos 0◦
0+1
sin 30◦ + cos 30◦
sin 45◦ + cos 45◦
1
2
1
√
+
3
2
1.366 (approx.)
1.414 (approx.)
sin 60◦ + cos 60◦
√1 + √1
2
2
√
3
1
2 + 2
sin 90◦ + cos 90◦
1+0
1
sin α + cos α, where α
is the smaller. . .
sin α + cos α, where α
is the larger. . .
1.366 (approx.)
3
5
+
4
5
1.4
4
5
+
3
5
1.4
2. If sin α = 1, then cos α = 0 and sin α + cos α = 1. If cos α = 1, then
sin α = 0 and sin α + cos α = 1. Otherwise, sin α < 1 and cos α < 1, so
sin α + cos α < 2.
11
3. First we will expand and simplify (sin α + cos α)2 :
(sin α + cos α)2 = sin2 α + 2 sin α cos α + cos2 α
= (sin2 α + cos2 α) + 2 sin α cos α
= 1 + 2 sin α cos α
We know that 0 ≤ sin α ≤ 1 and 0 ≤ cos α ≤ 1 because α is acute. So
2 sin α cos α is the product of three nonnegative numbers, and is itself a
nonnegative number. A nonnegative number added to 1 results in a number
≥ 1. Therefore, 1+2
√ sin α cos α ≥ 1. The square root of a number ≥ 1 is itself
≥
1.
Therefore,
1 +p2 sin α cos α ≥ 1. Rewriting the expression on the left:
√
1 + 2 sin α cos α = (sin α + cos α)2 = sin α + cos α. So, sin α + cos α ≥ 1.
√
√
1
1
2
2 2
2 2 √
◦
◦
4. sin 45 + cos 45 = √ + √ = √ = √ √ =
= 2
2
2
2
2
2 2
5. You should notice that the values for sin α+cos α increases with larger alpha
when 0◦ ≤ α < 45◦ , reaches a maximum value when α = 45◦ , then decreases
with larger α when 45◦ < α ≤ 90◦ .
Page 35
1.
(sin 0◦ )(cos 0◦ )
0·1
(sin 30◦ )(cos 30◦ )
1
2
(sin 45◦ )(cos 45◦ )
√
3
2
·
√1 · √1
2
2
√
3 1
2 · 2
(sin 60◦ )(cos 60◦ )
(sin α)(cos α), where α
is the smaller. . .
(sin α)(cos α), where α
is the larger. . .
0
0.433 (approx.)
0.5
0.433 (approx.)
3
5
·
4
5
0.48
4
5
·
3
5
0.48
How large can the product (sin α)(cos α) get? We can see from the table that
the maximum value of the product appears to be when α = 45◦ .
Page 37
1. cos α = 3/5, cos β = 4/5, sin α = 4/5, sin β = 3/5, tan α = 4/3, tan β = 3/4,
cot α = 3/4, cot β = 4/3.
2. We can show that this assumption is correct using the corollary of the
Pythagorean Theorem: a2 + b2 = 32 + 42 = 25 = c2 .
12
3. cos α = a/c, cos β = b/c, sin α = b/c, sin β = a/c, tan α = b/a, tan β = a/b,
cot α = a/b, cot β = b/a.
√
√
4. c = 122 + 52 = 169 = 13. cos α = 12/13. cos β = 5/13. cot α = 12/5.
cot β = 5/12.
5. First, we will use the Pythagorean Theorem to find the length of the longer
leg:
a2 + b2 = c2
a2 + 72 = 252
a2 + 49 = 625
a2 = 576
a = 24
We can now find the numerical values that were asked for: cos α = 24/25,
cos β = 7/25, cot α = 24/7, cot β = 7/24.
6.
a
c
b
c
a
b
b
a
= sin α = cos β
= cos α = sin β
= tan α = cot β
= cot α = tan β
7. First, we will use the Pythagorean Theorem to find the length of the other
leg:
a2 + b2 = c2
a2 + 32 = 52
a2 + 9 = 25
a2 = 16
a=4
We can now find the numerical values that were asked for: cos α = 4/5,
cot α = 4/3.
◦
8. If tan α = 1,
√ then a/b = 1, implying that a = b and α = 45 . cos α =
◦
cos 45 = 1/ 2. cot α = 1/1 = 1.
9. tan 45◦ = 1/1 = 1.
√
10. tan 30◦ = 1/ 3 ≈ 0.57735.
11. tan 45◦ + sin 30◦ = 1 +
numbers are rational.
1
2
=
3
2.
We don’t need a calculator because both
13
Chapter 2: Relations among Trigonometric Ratios
Page 43
s
1−
1. cos α =
tan α =
8
17
15
17
cot α =
15
8
=
8
17
2
r
=
64
1−
=
289
r
225
15
=
289
17
8
15
2. Let the length of the adjacent leg a be 37 and the length of the hypotenuse
be 1 (see the first triangle diagram on page 44).
s
r
√
√ √
2 r
p
3
9
2 10
40
4 10
=
sin α = 1 − a2 = 1 −
= 1−
=
= √
7
49
49
7
49
√
√
√
2 10
1 − a2
2 10
tan α =
= 73 =
a
3
7
3
a
= √
cot α = √
2 10
1 − a2
3. sin α =
√
√
1−
4. sin α = √
b2 ,
tan α =
1 − b2
b
, cot α = √
b
1 − b2
1
1
d
, cos α = √
, cot α =
2
2
d
1+d
1+d
5.
sin α
sin α
cos α
tan α
cot α
a
√
√
1 − a2
a
1 + a2
1
√
1 + a2
√
cos α
1 − a2
a
1
1 + a2
a
√
1 + a2
√
Page 45 (First)
1. Given in text
14
tan α
a
√
2
1
√ −a
1 − a2
a
a
1
a
cot α
√
1 − a2
a
a
√
1 − a2
1
a
a
2
◦
2. sin 45 =
1
√
2
2
=
1
2
3.
sin α
sin α
sin α
√
cos α
√
tan α
p
1 − sin2 α
1 − cos2 α
tan α
1 + tan2 α
1
√
1 + cot2 α
cot α
cos α
cos α
√
1
1 + tan2 α
cot α
√
1 + cot2 α
tan α
a
p
2
√ 1 − sin α
1 − cos2 α
cos α
tan α
Page 45 (Second)
1. tan α =
a
= cot β
b
b
= tan β
a
c
3. sec α = = csc β
a
c
4. csc α = = sec β
b
2. cot α =
Page 47
√ !2
2
1
3
1 3
1. (a) sin 30 + cos 30 =
+
= + =1
2
2
4 4
2 2
1
1
1 1
(b) sin2 45◦ + cos2 45◦ = √
+ √
= + =1
2 2
2
2
√ !2 2
3
1
3 1
(c) sin2 60◦ + cos2 60◦ =
+
= + =1
2
2
4 4
2
◦
2
◦
15
1
cot α
cot α
p
√
1 − sin2 α
sin α
cos α
1 − cos2 α
1
tan α
cot α
2.
sin2 α + cos2 α = 1
√ !2
5
+ cos2 α = 1
4
√ !2
5
5
11
=1−
cos α = 1 −
=
4
16
16
r
√
11
11
cos α =
=
16
4
2
3. sin2 α + cos2 α = 1
2
2
2
sin α +
=1
3
4
5
sin2 α = 1 − =
9
9
r
√
5
5
sin α =
=
9
3
4.
sin α
1
= tan α = √
cos α
3
sin2 α
1
=
cos2 α
3
sin2 α
1
=
3
1 − sin2 α
3 sin2 α = 1 − sin2 α
4 sin2 α = 1
1
sin2 α =
4
r
sin α =
1
1
=
4
2
16
sin α
1
= tan α = √
cos α
3
2
1
sin α
=
cos2 α
3
1 − cos2 α
1
=
cos2 α
3
1
1 − cos2 α
=3∗
3∗
2
cos α
3
3 ∗ (1 − cos2 α)
=1
cos2 α
2
3 ∗ (1 − cos α) = cos2 α
3 − 3 cos2 α = cos2 α
3 = 4 cos2 α
3
= cos2 α
4
√
3
= cos α
2
And then to check our solution we can calculate the fraction we are given
√1
3
1
√2
3
2
from our cos α and sin α fractions.
2
√
2∗ 3
1
√
3
5. (a) cot x sin x =
(b)
1
tan x
sin x =
sin x
sin x
·
tan x
= cos x = cos x
sin x
sin x
sin x ·
1
sin x
1
sin x
sin x cos x
sin x
sin x
= sin x =
= cos x
tan x
sin x
cos x
=
sin x
sin x cos x
1
=
sin x
1
=
sin x cos x
cos x
(c) cos2 α−sin2 α = cos2 α−(1−cos2 α) = cos2 α−1+cos2 α = 2 cos2 α−1
(d) This one is tricky. You might need to try a few different approaches
(squaring above and below, multiplying above and below by cos α sin α).
Eventually it becomes clear that you need to multiply above and below
by (1 − cos α) and find a way to cancel out the sin α factor in the
numerator:
sin α(1 − cos α)
sin α(1 − cos α)
sin α
=
=
1 + cos α
(1 + cos α)(1 − cos α)
1 − cos α + cos α − cos2 α
sin α(1 − cos α)
sin α(1 − cos α)
sin α(1 − cos α)
=
=
=
2
2
1 − cos α
1 − (1 − sin α)
1 − 1 + sin2 α
sin α(1 − cos α)
1 − cos α
=
=
2
sin α
sin α
17
(e)
sin2 α + 2 cos2 α − 1
1 − cos2 α + 2 cos2 α − 1
cos2 α
=
=
2
2
cos α 2
cot α
cot α
sin α
=
cos2 α sin2 α
=
cos2 α
cos2 α
cos2 α
sin2 α
2
= sin α
2
(f) cos2 α =
=
cos α
cos2 α
=
1
cos2 α + sin2 α
cos2 α
cos2 α
cos2 α+sinα
cos2 α
=
1
cos2 α
cos2 α
+
sin2 α
cos2 α
1
1 + tan2 α
cos2 α
sin2 α
=
(g) sin2 α =
2
1
cos α + sin2 α
=
=
sin2 α
sin2 α
cos2 α+sinα
sin2 α
=
1
cos2 α
sin2 α
+
sin2 α
sin2 α
1
cot α + 1
1 − cos α
(1 − cos α)(1 + cos α)
1 + cos α − cos α − cos2 α
(h)
=
=
1 + cos α
(1 + cos α)(1 + cos α)
(1 + cos α)2
=
2
1 − cos2 α
sin2 α
=
(1 + cos α)2
(1 + cos α)2
2
sin α
=
1 + cos α
(i) The key to solving this one is the formula for factoring a difference of
cubes: a3 − b3 = (a − b)(a2 + ab + b2 ).
=
(sin α − cos α)(sin2 α + sin α cos α + cos2 α)
sin3 α − cos3 α
=
sin α − cos α
sin α − cos α
2
= sin α + sin α cos α + cos2 α
= 1 + sin α cos α
6. (a) We can rewrite the LHS to show that sin4 α − cos4 α = cos2 α − sin2 α:
sin4 α − cos4 α = (sin2 α + cos2 α)(sin2 α − cos2 α) = 1(sin2 α − cos2 α)
= sin2 α − cos2 α
Answer: There are no angles α for which sin4 α−cos4 α > cos2 α−sin2 α
because the expressions on either side of the inequality are equivalent.
(b) sin4 α−cos4 α >= cos2 α−sin2 α for all angles α because the expressions
on either side of the inequality are equivalent.
18
7. If we rewrite 2 sin α cos α as a fraction, we can divide above and below by
cos α to convert the numerator and denominator into expressions in terms
of tan α:
2 sin α cos α =
2 sin α cos α
2 sin α cos α
=
1
sin2 α + cos2 α
=
2 sin α cos α
cos2 α
sin2 α+cos2 α
cos2 α
=
2 tan α
tan2 α + 1
2 sin α
cos α
sin2 α
cos2 α
cos2 α + cos2 α
=
Now we can plug in the given value for tan α to find the value of 2 sin α cos α
in this instance:
2 sin α cos α =
2( 52 )
2 tan α
=
=
tan2 α + 1
( 25 )2 + 1
4
5
4
25
+1
=
4
5
4
25
+
25
25
=
20
25
29
25
=
20
29
8. First, we will rewrite the expresssion cos2 α − sin2 α in terms of tan α:
cos2 α − sin2 α
cos2 α − sin2 α
cos α − sin α =
=
=
1
cos2 α + sin2 α
2
2
=
cos2 α−sin2 α
cos2 α
cos2 α+sin2 α
cos2 α
1 − tan2 α
1 + tan2 α
(a) To find the numerical value of cos2 α − sin2 α when tan α =
substitute 25 for tan α in the formula above:
cos2 α − sin2 α =
1−
1 − tan2 α
=
1 + tan2 α
1+
2 2
5
2 2
5
=
1−
1+
4
25
4
25
=
21
25
29
25
2
5
=
(b) Substituting r for tan α in the formula above:
cos2 α − sin2 α =
1 − r2
1 + r2
9. First, we will rewrite the expresssion in terms of tan α:
sin α − 2 cos α
=
cos α − 3 sin α
Next, we substitute
2
5
sin α−2 cos α
cos α
cos α−3 sin α
cos α
=
sin α
cos α
cos α
cos α
cos α
− 2cos
tan α − 2
α
=
3 sin α
1 − 3 tan α
− cos α
for tan α:
2
−2
tan α − 2
=
= 5
1 − 3 tan α
1 − 3 25
19
2
10
5 − 5
5
6
5 − 5
=
− 85
=8
− 15
we can
21
29
10. First, we will rewrite the expresssion in terms of tan α:
a sin α + b cos α
=
c cos α + d sin α
a sin α
cos α
c cos α
cos α
+
+
b cos α
cos α
d cos α
cos α
=
a tan α + b
c + d tan α
2
5
for tan α and simplify:
2a+5b
a 25 + b 55
a tan α + b
2a + 5b
5
= 5c+2d
=
=
5
2
c + d tan α
5c + 2d
c 5 +d 5
5
Next, we substitute
Now we can see why the problem included the restriction that 5c + 2d ̸= 0;
the value of the expression is undefined if the denominator is zero. The sum
of two rational numbers is a rational number. Therefore the numerator and
denominator in the expression are both rational numbers. The quotient of
two rational numbers is a rational number. Therefore, the entire expression
evaluates to a rational number for arbitratrary rational values of a, b, c and
d.
11. We can expand and simplify the expression:
(sin α + cos α)2 + (sin α − cos α)2
= sin2 α + 2 sin α cos α + cos2 α + sin2 α − 2 sin α cos α + cos2 α
= 2 sin2 α + 2 cos2 α
= 2(sin2 α + cos2 α)
= 2(1)
=2
As the expression evaluates to a constant, it is as large as possible for all
values of α.
Page 49
1. Rewriting any instances of sec α or csc α on either side of the identities:
(a) tan α csc α = sec α
1
1
tan α
=
sin α
cos α
tan α
1
=
sin α
cos α
(b) cot α csc α = sec α
1
1
cot α
=
cos α
sin α
cot α
1
=
cos α
sin α
20
1
csc α = cot α
sec α
1
1
= cot α
·
1
sin α
cos α
1
cos α
= cot α
sin α
cos α
= cot α
sin α
(d) tan2 α = (sec α + 1)(sec α − 1)
(c)
tan2 α = sec2 α − 1
1
tan2 α =
−1
cos2 α
(e) csc2 α = 1 + cot2 α
1
= 1 + cot2 α
sin2 α
2. Rewriting any instances of sin α or cos α on either side of the identities, and
eliminating fractions:
(a)
(b)
tan α
1
=
sin α
cos α
1
tan α
= sec α
sin α
tan α csc α = sec α
1
cos α = cot α
sin α
cos α
= cot α
sin α
cot α = cot α
1
cos2 α
tan2 α + 1 = sec2 α
(c) tan2 α + 1 =
(d)
1
= 1 + cot2 α
sin2 α
csc2 α = 1 + cot2 α
21
Page 50
1. First, we find the value of a2 + b2 :
a2 + b2 = (cos2 α − sin2 α)2 + (2 sin α cos α)2
= cos4 α − 2 cos2 α sin2 α + sin4 α + 4 sin2 α cos2 α
= cos4 α + 2 cos2 α sin2 α + sin4 α
= (cos2 α + sin2 α)2
= (1)2
=1
According to the lemma on Page 50, as a2 + b2 = 1, an angle θ exists such
that a = cos θ and b = sin θ.
2. First, we find the value of a2 + b2 :
!2
r
1
+
cos
α
a2 + b2 =
+
2
r
1 − cos α
2
!2
1 + cos α 1 − cos α
+
2
2
1 + cos α + 1 − cos α
=
2
2
=
2
=1
=
3. First, we will rewrite a and b to eliminate the cube exponents:
a = 4 cos3 α − 3 cos α
= 4 cos α cos2 α − 3 cos α
= 4 cos α(1 − sin2 α) − 3 cos α
= 4 cos α − 4 sin2 α cos α − 3 cos α
= cos α − 4 sin2 α cos α
b = 3 sin α − 4 sin3 α
= 3 sin α − 4 sin α sin2 α
= 3 sin α − 4 sin α(1 − cos2 α)
= − sin α + 4 sin α cos2 α
Next, we will expand a2 and b2 :
a2 = (cos α − 4 sin2 α cos α)2
= cos2 α − 8 sin2 α cos2 α + 16 sin4 α cos2 α
22
b2 = (− sin α + 4 sin α cos2 α)2
= sin2 α − 8 sin2 α cos2 α + 16 sin2 α cos4 α
Next, we add the expressions for a2 and b2 and simplify to 1:
a2 + b2 = cos2 α − 8 sin2 α cos2 α + 16 sin4 α cos2 α + sin2 α − 8 sin2 α cos2 α+
16 sin2 α cos4 α
= cos2 α + sin2 α − 16 sin2 α cos2 α + 16 sin4 α cos2 α + 16 sin2 α cos4 α
= cos2 α + sin2 α + 16 sin2 α cos2 α(−1 + sin2 α + cos2 α)
= 1 + 16 sin2 α cos2 α(0)
=1
According to the lemma on Page 50, as a2 + b2 = 1, an angle θ exists such
that a = cos θ and b = sin θ.
4. First, we find the value of a2 + b2 :
2 2
1 − t2
2t
2
2
a +b =
+
1 + t2
1 + t2
=
(2t)2
(1 − t2 )2
+
(1 + t2 )2
(1 + t2 )2
=
(1 − t2 )2 + (2t)2
(1 + t2 )2
=
1 − 2t2 + t4 + 4t2
(1 + t2 )(1 + t2 )
(1 + t2 )(1 + t2 )
(1 + t2 )(1 + t2 )
=1
=
According to the lemma on Page 50, as a2 + b2 = 1, an angle θ exists such
that a = cos θ and b = sin θ.
5. We expand (p2 − q 2 )2 + (2pq)2 and use the fact that p2 + q 2 = 1 to simplify
to 1:
(p2 − q 2 )2 + (2pq)2 = p4 − 2p2 q 2 + q 4 + 4p2 q 2
= p4 + 2p2 q 2 + q 4
= (p2 + q 2 )2
= (1)2
=1
This is similar to Exercise 1 above.
23
Page 51
1. sin α < 1 when α is acute, therefore 1−sin α > 0 when α is acute. 1−sin α =
0 when sin α = 1, i.e., α = 90◦ .
2. cos α < 1 when α is acute, therefore 1−cos α > 0 when α is acute. 1−cos α =
0 when cos α = 1, i.e., α = 0◦ .
3. Statement a) is always true. Statements b) and c) both include the case that
sin2 α + cos2 α = 1, which is always true.
4. Let x be the maximum cost of the items in a supermarket. In Supermarket
A, x ≤ $1. In Supermarket B, x < $1. In Supermarket C, x ≤ $1. In
Supermarket D, x > $1. We can see that Supermarkets A and C are offering
the same terms.
5. Inequality a) is correct. For b) to be correct, an angle α would have to
exist such that sin α + cos α = 2. We know that this is not the case. When
α = 90◦ , sin α = 1 and cos α = 0. When α = 0◦ , sin α = 0 and cos α = 1.
When 0◦ < α < 90◦ , sin α < 1 and cos α < 1. In all cases, sin α + cos α < 2.
6. The largest possible value of sin α is 1, and occurs when α = 90◦ . The largest
possible value of cos α is 1, and occurs when α = 0◦ . See Page 32.
Page 52
1. sin 30◦ = 0.5, sin 45◦ = 0.707, sin 60◦ = 0.866.
◦
2. By using
√ the tan button to calculate tan 60 , and the sqrt button to calculate 3, Betty can compare the results: both are 1.732.
3. Press tan, then enter the angle degree measure, then press 1/x
4.
in radical or rational form
α
sin α
30◦
1
2
1
√
2
√
3
2
45◦
60◦
cos α
√
3
2
1
√
2
1
2
24
tan α
cot α
1
√
3
√
1
1
√
1
√
3
3
3
in decimal form, from calculator
α
sin α
cos α
tan α
cot α
30◦
0.5
0.866
0.577
1.732
45◦
0.707
0.707
1
1
60◦
0.866
0.5
1.732
0.577
Page 53
1. The sine of the larger angle is 4/5 = .8. We can use the inverse sine function
to find the angle: arcsin .8 = 53.1301◦ . The sum of the three angles in the
triangles is: arcsin .6 + arcsin .8 + 90◦ = 36.8699◦ + 53.1301◦ + 90◦ = 180◦ .
2. (a) arcsin 1 = 90◦
(b) arccos 0.7071067811865 = 45◦
3. arccos 0.8 = 36.8699◦
4. arcsin 0.6 = 36.8699◦
5. Half of sin 30◦ (0.25) seems like a reasonable estimate. The actual value is
0.2588.
6. Entering arcsin 0.3 in calculator, we get 17.458◦
If we enter sin 17.458◦
7. With arcsin x = 53◦ , to find value of x, we do x = sin 53◦ , which gives us
x = 0.7986
If we take arcsin 0.7986, we will get 52.9966
√
8. x = 3/2
9. arcsin(sin 17◦) = 17◦
10. sin(arcsin 0.4) = 0.4
11. arcsin(sin 30◦ )) = 30◦
arcsin 1/2 = 30{ circ}. Since sin 30◦ = 0.5 as in question 5
12. cos2 20◦ + sin2 20◦ = 1
cos2 80◦ + sin2 80◦ = 1
13. tan 20◦ = 0.36397 and
sin 20◦
cos 20◦
Note tan 80◦ = 5.67128 and
0.34202
0.93969 ≈ 0.36397
sin 80◦
0.98481
cos 80◦ = 0.17365 ≈ 5.67128
=
25
y
1
0
π
2
π
x
3π
2
2π
−1
14.
Page 55
9
β
a
α = 72
b
1.
To find a
a
c
a
◦
sin 72 =
9
a = 9 × sin 72◦
sin α =
a ≈ 8.5595
To find b
b
c
b
cos 72◦ =
9
b = 9 × cos 72◦
cos α =
b ≈ 2.7812
c
β
7
α
2.
10
To find the hypotenuse c, use pythagoras theorem
p
c = a2 + b2
√
c = 149
26
To find α
b
a
7
tan α =
10
tan α =
α = arctan
7
10
α ≈ 34.992◦
To find β
a
b
10
tan β =
7
tan β =
β = arctan
10
7
β ≈ 55.008◦
c
β = 27
a
α
3.
12
To other acute angle will be 90 − 27 = 63 degrees
To find length of hypotenuse c
12
c
12
c=
sin 27
= 26.432
sin β =
To find length of the other leg a
12
a
12
a=
tan 27
= 23.551
tan β =
c
4.
β
20
α = 73
a
27
To other acute angle will be 90 − 73 = 17 degrees
To find length of hypotenuse c
20
c
20
c=
sin 73
= 20.914
sin β =
To find length of the other leg a
20
a
20
a=
tan 73
= 6.115
tan β =
Page 56
50
1.
α = 46
a
To find length of shadow a
50
a
50
a=
tan 46
a = 48.284
tan α =
Therefore the length of shadow is 48.284 feet
a
α = 32
35
2.
To find length of shadow a
a
35
a = 35 tan 32
tan α =
a = 21.87
Therefore the height of flagpole is 21.87 feet
28
3. Shadow is the longest in the morning as the sun is at lower altitude
Shadow is shortest during mid day as the sun is exactly above the pole
4. There will be no shadow when the sun is exactly above the flag pole, which
happens at a specific time during ’zero shadow day’ (search online for exact
time and day for your location)
Page 59
1. Draw point C on the left side of circle, by circle geometry, the angle in that
triangle vertex will be θ/2. Then apply the relation sin α = AB
2r , where AB
is the opposite side of angle α. Therefore sin θ/2 = AB since radius is 1/2
2. Draw point C on the left side of circle, by circle geometry, the angle in that
triangle vertex will be θ/2. Then apply the relation sin α = AB
2r , where AB
is the opposite side of angle α. Therefore sin θ/2 = AB
since
radius is 1,
2
rearranging the equation gives 2 sin θ/2 = AB
3. From question 1, for circle of diameter 1, we have the relation sin θ/2 = AB.
Therefore if angle ϕ > θ then the opposite side of angle ϕ will be longer too,
therefore sin ϕ > sin θ
4. Assuming the three points are on a circle, and realising that the sides 6, 8
and 10 form a right angle triangle, therefore the side of length 10 would have
been the diameter of the circle. Therefore the radius for the circle is 5
PB
5. Since sin α = P2rB , re-arranging the equation gives sin
α = 2r. That is, the
ratio of a : sin α is equal to the diameter of circle (double of radius)
6. let a be opposite side of angle α let b be opposite side of angle β let c be
opposite side of angle γ
Exercise shows that sina α = 2r, sinb β = 2r, sinc γ = 2r. Since all are equal to
2r then they are ratios with same value, that is
a
b
c
= =
sin α
β
γ
AB
7. Since sin α = 2×1/2
, then chord AB = sin α. Since arc AC is double of arc
AB, then chord AC = 2 sin α
But the diagram shows that in triangle ABC and using triangle inequality,
AB + BC ≥ AC. Therefore sin 2α < 2 sin α
if α = 60◦ and radius is 10/2=5
√
The chord AB would be 2 × 5 × sin 60◦ = 5 3 units long
8. Using sin α =
AB
2r ,
◦
9. Using sin α = AB
2r , if α = 30 (as central angle is double of angle on circle)
and radius is 10/2=5
The chord AB would be 2 × 5 × sin 30◦ = 5 units long
29
10. The diagonal of a square inside a circle will be the diameter of circle and the
angle of each triangle corner is 45◦
where 2r is 10 (diameter) and angle α = 45◦ , we get
Using sin α = AB
2r , √
AB = 10 sin 45◦ = 5 2 units long for each side of square inside a circle
11. Using sin α = AB
2r , then letting one side of pentagon AB = 2r × sin α.
Nothing that a pentagon can be cut into three distinct triangles, therefore
the internal angles add up to 180 × 3 = 540 making each internal angle
540 ÷ 5 = 108 degrees large
As each vertex is divided into three equal angles, it will give α = 108÷3 = 36
degrees. As diameter is 10, each side of pentagon will be 10 × sin 36
Page 62
1. The degree measure of a semicircle is 180◦ . The degree measure of a quarter
circle is 90◦ .
2. The measure of arc cut off by one side of regular pentagon inscribed in a
circle is 360◦ /5 = 72◦ . For a regular hexagon: 360◦ /6 = 60◦ . For a regular
octagon: 360◦ /8 = 45◦ .
Page 64
1. Let AB be an arc on the circle and points Q and R are two other points on
the circle
The angle in the centre will be double of angle AQB, which will be double
of angle ARB as well, therefore the both aangles AQB and ARB must be
equal
2. A regular pentagon can be divided into three equal circles, therefore a pentagon’s total internal angle is 180 × 3 = 540 degrees.
Since the intenal angles are equal size for a regular pentagon, then each angle
is 540 ÷ 5 = 108 degrees large
3. Let points ABCD be points on a circle, then draw line AC. Note that the
angle AOC will be double of ABC
Meawhile on the other side of AOC will be double of ADC, these two
sides makes a full revolution (360 degrees). Therefore we have 2̸ ABC +
2̸ ADC = 360, that is ̸ ABC + ̸ ADC = 180 degrees
Page 65
1. As it is larger than 60 degrees, then the points will be inside the circle
2. As it is 90 degrees, then the points will be on the circle with AB being the
diameter of the circle
30
3. Triangle AQB would have a larger inscribed circle of radius R, therefore the
AB
relation sin α = AB
2R that is 2R = sin α
Triangle APB is inside a smaller circle of radius r, therefore the relation
AB
sin Q = AB
2r . That is 2r = sin Q
The ratio r:R will be
AB
sin Q
÷
AB
sin α
=
sin α
sin Q .
That is
r
R
=
sin α
sin Q
if r < R then sin α < sin Q. That is α < Q for the left diagram if r > R
then sin α > sin Q. That is α > Q for the right diagram
4. Following above proof, let angle Q be 0.5α. Since sin α > sin 0.5α, then
r > R, so it will be inscribed in the circle of original arc
5. On the same side, the angle subtended will be α, where angle subtended in
the centre would be 2α. The other side of the centre angle would be 360−2α,
which gives 180 − α as size of subtended angle on the other arc side
Note, this is proof for opposite angles of a cyclic quadrilateral are supplementary
Chapter 3: Relationships in a Triangle
Page 68
1.
Data
Determine a triangle?
Restrictions?
1
ABa
Yes
0◦ < m̸ A, m̸ B, m̸ A + m̸ B < 180◦
2
ABb
Yes (Duplicate of 1)
0◦ < m̸ A, m̸ B, m̸ A + m̸ B < 180◦
3
ABc
Yes
0◦ < m̸ A, m̸ B, m̸ A + m̸ B < 180◦
4
AbC
Yes (Duplicate of 3)
0◦ < m̸ A, m̸ B, m̸ A + m̸ B < 180◦
5
ABC
No
One side length must also be given.
6
Abc
Yes
0◦ < m̸ A < 180◦
7
Bbc
No
b ≥ c sin B. Also, m̸ A must be given in some cases.
8
Cbc
No (Duplicate of 7)
c ≥ b sin C. Also, m̸ A must be given in some cases.
Page 70
1.
31
A
α
Q
hb
c
γ
a
B
C
A
α
R
hc
b
β
a
B
C
2.
hc = b sin α
= 12 sin 70◦
≈ 11.28
3. Let us consider a right triangle ABC with a right angle at A.
A
c
b
ha
γ
β
B
P
C
32
Analyzing the right triangle AP B in the above diagram gives us that ha =
c sin β. Similarly, analyzing the right triangle AP C in the above diagram
gives us that ha = b sin γ. This is consistent with our formulas for ha .
A
b = hc
c = hb
γ
β
a
B
C
Since A is a right angle, b and c are themselves altitudes, as shown by the
diagram above. Since the sine of a right angle is equal to 1, this is consistent
with the formulas hb = c sin α and hc = b sin α. Additionally, by analyzing
the sines at angles B and C, we find that hb = a sin γ and hc = a sin β,
which again agrees with the formulas for hb and hc .
4.
P
10
hp
30◦
12
R
33
Q
1
php
2
1
= pq sin R
2
1
= (12) (10) sin 30◦
2
1
1
= (12) (10)
2
2
A=
= 30
Page 71
1.
A
α
b
c
β
γ
C
a
B
hc
R
For the hb formulas, there is nothing new to check as α and γ are both acute
angles.
From the above diagram, since △ARC is a right triangle, we can see that
hc = b sin α. Also, since △BRC is a right triangle, we have that hc =
a sin (180◦ − β). Since we have asserted that the sine of an angle should be
equal to the sine of its supplement, the previous formula is equivalent to
hc = a sin β.
Page 73
1. We first substitute a with b, b with c, c with a, α with β, β with γ, γ with
α.
b
b
c
a
=
→
=
sin α
sin β
sin β
sin γ
34
We now substitute a with c, c with b, b with a, α with γ, γ with β, β with
α.
a
b
c
a
=
→
=
sin α
sin β
sin γ
sin α
Combining these equalities together gives the Law of Sines:
a
b
c
=
=
.
sin α
sin β
sin γ
√
2. Let A = 30◦ , B = 60◦ , C = 90◦ . We know that b = a 3 and c = 2a by the
properties of 30-60-90 triangles.
a
a
=
= 2a
sin α
1/2
√
b
a 3
=√
= 2a
sin β
3/2
c
2a
=
= 2a
sin γ
1
Since all of the above ratios are equal to 2a, we have shown that the Law of
Sines holds for 30-60-90 triangles.
3. In the triangle on the left, the angle opposite the side of length 10 measures
60◦ . If x is the side length opposite the 50◦ angle and y is the side length
opposite the 70◦ angle, then
x
10
10 sin 50◦
=
=⇒ x =
≈ 8.85
◦
◦
sin 50
sin 60
sin 60◦
y
10
10 sin 70◦
=
=⇒
y
=
≈ 10.85
sin 70◦
sin 60◦
sin 60◦
In the triangle on the right, the missing angle measures 60◦ . If x is the side
length opposite the 60◦ angle and y is the side length opposite the 65◦ angle,
then
12
12 sin 60◦
x
=
=⇒
x
=
≈ 12.69
sin 60◦
sin 55◦
sin 55◦
y
12
12 sin 65◦
=
=⇒
y
=
≈ 13.28
sin 65◦
sin 55◦
sin 55◦
4. As discussed in Section 4 of this chapter, the formulas for the altitudes using
the sine function remain unchanged for obtuse triangles if we assert that the
sine of an angle is equal to the sine of its supplement. Therefore, the proof of
the Law of Sines for obtuse triangles is unchanged from the proof presented
in the beginning of this section.
35
5. In the triangle on the left, the angle opposite the side of length 6 measures
45◦ . If x is the side length opposite the 15◦ angle and y is the side length
opposite the 120◦ angle, then
6
6 sin 15◦
x
=
=⇒ x =
≈ 2.20
◦
◦
sin 15
sin 45
sin 45◦
y
6
6 sin 120◦
=
=⇒ y =
≈ 7.35
◦
◦
sin 120
sin 45
sin 45◦
In the triangle on the right, the missing angle measures 115◦ . If x is the
side length opposite the 40◦ angle and y is the side length opposite the 115◦
angle, then
x
14
14 sin 40◦
=
=⇒
x
=
≈ 21.29
sin 40◦
sin 25◦
sin 25◦
14
14 sin 115◦
y
=
=⇒
y
=
≈ 30.02
sin 115◦
sin 25◦
sin 25◦
6. We first solve for the angle opposite the side of length 10.
8
10
10
=
=⇒ sin α =
sin 50◦ = 0.9576 =⇒ α = 73.2◦
sin 50◦
sin α
8
We now solve for the third angle by subtracting the other two angles from
180◦ .
β = 180◦ − 50◦ − 73.2◦ = 56.8◦
7. Since the sine of an angle is equal to the sine of its supplement, based on
our calculations in the previous exercise, α could equal 73.2◦ or 106.8◦ . This
implies that β can equal 56.8◦ or 23.2◦ . The below diagram shows the two
triangles that satisfy the given constraints overlaid on top of each other.
73.2◦
8
106.8◦
8
50◦
10
36
8. This result follows immediately from the theorem proven in Section 12 of
Chapter 2.
Page 75 (First)
1.
2R =
10
10
10
=⇒ R =
=
= 10
sin 30◦
2 sin 30◦
2(1/2)
A
30◦
10
60◦
P
O
B
Alternatively, we can find the radius from first principles. Let AP B be a
triangle inscribed in a circle O with AB = 10 and m̸ AP B = 30◦ . Since
̸ AP B is an inscribed angle, the corresponding central angle ̸ AOB has a
measure of 60◦ . We also know that segments OA and OB are congruent
because they are both radii of the circle O. Therefore, △AOB is isosceles.
This combined with the fact that the measure of ̸ AOB is 60◦ implies that
△AOB is equilateral. Because AB is equal to 10, we know that the the
radius of the circle is also 10 because the other two sides of the equilateral
triangle are radii.
2.
2R =
8
8
8
=⇒ R =
=
=4
◦
◦
sin 90
2 sin 90
2(1)
Alternatively, we can apply Thales’s theorem (see the appendix of Chapter
2) to show that the hypotenuse of an inscribed triangle is a diameter of the
circumscribing circle. If the diameter of the circle is 8, then the radius would
be 4.
37
Page 75 (Second)
1.
A=
1
(8) (11) sin 40◦ = 44 sin 40◦ ≈ 28.3
2
2. (a)
A=
1
(10) (9) sin 23◦ = 45 sin 23◦ ≈ 17.6
2
(b)
A=
1
(3) (7) sin 130◦ = 10.5 sin 130◦ ≈ 8.0
2
(c)
1
(3) (7) sin 90◦ = 10.5 sin 90◦ = 10.5
2
Since this is a right triangle, the sides adjacent to the 90◦ angle are
altitudes, so we can use the more elementary formula for the area of a
triangle: A = 12 bh.
A=
3. Let b be the length of AC.
40 =
1
40
b (6) sin 40◦ =⇒ b =
≈ 20.7
2
3 sin 40◦
4. Let α be the measure of angle P .
9=
9
1
(5) (6) sin α =⇒ sin α =
=⇒ α ≈ 36.9◦ or α ≈ 143.1◦
2
15
5. The area of this triangle is 21 ab sin γ, where γ is the angle included between
the sides of length a and b. Because sin γ has a maximum value of 1 when
γ is 90◦ , the area of the triangle is maximized when γ is 90◦ . This gives an
area of 12 ab. Since γ = 90◦ , the triangle is a right triangle.
In this triangle, a and b are both legs of the right triangle. We can let one
of a or b be the hypotenuse of a right triangle to get another non-congruent
right triangle with side lengths of a and b. Without loss of generality, we
can choose b to be the hypotenuse.
By the Pythagorean theorem, the other
√
leg of the triangle has length b2 − a2 . Therefore, the area of this triangle
is
1 p 2
a b − a2 .
2
6. As in the previous exercise, for two given side lengths of a triangle, the area
of the triangle is maximized when the angle included between these two sides
is equal to 90◦ . Thus, the maximum area of an isosceles triangle with a leg
length of x is
1
1
(x) (x) sin 90◦ = x2 .
2
2
38
7.
a
A
B
b
b
D
C
a
Because opposite angles of a parallelogram are congruent, the area of the
above parallelogram is given by
1
1
1
1
ab sin A + ab sin C = ab sin C + ab sin C = ab sin C.
2
2
2
2
Notice that A and C are congruent, B and C are supplements, and C and
D are supplements by the properties of parallelograms. This means that the
above area formula can be used with any angle of the parallelogram since
sin A = sin B = sin C = sin D.
8.
z
y
α
180◦ − α
◦
α 180 − α
w
x
39
A=
=
=
=
=
1
1
1
1
wx sin α + xy sin (180◦ − α) + yz sin α + wz sin (180◦ − α)
2
2
2
2
1
1
1
1
wx sin α + xy sin α + yz sin α + wz sin α
2
2
2
2
1
(wx + xy + yz + wz) sin α
2
1
(w (x + z) + y (x + z)) sin α
2
1
(w + y) (x + z) sin α
2
w + y and x + z are the lengths of the diagonals in the quadrilateral in the
diagram above. The above derivation shows that the choice of α does not
matter since sin α = sin (180◦ − α).
9.
y
180◦ − α
α
z
x
d
A=
=
=
=
=
10.
1
1
1
1
(d + z) x sin α + (d + z) y sin (180◦ − α) − xz sin α − yz sin (180◦ − α)
2
2
2
2
1
1
1
1
(d + z) x sin α + (d + z) y sin α − xz sin α − yz sin α
2
2
2
2
1
[(d + z) x + (d + z) y − xz − yz] sin α
2
1
[(d + z) (x + y) − z (x + y)] sin α
2
1
d (x + y) sin α
2
1
(4) (6) sin α
4·6
3
2
=
=
1
8 · 10
10
(8) (10) sin α
2
40
11. See solution provided in the textbook
12.
A
B
w
x
α
180◦ − α
◦
α 180 − α
z
y
D
C
1
1
|AP B| × |CP D| =
yz sin α = wxyz sin2 α
2
4
1
1
1
|BP C|×|DP A| =
xy sin (180◦ − α)
wz sin (180◦ − α) = wxyz sin2 α
2
2
4
1
wx sin α
2
This identity is true even if the diagonals intersect outside of the quadrilateral. Using the diagram above in Exercise 9, we can show that |AP B| ×
|CP D| and |BP C| × |DP A| are both equal to
1
xyz (d + z) sin2 α.
4
13.
C
b
a
B
P
A
41
c = AB = AP + P B = b cos A + a cos B
Let’s suppose angle A is obtuse. Then we need to subtract . . .
C
b
a
P
B
A
c = AB = AP − BP = b cos A − a cos (180◦ − B) = b cos A + a cos B
As shown above, no changes need to be made to this result for obtuse angles.
Page 79
1.
C
h
b
a
x = a cos B ′
P
c
B
b2 = h2 + (c + x)
2
= a2 − x2 + c2 + 2cx + x2
= a2 + c2 + 2cx
= a2 + c2 + 2ac cos B ′
42
A
Page 80
1.
2. (a)
x2 = y 2 + z 2 − 2yz cos X
(b)
x2 = y 2 + z 2 − 2yz cos X
(c)
a2 = b2 + c2 − 2bc cos A
(d)
b2 = a2 + c2 − 2ac cos B
(e)
c2 = a2 + b2 − 2ac cos C
3. Given two sides and the angle between them, the Law of Cosines can be
used to find the length of the third side. Another application of the Law of
Cosines can be used to solve for one of the unknown angles since the three
sides of the triangle are now known. The final angle of the triangle can be
determined by subtracting the measures of the two known angles from 180◦ .
4. (a)
x=
p
122 + 152 − 2 (12) (15) cos 50◦ ≈ 11.7
x=
p
(b)
102 + 162 − 2 (10) (16) cos 110◦ ≈ 21.6
(c)
cos x◦ =
19
62 + 82 − 92
=
=⇒ x ≈ 78.6
2 (6) (8)
96
(d)
cos x◦ =
5.
102 + 52 − 122
19
=−
=⇒ x ≈ 101.0
2 (10) (5)
100
cos A =
102 + 72 − 62
113
=
=⇒ A ≈ 36.2◦
2 (10) (7)
140
cos B =
102 + 62 − 72
29
=
=⇒ B ≈ 43.5◦
2 (10) (6)
40
cos C =
72 + 62 − 102
5
=−
=⇒ C ≈ 100.3◦
2 (7) (6)
28
6.
43
12
A
B
3
3
D
C
12
In the above diagram of a parallelogram with side lengths of 3 and 12, let
d1 be the length of the diagonal AC and let d2 be the length of the diagonal
BD. By the Law of Cosines, we know that
d21 = 32 + 122 − 2(3)(12) cos B
and
d22 = 32 + 122 − 2(3)(12) cos A.
However, since adjacent angles in a parallelogram are supplementary, cos B =
− cos A. Therefore,
d21 + d22 = 32 + 122 − 2(3)(12) cos B + 32 + 122 − 2(3)(12) cos A
= 32 + 122 + 2(3)(12) cos A + 32 + 122 − 2(3)(12) cos A
= 32 + 122 + 32 + 122
= 306
Thus, the answer does not depend on the specific shape of the parallelogram.
Only the side lengths are pertinent.
7.
a
A
B
b
D
b
C
a
We can generalize the argument from the previous exercise to an arbitrary
parallelogram. As before, let d1 be the length of the diagonal AC and let d2
be the length of the diagonal BD. Then,
d21 + d22 = a2 + b2 − 2ab cos B + a2 + b2 − 2ab cos A
= a2 + b2 + 2ab cos A + a2 + b2 − 2ab cos A
= a2 + b2 + a2 + b2 ,
44
which shows that the sum of the squares of the diagonals of a parallelogram
is equal to the sum of the squares of the four sides.
8.
B
A
M
D
C
We can create a parallelogram from triangle ABC by creating a copy of
the triangle and reflecting it through the point M . This allows us to apply
the result from the previous exercise about the sum of the squares of the
diagonals.
2
AD2 + BC 2 = AB 2 + AC 2 + BD2 + CD2 =⇒ (2AM ) + BC 2 = AB 2 + AC 2 + AC 2 + AB 2
=⇒ 4AM 2 + BC 2 = 2AB 2 + 2AC 2
=⇒ 4AM 2 = 2AB 2 + 2AC 2 − BC 2
9. We can get a formula for the length of each median in a triangle by applying
cyclic substitutions to the result we derived in the previous exercise.
4AM 2 = 2AB 2 + 2AC 2 − BC 2
4BM 2 = 2BC 2 + 2AB 2 − AC 2
4CM 2 = 2AC 2 + 2BC 2 − AB 2
Adding these three equations together gives
4AM 2 + 4BM 2 + 4CM 2 = 3AB 2 + 3AC 2 + 3BC 2 .
Factoring out 3 on the right-hand side of the equation and dividing both
sides by 4 gives the result:
AM 2 + BM 2 + CM 2 =
3
AB 2 + AC 2 + BC 2 .
4
10.
11.
√
1
c = 1 + 4 − 2 (1) (4) cos 60 = 17 − 8
= 13 =⇒ c = 13
2
2
2
2
◦
45
12.
c2 = a2 + b2 − 2ab cos 60◦ = a2 + b2 − 2ab
1
= a2 + b2 − ab
2
Since 120◦ is the supplement of 60◦ , the analogous result is that c2 = a2 +
b2 + ab.
13.
B
X
A
C
Let A, B, and C be the positions of the three riders at some time t, where t
is the number of elapsed hours since departure. Then AX = 60t, BX = 40t,
and CX = 20t. Applying the result from the previous exercise concerning
the Law of Cosines for triangles with a 120◦ angle,
q
2
2
AB = (60t) + (40t) + (60t) (40t) ≈ 87.2t
q
2
2
BC = (40t) + (20t) + (40t) (20t) ≈ 52.9t
q
2
2
AC = (60t) + (20t) + (60t) (20t) ≈ 72.1t
Therefore, after 1 hour, riders A and B are 87.2 miles apart, riders B and
C are 52.9 miles apart, and riders A and C are 72.1 miles apart. After 2
hours, riders A and B are 174.4 miles apart, riders B and C are 105.8 miles
apart, and riders A and C are 144.2 miles apart.
Page 83
1.
45-45-90 triangle:
46
2. The triangles which can be placed on their mirror images without reflection
are the isosceles triangles.
3.
Rectangle:
Rhombus:
Isosceles Trapezoid:
Page 84
1.
C
γ
a
B
β
c
2S
bc
b
=
=
ac
ac
a
2S
bc
c
sin γ =
=
=
ab
ab
a
sin β =
47
b
A
2.
c2 = a2 + b2 − 2ab cos γ
2S
= a2 + b2 − 2
cos γ
sin γ
= a2 + b2 − 4S cot γ
Page 85
1.
b2 = a2 + c2 − 2ac cos β
a2 = b2 + c2 − 2bc cos α
2. These three quantities are all equal to twice the circumradius of the triangle.
Page 86
1. (a) The dimensions of the left-hand side and right-hand side of the equation
are both [L]/[∅], where we use ∅ to represent a dimensionless quantity.
(b) The dimensions of the left-hand side are [L]/[L], which is a dimensionless quantity. The dimensions of the right-hand side are [∅]/[∅], which
is also a dimensionless quantity. Thus, the dimensions in this equation
are consistent.
(c) The dimensions of the left-hand side are [L]2 , and the dimensions of the
right-hand side are [L] · [L] · [∅], which is equivalent to the dimensions
of [L]2 .
(d) The dimensions of the left-hand side are [L]2 . The dimensions of each
term on the right-hand side are [L]2 (recall that cos γ is dimensionless),
so the overall dimensions of the right-hand side are also [L]2 .
Page 88
1. The semiperimeter of this triangle is (3 + 4 + 5)/2 = 6.
p
S = 6 (6 − 3) (6 − 4) (6 − 5) = 6
2. The semiperimeter of this triangle is (5 + 12 + 13)/2 = 15.
p
S = 15 (15 − 5) (15 − 12) (15 − 13) = 30
3. The semiperimeter of an equilateral triangle with side length l is 3l/2
r
r
√
3l l l l
3 4
3
S=
· · · =
l = l2
2 2 2 2
16
4
48
4. Each angle in an equilateral triangle is 60◦ .
S=
√
1 2
3
l sin 60◦ = l2
2
4
5. The semiperimeter of a 13-14-15 triangle is (13 + 14 + 15)/2 = 21.
p
S = 21 (21 − 13) (21 − 14) (21 − 15) = 84
The semiperimeter of a 25-39-56 triangle is (25 + 39 + 56)/2 = 60.
p
S = 60 (60 − 25) (60 − 39) (60 − 56) = 420
The semiperimeter of a 25-39-16 triangle is (25 + 39 + 16)/2 = 40.
p
S = 40 (40 − 25) (40 − 39) (40 − 16) = 120
Chapter 4: Angles and Rotations
Page 93
160◦
A
1. (a)
A
190◦
(b)
49
400◦
A
(c)
A
600◦
(d)
1200◦
A
(e)
A
−70◦
(f)
50
A
−400◦
(g)
360◦
A
(h)
−270◦
A
(i)
Page 98
30◦
1. (a)
Because 390◦ = 360◦ + 30◦ , a point rotated through an angle of 390◦
would end up at the same location as a point rotated through an angle
51
of 30◦ . Therefore,
sin 390◦ = sin 30◦ =
1
.
2
120◦
60
◦
(b)
Because 3720◦ = 10 · 360◦ + 120◦ , a point rotated through an angle of
3720◦ would end up at the same location as a point rotated through an
angle of 120◦ . Therefore,
1
cos 3720◦ = cos 120◦ = − cos 60◦ = − .
2
45◦
(c)
Because 1845◦ = 5 · 360◦ + 45◦ , a point rotated through an angle of
1845◦ would end up at the same location as a point rotated through an
angle of 45◦ . Therefore,
√
sin 45◦
2/2
◦
◦
= 1.
tan 1845 = tan 45 =
=√
cos 45◦
2/2
52
45◦
315◦
(d)
Because 315◦ = 360◦ − 45◦ , a point rotated through an angle of 315◦
would end up at the same location as a point rotated through an angle
of −45◦ . Therefore,
√
2
◦
◦
◦
sin 315 = sin (−45 ) = − sin 45 = −
2
60◦
(e)
Because 420◦ = 360◦ + 60◦ , a point rotated through an angle of 420◦
would end up at the same location as a point rotated through an angle
of 60◦ . Therefore,
cot 420◦ = cot 60◦ =
−30◦
(f)
53
cos 60◦
1/2
1
=√
=√ .
sin 60◦
3/2
3
tan (−30◦ ) =
1/2
− sin 30◦
1
sin (−30◦ )
= −√
=
= −√
◦
◦
cos (−30 )
cos 30
3/2
3
2. (a)
tan 360◦ = tan 0◦ =
0
sin 0◦
= =0
◦
cos 0
1
(b)
sin 180◦ = sin 0◦ = 0
(c)
cos 180◦ = − cos 0◦ = −1
(d)
cot 90◦ =
(e)
cot 360◦ = cot 0◦ =
0
cos 90◦
= =0
sin 90◦
1
cos 0◦
1
=
=⇒ cot 360◦ is undefined.
◦
sin 0
0
(f)
tan (−270◦ ) = tan 90◦ =
sin 90◦
1
=
=⇒ tan (−270◦ ) is undefined.
cos 90◦
0
Page 100
1. 400◦ = 360◦ + 40◦ , so P will lie in the first quadrant.
3600◦ = 10 · 360◦ + 0◦ , so P will lie in the first quadrant (on the positive
x-axis).
1845◦ = 5 · 360◦ + 45◦ , so P will lie in the first quadrant.
−30◦ = −360◦ + 330◦ , so P will lie in the fourth quadrant.
−359◦ = −360◦ + 1◦ , so P will lie in the first quadrant.
2.
sin 30◦
sin (−30◦ )
−1/2
2/2
sin (−135◦ )
√
− 2/2
sin 210◦
−1/2
sin (−210◦ )
1/2
sin 300◦
√
− 3/2
sin (−300◦ )
sin 390◦
1/2
sin (−390◦ )
−1/2
sin (−480◦ )
√
− 3/2
sin 135◦
sin 480◦
1/2
√
√
3/2
54
√
3/2
From the table, we can see that sin (−α) = − sin α.
3. (a)
sin α = 0 =⇒ α = 180◦
(b)
cos α = 0 =⇒ α = 90◦ , 270◦
(c)
sin α = 1 =⇒ α = 90◦
(d) cos α = 1 is true for α = 0◦ and α = 360◦ , but these are not in the
interval 0 < α < 360◦ .
(e)
sin α = −1 =⇒ α = 270◦
(f)
cos α =
1
=⇒ α = 60◦ , 300◦
2
(g)
sin α = −
1
=⇒ α = 210◦ , 330◦
2
(h)
√
1
2
sin α =
=⇒ sin α = ±
=⇒ α = 45◦ , 135◦ , 225◦ , 315◦
2
2
(i) This equation has no solutions because the square of a real number
cannot be negative.
2
4. (a) If sin α = 5/13, then α can lie in either the first or second quadrant.
cos α = 12/13 when α is in the first quadrant, and cos α = −12/13
when α is in the second quadrant.
(b) If sin α = −5/13, then α can lie in either the third or fourth quadrant.
cos α = 12/13 when α is in the fourth quadrant, and cos α = −12/13
when α is in the third quadrant.
5.
P = (a, b)
θ
55
Let P = (a, b) be a point on the coordinate plane. Since a2 + b2 = 1, P
lies on a circle of radius 1 centered at the origin (the unit circle). Consider
an angle θ between the positive x-axis and the line segment connecting P
to the origin. By the extended definitions of the sine and cosine functions,
sin θ = b and cos θ = a.
Page 102
1. (a) Even
6
2
f (−x) = (−x) − (−x) + 7 = x6 − x2 + 7 = f (x)
(b) Odd
3
f (−x) = (−x) − sin (−x) = −x3 + sin x = −f (x)
(c) Neither
f (−x) =
1
−x + 1
(d) Even
f (−x) = sec (−x) =
1
1
=
= sec x = f (x)
cos (−x)
cos x
(e) Odd
f (−x) = csc (−x) =
1
1
=−
= − csc x = −f (x)
sin (−x)
sin (x)
(f) Odd
f (−x) = 2 sin (−x) cos (−x) = −2 sin (x) cos (x) = −f (x)
(g) Even
2
f (−x) = sin2 (−x) = (− sin x) = sin2 x = f (x)
(h) Even
2
f (−x) = cos2 (−x) = (cos x) = cos2 x = f (x)
(i) Even
f (−x) = sin2 (−x) + cos2 (−x) = sin2 x + cos2 x = f (x)
2.
1
1
[f (−x) + f (− (−x))] = [f (−x) + f (x)] = g (x) ,
2
2
g (−x) =
so g (x) is even.
h (−x) =
1
1
[f (−x) − f (− (−x))] = [f (−x) − f (x)] = −h (x) ,
2
2
56
so h (x) is odd.
For any function f (x),
f (x) =
1
1
[f (x) + f (−x)] + [f (x) − f (−x)] = g (x) + h (x) ,
2
2
so f (x) can be written as the sum of an even function and an odd function.
3. Following the notation of the previous question, we denote the even part of
f (x) as g (x) and the odd part of f (x) as h (x).
(a) Since we know cos x is an even function and sin x is an odd function,
g (x) = cos x, h (x) = sin x
(b) Since a polynomial is an even function when all variables are raised to
even exponents and an odd function when all variables are raised to
odd exponents,
g (x) = x2 + 1, h (x) = x3 + x
(c) Applying the result from the previous exercise,
g (x) =
1 x
1 x
2 + 2−x , h (x) =
2 − 2−x
2
2
(d) Applying the result from the previous exercise,
1 1 − sin x 1 + sin x
1 + sin2 x
1 2 + 2 sin2 x
g (x) =
+
=
=
2
2 1 + sin x 1 − sin x
2 1 − sin x
1 − sin2 x
1 1 − sin x 1 + sin x
1
−4 sin x
−2 sin x
−
h (x) =
=
=
2 1 + sin x 1 − sin x
2 1 − sin2 x
1 − sin2 x
(e) Applying the result from the previous exercise,
1
1
1
1
4
2
g (x) =
+
=
=
2 x + 2 −x + 2
2 4 − x2
4 − x2
h (x) =
1
2
1
1
−
x + 2 −x + 2
=
1
2
−2x
4 − x2
=−
x
4 − x2
Chapter 5: Radian Measure
Page 107
1. 180 degrees is equal to π radians. 90 degrees is equal to π/2 radians.
2.
π r = 180◦ =⇒ 1r = (180/π)◦ =⇒ 2r = (360/π)◦ ≈ 114.6◦
57
3. Since a full rotation is 2π radians, 1/4 of a full rotation will be π/2 radians.
4. Since 45 degrees is 1/8 of a full rotation, 1/8 of a full rotation will be π/4
radians.
5. Filled table below
Degree Measure
90
180
270
360
90
180
270
360
Radian Measure
π/2
π
3π/2
2π
π/2
π
3π/2
2π
Degree Measure
0
30
72
120
135
30
36
45
60
120
126
Radian Measure
0
π/6
2π/5
2π/3
3π/4
π/6
π/5
π/4
π/3
2π/3
7π/10
Degree Measure
198
210
216
225
240
198
200
210
216
225
240
Radian Measure
11π/10
7π/6
6π/5
5π/4
4π/3
11π/10
10π/9
7π/6
6π/5
5π/4
4π/3
6. Filled tables below
58
7. Since 360 degrees is equal to 2π radians, then 1 degree is equal to 2π/360,
or π/180, radians.
8. (a) sin(1r ) ≈ 0.8415
(b) sin(1◦ ) ≈ 0.0175
9. Filled table below
α (in radian)
sin α
π/6
1/2
√
3/2
π/3
π/2
√
2π/3
1/2
1
3/2
5π/4
−1/2
√
− 2/2
3π/2
−1
11π/6
−1/2
7π/6
cos α
√
3/2
0
−1/2
√
− 3/2
√
− 2/2
√
0
3/2
10. With an angle of 2 radians and a radius of 1, the length of the arc is 2×1 = 2.
With an angle of 3 radians and a radius of 1, the length of the arc is 3×1 = 3.
With an angle of π radians and a radius of 1, the length of the arc is π×1 = π.
11. With an angle of 2 radians and a radius of 3, the length of the arc is 2×3 = 6.
With an angle of 3 radians and a radius of 3, the length of the arc is 3×3 = 9.
With an angle of π radians and a radius of 3, the length of the arc is π × 3 =
3π.
12. Using the fact that sine and cosine are cofunctions (see Section 4 of Chapter
1),
π π
π
7π
sin = cos
−
= cos
,
9
2
9
18
so α = 7π/18.
13. Again, applying the properties of cofunctions, we know that sin α = cos (π/2 − α).
14. A complete rotation around a circle corresponds to an angle of 2π radians,
so each of the six sectors in the diagram is an angle of 2π/6, which is a bit
more than 1 radian (since 2π > 6).
However, 2π/6 radians is equal to 60 degrees, so if 2π/6 is a bit more than
1, then 1 radian is less than 60 degrees
Geometric solution:
59
C
B
A
D
O
E
F
In the diagram above, we place the six points A through F so that they are
equally spaced across the circle O. This means that adjacent points on the
circle are separated by 360◦ /6 = 60◦ . Focusing on △AOB, we notice that
this triangle is isosceles because AO and BO are both equal to the radius
of the circle. This implies that ̸ OAB is congruent to ̸ OBA. Because the
three angles in a triangle total to 180◦ , we have that
m̸ AOB = m̸ OAB = m̸ OBA = 60◦ .
In other words, △AOB is an equilateral triangle. This means that the side
AB is equal to the radius of O. Therefore, the arc that is intercepted by the
chord AB must be longer than the radius of O because the shortest path
between two points is a straight line. This implies that the central angle
̸ AOB is greater than 1 radian. Since ̸ AOB is a 60◦ angle, we have shown
that 60◦ is greater than 1 radian.
Page 111
1. Since a circle of radius 1 traveling 1 foot corresponds to a rotation of 1
radian, a circle of radius 1 traveling 5 feet down a road corresponds to a
rotation of 5 radians.
2. From the previous question, the circle has rotated 5 radians, which is equal
to 5 · 180◦ /π ≈ 286.48◦ (two decimal places).
3. Since a circle of radius 1 rotates 1 radian to travel 1 foot, if it rotates 4
radians, then it has traveled 4 feet.
4. Since 120◦ = 2π/3 radians, the wheel has traveled 2π/3 feet down the road.
5. A sector of radius 1 with angle α radians has an arc length of 1 × α, so:
An angle of 1/2 radian has an arc length of 1/2.
60
An angle of π/2 radian has an arc length of π/2.
An angle of α radian has an arc length of α.
6. Since 2π radians equal to 360◦ then
720◦ = 4π radians
1440◦ = 8π radians
3600◦ = 20π radians
15120◦ = 84π radians
12π radians = 2160◦
12π radians = 2160◦
15π radians = 2700◦
100π radians = 18000◦
7. A sector of radius 3 with angle α radians has an arc length of 3α, so:
An angle of 1/2 radian has an arc length of 3/2.
An angle of π/2 radian has an arc length of 3π/2.
An angle of α radian has an arc length of 3α.
8. A sector of radius 3 and angle α has arc length of 3α.
An angle of 1.5 radians has an arc length of 4.5.
9. A sector of radius 5 and angle α has arc length of 5α.
An angle of 80 degrees is equal to 4π/9 radians. Thereforem its arc length
is 5 × 4π/9 = 20π/9.
10. A sector of radius 2 and angle α has an arc length of 2α. This means that
for an arc length of α, its angle would be α/2. If the arc length is 3, then
its central angle is 3/2 radians.
11. A sector of radius 6 and angle α has an arc length of 6α. This means that
for arc length of α, its angle would be α/6. If the arc length is 2, then its
◦
central angle is 1/3 radians, which is equal to (60/π) ≈ 19.1◦ .
12. A circle of radius 7 units will travel 7 units with a rotation of 1 radian. That
is, each unit of travel requires a rotation of 1/7 radians. Therefore, 20 units
of travel require 20/7 radians of rotation.
13. A circle of radius 8 units will travel 8 units with a rotation of 1 radian. Since
150 degrees is equal to 5π/6 radians, the circle has rolled 8 × 5π/6 = 20π/3
units.
14. In twelve hours, the hour hand makes a complete rotation around the watch
face (e.g., 12:00 AM to 12:00 PM), so in one hour, the hour hand makes
1/12 of a full rotation. Because the hands of a clock rotate in a clockwise
61
sense, the angle of rotation will be negative. Thus, the angle through which
the hour hand rotates in one hour is
−
1
π
· 2π = − .
12
6
15. In one hour, the minute hand makes one complete rotation around the watch
face, so it rotates through an angle of −2π. In the same time, the second
hand makes 60 complete rotations around the watch face since it completes
one rotation in one minute. Therefore, the second hand rotates through an
angle of −60 · 2π = −120π.
16. Because the hands of a watch travel clockwise, Joe’s angle should be negative
since counterclockwise rotations are positive by convention.
17.
1000 in.
≈ 159.2
2π in.
This trip consists of about 159 full rotations of the hour hand. Multiplying
by 12 gives the number of hours the trip takes since a complete rotation of
the hour hand takes 12 hours.
1000
· 12 hours ≈ 1910 hours
2π
18. The angular speed of the hour hand on a pocket watch and the hour hand on
Big Ben should be the same because in one hour, the hour hand travels the
same angular distance on both clocks. Since in both this exercise and the
previous exercise the clocks are traveling through an angle of 1000 radians,
both trips should take the same length of time: 1910 hours.
19. The wheel turns 2π radians before the spoke returns to the same position.
After the wheel turns π radians, the spoke will go from pointing vertically
downwards to pointing vertically upwards.
20. (a) A wheel of radius one meter will roll 2π meters each revolution, so the
blue marks will be 2π meters apart.
(b) The wheel will make a full revolution in this time, so it has rolled
through an angle of 2π radians.
(c) Not more than once. Suppose we have a red mark and a blue mark
coinciding. When they next coincide, the distance between the two
coincidences will be equal to 3n (measuring using the gaps between red
dots) or 2πm (measuring using the gaps between blue dots), where n
and m are positive integers. Because these are two equivalent ways to
measure the same distance, we can say that 3n = 2πm. Solving for π,
3n
we find that π = 2m
. However, this is not possible as π is irrational,
so it cannot be expressed as the ratio of integers. This contradiction
means that if a red dot and blue dot do coincide, then they cannot
coincide again.
62
(d)
Let’s consider how the red marks hit the wheel as it rolls. Hitting three
red marks in a row corresponds to traveling 6 meters (count the space in
between the marks). Because the radius of the wheel is 1 meter, when
the wheel hits a triple of red marks, the last mark will hit the wheel in
a position that is 6 radians ahead of the position where the first mark
was hit. Alternatively, we can say that the last mark hits the wheel in
a position that is 2π − 6 radians behind the position of the first mark.
In the above diagram, the red marks are separated by 2π − 6 radians,
so they represent the positions of every other red mark that hits the
wheel in some period of time. Notice that the position of the red marks
on the wheel implies that the blue mark cannot be more than 2π − 6
radians away from a red mark. This means that no matter how far the
wheel travels, the blue mark will eventually be placed within 2π − 6
meters of a red mark.
Another perspective:
Compare the location of marks by plotting the marks’ locations
5
10
15
20
25
30
35
40
Note the distance between blue and red markers will be different but
the closest they come together are
Red Blue Difference (approx)
6
2π
0.283
12
4π
-0.566
18
6π
-0.850
24
8π
-1.133
30
10π
-1.416
(e) We assume that the blue dot and red dot coincide to begin with. 100
rotations corresponds to the wheel travelling 200π meters. Notice that
63
200π ≈ 209 · 3 + 1.3. Because 14 · 3 < 1.3 < 12 · 3, the blue dot will be
between two pink dots: one of the pink dots will be at the midpoint of
two red dots, and the second pink dot will be directly behind the first.
Page 114
1. The calculator was in radian mode, since sin(1r ) ≈ 0.8415, while sin(1◦ ) ≈
0.0175.
2.
x (in radians)
sin x
Difference (approx.)
0.2
0.19867
1.33 × 10−3
0.15
0.14944
5.62 × 10−4
0.05
0.04998
2.08 × 10−5
In all cases, x > sin x.
3. In the below table, the difference being calculated is x − x3 /6 − sin x.
x (in radians)
sin x
x − x3 /6
Difference (approx.)
0.2
0.1986693
0.1986667
−2.66 × 10−6
0.15
0.1494381
0.1494375
−6.32 × 10−7
0.05
0.0499792
0.0499792
−2.60 × 10−9
4. sin 10◦ ≈ 0.174, and 10/60 ≈ 0.167, so the error in the approximation is
0.007.
5. (a)
0.13
1
1
599
=
−
=
6
10 6000
6000
Error: sin 0.1 is greater than 599/6000 by 8.33 × 10−8 .
sin 0.1 ≈ 0.1 −
(b)
π
3.14
157
π
≈
≈
=
1800
1800
1800
90000
Error: sin 0.1◦ is greater than 157/90000 by 8.84 × 10−7 .
sin 0.1◦ = sin
6. (a) sin 1000◦ ≈ −0.9848
(b) sin 1000r ≈ 0.8269
7. (a) sin(sin 1000r ) ≈ 0.7358
(b) sin 3.14r ≈ 0.00159
8. Let ε = π/2 − 1.5707 ≈ 9.6 × 10−5 .
π
cos 1.5707 = cos
− ε = sin ε < ε < 0.0001
2
64
Page 116
1. Using the graph below
y
1
−3π/7
−1π
x
0
7π/5
1π
2π
−1
(a) sin 7π/5 is negative because π < 7π/5 < 2π.
Estimated x ≈ −1 (actual is −0.9511)
(b) sin −3π/7 is negative because −π < −3π/7 < 0.
Estimated x ≈ −1 (actual is −0.9749)
(c) By symmetry, sin (π − π/6) = sin 5π/6 = 1/2. By periodicity, sin (2π + π/6) =
sin 13π/6 = 1/2. In general, sin x = 1/2 when x = π/6 + 2πn or
x = 5π/6 + 2πn, where n is an integer.
(d) Draw a line at y = sin(π/12) ≈ 0.26 and then estimate the location of
the intersections with the sine wave.
x = π/12, 11π/12, 25π/12, . . .
≈ 0.26, 2.88, 6.54, . . .
y
1
0
−1π
0.26
2.88
1π
x
2π
−1
(e) Draw a line at y = 0.8 and then estimate the location of the intersections with the sine wave. x ≈ 0.93, 2.21, 7.21, . . .
y
1
0
−1π
0.93
−1
65
2.21
x
1π
2π
√
√
Alternatively, notice that sin π/4 = 2/2 ≈ 0.70 and sin π/3 = 3/2 ≈
0.85. Taking a weighted average, we can estimate that the sine of
1 π 2 π
11π
· + · =
3 4
3 3
36
is approximately 0.8. Then, based on the properties of the sine function,
25π/36 and 83π/36 should also have sines of approximately 0.8.
Note: For parts (a) and (b) of the above exercise, we can use the approximation cos x ≈ 1 − 12 x2 to get better estimates. This approximation can be
derived from the approximation sin x ≈ x using the identity sin2 x + cos2 x = 1.
Let us suppose that x is a positive angle close to
√ zero. Replacing sin x with
x in the aforementioned identity, we get cos x ≈ 1 − x2 . Using the binomial
α
approximation, (1 + z) ≈ 1 + αz, we arrive at the desired approximation for
cos x:
p
1/2
1
cos x ≈ 1 − x2 = 1 + −x2
≈ 1 − x2 .
2
For part (a), we get the estimate
sin
π
7π
1
2π
π
π
1 π 2
−1 ≈
= − sin
= − sin
−
= − cos
≈
−1 = −0.95.
5
5
2
10
10
2 10
20
For part (b), we get the estimate
sin
π
−3π
3π
π
1 π 2
1
π
= − sin
= − sin
−
≈
−1 ≈
−1 = −0.975.
= − cos
7
7
2
14
14
2 14
40
Both of these estimates improve upon the first-order approximation of −1.
Page 120
1. Results are written to 5 decimal places.
α (radians)
α (degrees)
sin α
1
57.29578
0.84147
0.5
28.64789
0.47943
0.2
11.45916
0.19867
0.1
5.72958
0.09983
0.01
0.57296
0.01000
0.02
1.14592
0.02000
0.001
0.05730
0.00100
0.002
0.11459
0.00200
0.005
0.28648
0.00500
66
2. Since sin x ≈ x for angles close to zero (when x is in radians), sin 0.00123456 ≈
0.00123456. Since x > sin x for positive angles, we know this is an overestimate. The calculator result is 0.0012345597, so our estimate is accurate to
7 decimal places.
3. (a) Results are written to 5 decimal places.
α
α − α3 /6
sin α
1
0.83333
0.84147
0.5
0.47917
0.47943
0.2
0.19867
0.19867
0.1
0.09983
0.09983
0.05
0.04998
0.04998
0.01
0.00999
0.00999
0.001
0.00099
0.00099
(b) Recall that we can multiply the degree measure of an angle by π/180
to get the radian measure, so we simply replace α by πD/180 in our
approximation. This gives
3
πD
1 πD
π3 D3
πD
sin D ≈
−
−
=
180 6 180
180 34992000
For D = 1◦ , the above approximation gives sin 1◦ ≈ 0.017421, while the
calculator result is 0.017452. This is an underestimate by 3.1 × 10−5 .
4. Using the fact that α = πD/180 from the previous part, the largest possible
error for an angle measured in degrees is given by
5
1
πD
π5 D5
=
.
120 180
22674816000000
5.
x5 /120 < 0.001
x5 < 0.12
√
5
x < 0.12
x < 0.65438
6. The signs are alternating between terms, so the third term should be positive.
The exponent of x increases by two in each successive term, so the third
term should have an exponent of 5.
The denominator of first term is 1! and the denominator of second term is
3!. Therefore the denominator of the fifth term should be 5! = 120.
67
Therefore, the third term is
+
x5
.
120
Hint: This is the Taylor Series for sin x.
Note: ! is used to denote the factorial function, where the factorial of a
positive integer n is the product of all positive integers less than or equal to
n.
7. The aliens appear to take clockwise angles to be positive, which is the opposite of the convention we typically use. Furthermore, it is not clear that
the aliens even use counterclockwise angles since all of the examples only
feature clockwise angles. This could mean that the aliens only work with
non-negative (or in our system, non-positive) angles.
The aliens do consider angles corresponding to rotations greater than a full
rotation, as we do. φ seems to be used to denote one complete rotation.
Comparing to our system, φ = −2π radians.
8. It is easier to rewrite each of the radian angle measurements as multiples of
π since the first quadrant contains angles between 0 and 0.5π, the second
quadrant contains angles between 0.5π and π, etc.
Angle in radians
Angle in terms of π
Quadrant
1
0.318π
1
2
0.636π
2
3
0.954π
2
4
1.273π
3
5
1.591π
4
6
1.909π
4
1000 radians is approximately equal to 318.31π, which represents 159 full
rotations with a remainder of 0.31π radians. Thus an angle of 1000 radians
lies in the first quadrant.
1000 degrees corresponds to 1000/360 ≈ 2.78 rotations, which represents 2
whole rotations and a further 0.78 of a rotation, which is slightly more than
a 3/4 rotation. Thus, an angle of 1000 degrees lies in the fourth quadrant.
9. Approximately 1/4 of the angles lie in each quadrant. See the solution below
the question in the textbook.
Chapter 6: The Addition Formulas
Page 123
1.
68
α
β
sin α
60◦
30◦
π/4
π/4
π/6
π/3
√
√
sin β
3/2
sin α + sin β
√
1/2
√
2/2
√
1/2
√
2/2
2+
3/2
3 + 1 /2
sin 90◦ = 1
√ √
2 /2 = 2
sin π/2 = 1
1+
√ 3 /2
2. For these values of α and β, sin α and sin β are both at least 1/2. Furthermore, at least one of sin α and sin β is strictly greater than 1/2. Therefore,
1 1
+ = 1 = sin (α + β) .
2 2
sin α + sin β >
3. (a)
√
3 1
+
2
2
◦
◦
◦
sin (60 + 30 ) = sin 90 = 1
◦
◦
sin 60 + sin 30 =
This identity is not correct.
(b)
1
sin (60◦ − 30◦ ) = sin 30◦ =
2
√
3 1
−
sin 60◦ − sin 30◦ =
2
2
This identity is not correct.
(c)
√ !2 2
1
3 1
1
3
−
= − =
2
2
4 4
2
sin2 60◦ − sin2 30◦ =
sin (60◦ + 30◦ ) sin (60◦ − 30◦ ) = sin 90◦ · sin 30◦ = 1 ·
1
1
=
2
2
This identity is correct for the given angles.
4. See Chapter 2, Section 12 and the Appendix of Chapter 2 to review some of
the geometry used in this solution.
(a) Because ̸ ABC is subtended by the diameter AC, ̸ ABC is a right
angle and △ABC is a right triangle (this fact is known as Thales’s
Theorem). Therefore, sin α is equal to the length of the opposite side
(BC) divided by the length of the hypotenuse (AC). AC is a diameter
of the circle, so it has length 1. Thus, we have that sin α is simply equal
to BC.
A similar argument shows that △ADC is a right triangle with hypotenuse AC of length 1, which implies that sin β = DC
69
sin (α + β)
sin π/2 = 1
(b) Recall that chords of congruent circles which subtend equal angles are
themselves equal. This implies that BC in the diagram of part (a) is
equal to BC in the diagram of part (b) because in both diagrams, the
chord BC subtends an angle of measure α. Similarly, DC is the same
in both diagrams because in both diagrams, the chord DC subtends an
angle of measure β. Therefore, BC is still equal to sin α, and DC is
still equal to sin β.
(c) From part (b) above, we can conclude that a chord which subtends an
inscribed angle with measure α in a circle with diameter 1 has length
sin α. Thus, we draw BD, the chord which subtends ̸ BAD in both
figures and which consequently has length sin (α + β).
Note that the above reasoning implies that the sine of an angle with measure
less than 180◦ cannot exceed 1 since the diameter is the longest chord in a
circle.
5. Recall that the sine of any angle is at most 1. Therefore,
sin 105◦ ≤ 1 =
1 1
+ < sin 45◦ + sin 60◦ ,
2 2
which shows that sin 105◦ cannot equal sin 45◦ + sin 60◦ .
Page 125
1. Addition formula for sine:
sin (60◦ + 30◦ ) = sin 60◦ cos 30◦ + cos 60◦ sin 30◦
√ √
3
3 1 1
=
·
+ ·
2
2
2 2
3 1
= +
4 4
=1
= sin 90◦
Addition formula for cosine:
cos (60◦ + 30◦ ) = cos 60◦ cos 30◦ − sin 60◦ sin 30◦
√
√
1
3
3 1
= ·
−
·
2√ 2 √ 2 2
3
3
=
−
4
4
=0
= cos 90◦
70
Difference formula for sine:
sin (60◦ − 30◦ ) = sin 60◦ cos 30◦ − cos 60◦ sin 30◦
√ √
3
3 1 1
=
·
− ·
2
2
2 2
3 1
= −
4 4
1
=
2
= sin 30◦
Difference formula for cosine:
cos (60◦ − 30◦ ) = cos 60◦ cos 30◦ + sin 60◦ sin 30◦
√
√
1
3
3 1
= ·
+
·
2√ 2 √ 2 2
3
3
=
+
4
4
√
3
=
2
= cos 30◦
2. Addition formula for sine (α = 0):
sin (0 + β) = sin 0 cos β + cos 0 sin β
= 0 · cos β + 1 · sin β
= sin β
Addition formula for cosine (α = 0):
cos (0 + β) = cos 0 cos β − sin 0 sin β
= 1 · cos β − 0 · sin β
= cos β
Difference formula for sine (α = 0):
sin (0 − β) = sin 0 cos β − cos 0 sin β
= 0 · cos β − 1 · sin β
= − sin β
Notice that this demonstrates that the sine function is odd.
Difference formula for cosine (α = 0):
cos (0 − β) = cos 0 cos β + sin 0 sin β
= 1 · cos β + 0 · sin β
= cos β
71
Notice that this demonstrates that the cosine function is even.
Addition formula for sine (β = 0):
sin (α + 0) = sin α cos 0 + cos α sin 0
= sin α · 1 + cos α · 0
= sin α
Addition formula for cosine (β = 0):
cos (α + 0) = cos α cos 0 − sin α sin 0
= cos α · 1 − sin α · 0
= cos α
Difference formula for sine (β = 0):
sin (α − 0) = sin α cos 0 − cos α sin 0
= sin α · 1 − cos α · 0
= sin α
Difference formula for cosine (β = 0):
cos (α + 0) = cos α cos 0 + sin α sin 0
= cos α · 1 + sin α · 0
= cos α
3. Following the hint, we notice that in a right triangle, the side opposite one
of the acute angles is the side adjacent to the other acute angle. Thus, if
α + β = π/2, then sin α = cos β and sin β = cos α (see also Chapter 1,
Section 4).
sin (α + β) = sin α cos β + cos α sin β
= sin α sin α + cos α cos α
= sin2 α + cos2 α
=1
4. Addition formula for sine:
π π
π
π
π
π
sin
+
= sin cos + cos sin
4
4
4
4
√ 4√ 4 √ √
2
2
2
2
=
·
+
·
2
2
2
2
1 1
= +
2 2
=1
π
= sin
2
72
Addition formula for cosine:
π π
π
π
π
π
= cos cos − sin sin
cos
+
4
4
4
4
4
4
√ √
√ √
2
2
2
2
=
·
−
·
2
2
2
2
1 1
= −
2 2
=0
π
= cos
2
2
5. Recall that (A ± B) = A2 ± 2AB + B 2 .
2
2
(sin α cos β + cos α sin β) + (cos α cos β − sin α sin β)
= sin2 α cos2 β + 2 sin α cos β cos α sin β + cos2 α sin2 β+
cos2 α cos2 β − 2 cos α cos β sin α sin β + sin2 α sin2 β
= sin2 α cos2 β + cos2 α sin2 β + cos2 α cos2 β + sin2 α sin2 β
= sin2 α cos2 β + sin2 β + cos2 α sin2 β + cos2 β
= sin2 α · 1 + cos2 α · 1
=1
6. After expanding using the identity (A + B) (A − B) = A2 − B 2 , we cleverly
“add by zero” to get the desired result.
(sin α cos β + cos α sin β) (sin α cos β − cos α sin β)
= sin2 α cos2 β − cos2 α sin2 β
= sin2 α cos2 β + sin2 α sin2 β − sin2 α sin2 β − cos2 α sin2 β
= sin2 α cos2 β + sin2 β − sin2 β sin2 α + cos2 α
= sin2 α · 1 − sin2 β · 1
= sin2 α − sin2 β
Page 129
1.
2.
73
Page 131
1.
sin (30◦ + 30◦ ) = sin 30◦ cos 30◦ + cos 30◦ sin 30◦
√
√
1
3
3 1
= ·
+
·
2 2 √ 2 2
√
3
3
=
+
4
4
√
3
=
2
= sin 60◦
cos (30◦ + 30◦ ) = cos 30◦ cos 30◦ − sin 30◦ sin 30◦
√ √
3
3 1 1
·
− ·
=
2
2
2 2
3 1
= −
4 4
1
=
2
= cos 60◦
2. Assuming α and β are acute angles:
sin α =
sin β =
p
3
=⇒ cos α = 1 − sin2 α =
5
5
=⇒ cos β =
13
q
1 − sin2 β =
q
4
5
q
12
2
1 − (5/13) =
13
sin (α + β) = sin α cos β + cos α sin β
3 12 4 5
= ·
+ ·
5 13 5 13
36 20
=
+
65 65
56
=
65
74
2
1 − (3/5) =
cos (α + β) = cos α cos β − sin α sin β
4 12 3 5
− ·
= ·
5 13 5 13
48 15
=
−
65 65
33
=
65
3.
sin 75◦ = sin (45◦ + 30◦ )
= sin 45◦ cos 30◦ + cos 45◦ sin 30◦
√
√ √
2
3
2 1
·
+
·
=
2
2
2
2
√
√
6+ 2
=
4
cos 75◦ = cos (45◦ + 30◦ )
= cos 45◦ cos 30◦ − sin 45◦ sin 30◦
√ √
√
2
3
2 1
=
·
−
·
2
2 2
√2
√
6− 2
=
4
4.
sin 15◦ = sin (45◦ − 30◦ )
= sin 45◦ cos 30◦ − cos 45◦ sin 30◦
√ √
√
2
3
2 1
=
·
−
·
2
2 2
√2
√
6− 2
=
4
cos 15◦ = cos (45◦ − 30◦ )
= cos 45◦ cos 30◦ + sin 45◦ sin 30◦
√ √
√
2
3
2 1
=
·
+
·
2
2
2
2
√
√
6+ 2
=
4
75
Notice that 75◦ and 15◦ are complementary angles, so we know sin 75◦ =
cos 15◦ and sin 15◦ = cos 75◦ .
5. (a) Yes, let α = β = π/4.
π π
π
π
π
π
= cos cos − sin sin
cos
+
4
4
4
4
4
4
√ √
√ √
2
2
2
2
=
·
−
·
2
2
2
2
1 1
= −
2 2
=0
More generally, we could suppose α + β = π/2 and follow the approach
in Exercise 3 of Section 2 earlier in this chapter.
(b) If α and β are acute angles, then 0 < α + β < π. Using the unit circle,
we can see that sin (α + β) must be positive since the angle α + β lies
in the upper-half of the plane, where the sine function is positive.
(c) sin (α + β) = sin α cos β + cos α sin β is positive when α and β are acute
angles since the sum and product of positive real numbers is also positive.
cos (α + β) need not be positive. As shown in part (a) of this exercise,
cos (α + β) can equal 0. Furthermore, cos (α + β) can be negative. Let
α = β = π/3. Then, assuming that we can extend the cosine addition
formula to angles α and β such that α + β is obtuse,
π π
π
π
π
π
cos
+
= cos cos − sin sin
3
3
3 √3 √ 3
3
1 1
3
3
= · −
·
2 2
2
2
1 3
= −
4 4
1
=−
2
6. As we saw in Exercise 1 of Section 1 of this chapter, sin α + sin β does
not equal sin (α + β) in general. A similar table can be used to show that
sin α − sin β does not equal sin (α − β) in general.
7. This is not a coincidence. The identity holds true even when substituting
more “arbitrary” values in for α and β. For example, using α = 37◦ and
β = 19◦ , we find that both sin2 α − sin2 β and sin (α + β) sin (α − β) are
equal to approximately 0.2562.
76
8. You may also refer to the proof in Exercise 6 of Section 2 of this chapter.
sin (α + β) sin (α − β) = (sin α cos β + cos α sin β) (sin α cos β − cos α sin β)
= sin2 α cos2 β − cos2 α sin2 β
= sin2 α cos2 β + sin2 α sin2 β − sin2 α sin2 β − cos2 α sin2 β
= sin2 α cos2 β + sin2 β − sin2 β sin2 α + cos2 α
= sin2 α · 1 − sin2 β · 1
= sin2 α − sin2 β
9. This proof is nearly identical to the one in the previous part. We just make a
small modification in how we “add by zero” in order to obtain to the desired
result.
sin (α + β) sin (α − β) = (sin α cos β + cos α sin β) (sin α cos β − cos α sin β)
= sin2 α cos2 β − cos2 α sin2 β
= sin2 α cos2 β + cos2 α cos2 β − cos2 α cos2 β − cos2 α sin2 β
= cos2 β sin2 α + cos2 α − cos2 α cos2 β + sin2 β
= cos2 β · 1 − cos2 α · 1
= cos2 β − cos2 α
10. We apply the sine addition formula in reverse.
sin 18◦ cos 12◦ + cos 18◦ sin 12◦ = sin (18◦ + 12◦ )
= sin 30◦
1
=
2
11. (a) Since we have not proved that the sine addition formula works for all
angles α and β, we use properties of the sine and cosine functions to
avoid working with angles larger than 90◦ .
sin 113◦ cos 307◦ + cos 113◦ sin 307◦
= sin (180◦ − 67◦ ) cos (360◦ − 53◦ ) + cos (180◦ − 67◦ ) sin (360◦ − 53◦ )
= sin 67◦ cos 53◦ + (− cos 67◦ ) (− sin 53◦ )
= sin (67◦ + 53◦ )
= sin 120◦
= sin 60◦
√
3
=
2
(b) Plugging into a calculator,
√
◦
◦
◦
◦
sin 113 cos 307 + cos 113 sin 307 ≈ 0.866 ≈
77
3
= sin 60◦
2
(c) Assuming that the sine addition formula does work for non-acute angles,
we arrive at the same result.
sin 113◦ cos 307◦ + cos 113◦ sin 307◦ = sin (113◦ + 307◦ )
= sin 420◦
= sin 60◦
√
3
=
2
12. We can use the addition formulas for sine and cosine by rewriting 2α as
α + α.
sin 2α cos α − cos 2α sin α = sin (α + α) cos α − cos (α + α) sin α
= (sin α cos α + cos α sin α) cos α − (cos α cos α − sin α sin α) sin α
= sin α cos2 α + sin α cos2 α − sin α cos2 α + sin3 α
= sin α cos2 α + sin3 α
= sin α cos2 α + sin2 α
= sin α
13.
sin (α + β) sin β + cos (α + β) cos β = (sin α cos β + cos α sin β) sin β + (cos α cos β − sin α sin β) cos β
= sin α sin β cos β + sin2 β cos α + cos α cos2 β − sin α sin β cos β
= sin2 β cos α + cos α cos2 β
= cos α sin2 β + cos2 β
= cos α
14.
sin α cos β + cos α sin β − cos α sin β
sin (α + β) − cos α sin β
=
cos (α + β) + sin α sin β
cos α cos β − sin α sin β + sin α sin β
sin α cos β
=
cos α cos β
sin α
=
cos α
= tan α
15.
π
π
π
sin α +
= sin α cos + cos α sin
4
√4
√4
2
2
= sin α ·
+ cos α ·
2
2
√
2
=
(sin α + cos α)
2
78
16.
cos (α + β)
cos α cos β − sin α sin β
=
cos α cos β
cos α cos β
sin α sin β
=1−
cos α cos β
sin α sin β
=1−
·
cos α cos β
= 1 − tan α tan β
2
17. Applying the law of cosines, we have that (b1 + b2 ) = c21 +c22 −2c1 c2 cos (α + β).
Solving for cos (α + β), we get
2
cos (α + β) =
c21 + c22 − (b1 + b2 )
.
2c1 c2
Before proceeding further, let’s establish some relationships between the
variables in the diagram. First, by the Pythagorean theorem, we have that
h2 = c21 − b21 = c22 − b22 . Additionally, we can compute the sines and cosines
for the angles α and β:
sin α =
b1
b2
h
h
, sin β = , cos α = , cos β = .
c1
c2
c1
c2
We can now simplify our expression for cos (α + β).
2
c21 + c22 − (b1 + b2 )
2c1 c2
2
2
c + c2 − b21 − 2b1 b2 − b22
= 1
2c1 c2
2
2h − 2b1 b2
=
2c1 c2
h2 − b1 b2
=
c1 c2
h h
b1 b2
=
·
−
·
c1 c2
c1 c2
= cos α cos β − sin α sin β
cos (α + β) =
Chapter 7: Trigonometric Identities
Page 141
1. (a) Yes
(b) No
79
(c) Yes
(d) Yes
(e) No
(f) No
2. (a)
tan α =
sin α
cos α
(b)
(1 + tan α) (1 − tan α) = 1 − tan2 α
sin2 α
cos2 α
2
cos α − sin2 α
=
cos2 α
=1−
(c)
tan α + tan β
tan α + tan β cos α cos β
=
·
1 − tan α tan β
1 − tan α tan β cos α cos β
sin α cos β + cos α sin β
=
cos α cos β − sin α sin β
(d)
sin2 α
cos2 α
+
cos2 α
sin2 α
4
sin α + cos4 α
=
sin2 α cos2 α
tan2 α + cot2 α =
(e)
sin α cos α
·
cos α sin α
=1
tan α cot α =
(f)
sin2 α
cos2 α
2
cos α + sin2 α
=
cos2 α
1
=
cos2 α
1 + tan2 α = 1 +
80
p
3. The Principle of Analytic Continuation does not apply because 1 − sin2 α
is not a rational trigonometric function.
The identity is incorrect for α =
q
2π/3 as cos (2π/3) = −1/2, while
1 − sin2 (2π/3) = 1/2.
4. The Principle of Analytic Continuation does apply because both sin2 α +
cos2 α and 1 are rational trigonometric functions. The identity is correct for
α = 2π/3.
2π
2π
sin
+ cos2
=
3
3
2
√ !2 2
3
1
3 1
+
= + =1
2
2
4 4
Page 142
1. Since α and β are acute angles,
p
3
=⇒ cos α = 1 − sin2 α =
sin α =
5
5
sin β =
=⇒ cos β =
13
q
1 − sin2 β =
s
2
3
4
1−
=
5
5
s
1−
5
13
2
=
12
13
sin (α + β) = sin α cos β + cos α sin β
3 12 4 5
+ ·
= ·
5 13 5 13
36 20
=
+
65 65
56
=
65
cos (α + β) = cos α cos β − sin α sin β
4 12 3 5
= ·
− ·
5 13 5 13
48 15
=
−
65 65
33
=
65
α + β lies in the first quadrant because sin (α + β) and cos (α + β) are both
positive.
2. Since α and β are acute angles,
p
4
sin α =
=⇒ cos α = 1 − sin2 α =
5
81
s
1−
2
4
3
=
5
5
12
sin β =
=⇒ cos β =
13
q
1 − sin2 β =
s
1−
12
13
2
=
5
13
sin (α + β) = sin α cos β + cos α sin β
4 5
3 12
= ·
+ ·
5 13 5 13
20 36
=
+
65 65
56
=
65
cos (α + β) = cos α cos β − sin α sin β
4 12
3 5
− ·
= ·
5 13 5 13
15 48
=
−
65 65
33
=−
65
α+β lies in the second quadrant because sin (α + β) is positive and cos (α + β)
is negative.
3.
s
2
p
3
3
4
2
sin α =
=⇒ cos α = ± 1 − sin α = ± 1 −
=±
5
5
5
s
q
2
5
12
5
=⇒ cos β = ± 1 − sin2 β = ± 1 −
=±
sin β =
13
13
13
cos α > 0, cos β > 0:
sin (α + β) = sin α cos β + cos α sin β
3 12 4 5
= ·
+ ·
5 13 5 13
36 20
=
+
65 65
56
=
65
cos α > 0, cos β < 0:
sin (α + β) = sin α cos β + cos α sin β
3
12
4 5
−
+ ·
=
5
13
5 13
36 20
=− +
65 65
16
=−
65
82
cos α < 0, cos β > 0:
sin (α + β) = sin α cos β + cos α sin β
3 12
4
5
= ·
+ −
·
5 13
5
13
36 20
=
−
65 65
16
=
65
cos α < 0, cos β < 0:
sin (α + β) = sin α cos β + cos α sin β
3
12
4
5
=
−
+ −
·
5
13
5
13
36 20
=− −
65 65
56
=−
65
There are four possible answers for sin (α + β).
4. (a)
√
√
π
2π
π
1
2π
3 1
3
cos − cos
sin =
· − −
sin
3
3
3
3
2 2
2
2
√
√
3
3
=
+
4
√4
3
=
2 π
= sin
3
2π π
= sin
−
3
3
(b)
√
3π
π
3π
2
π
− cos sin
=
sin cos
4
4
4
4
2
√ ! √ √
2
2
2
−
−
·
2
2
2
1 1
=− −
2 2
= −1
π
= sin −
2
π 3π
= sin
−
4
4
83
(c)
√
π
π
3π
3π
1
3
cos
sin
sin −
− cos −
=− ·0−
(−1)
6
2
6
2
2
2
√
3
=
2 π
= sin
3
5π
= sin −
3
π 3π
= sin − −
6
2
5. Applying the identity
(A − B)2 + (A + B)2 = A2 − 2AB + B 2 + A2 + 2AB + B 2 = 2A2 + 2B 2 ,
we have that,
2
2
cos2 (γ + δ) + cos2 (γ − δ) = (cos γ cos δ − sin γ sin δ) + (cos γ cos δ + sin γ sin δ)
= 2 cos2 γ cos2 δ + 2 sin2 γ sin2 δ.
Therefore,
2
cos α + cos
2
2π
2π
2π
2π
2
+ α + cos
− α = cos2 α + 2 cos2
cos2 α + 2 sin2
sin2 α
3
3
3
3
1
3
2
2
= cos α + 2
cos α + 2
sin2 α
4
4
3
cos2 α + sin2 α
=
2
3
=
2
6.
sin (x + y) + sin (x − y) = sin x cos y + cos x sin y + sin x cos y − cos x sin y
= sin x cos y + sin x cos y
= 2 sin x cos y
7.
cos (x + y) + cos (x − y) = cos x cos y − sin x sin y + cos x cos y + sin x sin y
= cos x cos y + cos x cos y
= 2 cos x cos y
84
8. Since (A − B)(A + B) = A2 − B 2 ,
cos (x + y) cos (x − y) = (cos x cos y − sin x sin y) (cos x cos y + sin x sin y)
= cos2 x cos2 y − sin2 x sin2 y
9. Since (A + B)(A − B) = A2 − B 2 ,
sin (x + y) sin (x − y) = (sin x cos y + cos x sin y) (sin x cos y − cos x sin y)
= sin2 x cos2 y − cos2 x sin2 y
10.
cos (x + y) cos (x − y) − sin (x + y) sin (x − y) = cos2 x cos2 y − sin2 x sin2 y − sin2 x cos2 y − cos2 x sin2 y
= cos2 x cos2 y + sin2 y − sin2 x sin2 y + cos2 y
= cos2 x − sin2 x
11.
cos 2x = cos (x + x)
= cos x cos x − sin x sin x
= cos2 x − sin2 x
There is no error, because cos 2x = cos2 x − sin2 x.
12.
cos (α + β) cos β + sin (α + β) sin β = (cos α cos β − sin α sin β) cos β + (sin α cos β + cos α sin β) sin β
= cos α cos2 β − sin α sin β cos β + sin α sin β cos β + sin2 β cos α
= cos α cos2 β + sin2 β cos α
= cos α cos2 β + sin2 β
= cos α
Alternatively, by applying the cosine difference formula in reverse,
cos (α + β) cos β + sin (α + β) sin β = cos (α + β − β) = cos α
85
Page 144
1.
tan
7π 5π
+
6
3
7π
5π
+ tan
6
3
=
7π
5π
tan
1 − tan
6
√
√ 3
1/ 3 − 3
√ √ =
1 − 1/ 3 − 3
√
√
1/ 3 − 3
=
√2 √ √
3
1/ 3 − 3
=
·√
2
3
1−3
= √
2 3
1
= −√
3
π
= tan −
6 17π
= tan
6
7π 5π
+
= tan
6
3
tan
2. Because the tangent function is odd, we know tan −β = − tan β.
tan (α − β) = tan (α + (−β))
tan α + tan −β
=
1 − tan α tan −β
tan α − tan β
=
1 + tan α tan β
3.
π
+ tan α
4
tan
+α =
π
4
1 − tan tan α
4
1 + tan α
=
1 − tan α
π
tan
86
4.
π
− tan α
4
tan
−α =
π
4
1 + tan tan α
4
1 − tan α
=
1 + tan α
π
tan
5. Since β = π/4 − α,
π
(1 + tan α) (1 + tan β) = (1 + tan α) 1 + tan
−α
4 1 − tan α
= (1 + tan α) 1 +
1 + tan α
= 1 + tan α + 1 − tan α
=2
Alternatively, since tan (α + β) = tan π/4 = 1,
(1 + tan α) (1 + tan β) = 1 + tan α + tan β + tan α tan β
1 − tan α tan β
+ tan α tan β
= 1 + (tan α + tan β)
1 − tan α tan β
tan α + tan β
+ tan α tan β
= 1 + (1 − tan α tan β)
1 − tan α tan β
= 1 + (1 − tan α tan β) tan (α + β) + tan α tan β
= 1 + 1 − tan α tan β + tan α tan β
=2
6.
tan (α + β + γ) =
=
=
=
=
tan (α + β) + tan γ
1 − tan (α + β) tan γ
tan α + tan β
+ tan γ
1 − tan α tan β
tan α + tan β
1−
tan γ
1 − tan α tan β
tan α + tan β
+ tan γ
1 − tan α tan β
1 − tan α tan β
·
tan α + tan β
1 − tan α tan β
1−
tan γ
1 − tan α tan β
tan α + tan β + tan γ (1 − tan α tan β)
1 − tan α tan β − (tan α + tan β) tan γ
tan α + tan β + tan γ − tan α tan β tan γ
1 − tan α tan β − tan α tan γ − tan β tan γ
87
7. Since tan (α + β + γ) = tan π = 0, from the previous part, we have
tan α+tan β+tan γ−tan α tan β tan γ = 0 =⇒ tan α+tan β+tan γ = tan α tan β tan γ.
This is because if a fraction equals zero, its numerator must be zero.
8. First, applying the tangent addition formula,
tan 3α = tan (2α + α) =
tan 2α + tan α
=⇒ tan 2α+tan α = tan 3α (1 − tan 2α tan α) .
1 − tan 2α tan α
Therefore,
tan 3α−tan 2α−tan α = tan 3α−tan 3α (1 − tan 2α tan α) = tan 3α tan 2α tan α.
tan α is not defined for α = π2 + nπ. Therefore, tan 2α is not defined for
α = π4 + n π2 and tan 3α is not defined for α = π6 + n π3 .
Page 147
q
7 2
) (and cannot be
1. (a) Since sin2 α + cos2 α = 1, we know cos α = 1 − ( 25
the negative version because cos α isqgiven as positive).
7
7 2
336
Thus sin 2α = 2 sin α cos α = 2 · 25
· 1 − ( 25
) = 625
.
7 2
) = 527
And cos 2α = 1 − 2 sin2 α = 1 − 2( 25
625 .
(b) This part is similar
q except that we use the negative version of cosine,
namely cos α = −
7 2
) .
1 − ( 25
336
Thus sin 2α = 2 sin α cos α = − 625
.
However, cosine value remains the same: cos 2α = 1 − 2 sin2 α = 1 −
7 2
2( 25
) = 527
625 .
2. Firstly, sin 2α = 2 sin α cos α. In other words, the sine value will be a product
of rational numbers, so it will also be rational. Similarly, cos 2α = 2 cos2 α−1
will be rational because cos α is rational. Exercise 1 confirms this result.
3. Let’s use the double angle formula cos 2α = 2 cos2 α − 1.
cos 2α = cos2 α =⇒ 2 cos2 α − 1 = cos2 α =⇒ cos2 α = 1 =⇒ cos α = ±1
cos α has a magnitude of 1 precisely when α is an integer multiple of π, so
the student’s angle must have also been an integer multiple of π.
4. We start with the given equation:
sin α + cos α = 0.2
2
sin α + 2 sin α cos α+ cos2 α = 0.04
1 + 2 sin α cos α = 0.04
2 sin α cos α = −0.96
Note that that is simply sin 2α. Hooray!
88
(Squared both sides.)
5. We can follow a very similar strategy here to find 1 − 2 sin α cos α = 0.09.
Then sin 2α = 2 sin α cos α = 0.91.
6.
cos 2α cos α + sin 2α sin α = 2 cos2 α − 1 cos α + 2 sin α cos α sin α
= cos α 2 cos2 α − 1 + 2 sin2 α
= cos α (2 − 1)
= cos α
Alternatively, we may use the cosine difference formula.
cos 2α cos α + sin 2α sin α = cos (2α − α) = cos α
7. Applying the sine addition formula,
sin 2α cos α + cos 2α sin α = sin (2α + α) = sin 3α
Applying the sine subtraction formula,
sin 4α cos α − cos 4α sin α = sin (4α − α) = sin 3α
Since both sides of the identity are equal to sin 3α, the identity is true.
8. Yes, the book asked you to prove something incorrect! For counterexample,
consider that cos 2α can be negative but cos2 α is never negative. However,
we can prove a relationship.
Recall that cos 2α = cos2 α − sin2 α. However, sin2 α ≥ 0 since it’s a square.
sin2 α ≥ 0
− sin2 α ≤ 0
cos2 α − sin2 α ≤ cos2 α
cos 2α ≤ cos2 α
9.
sin
α
α 2
α
α
α
α
− cos
= sin2 − 2 sin cos + cos2
2
2
2
2
2
2
= 1 − sin α
89
10. Following the hint, we compute the value of cos 10◦ sin 10◦ sin 50◦ sin 70◦ .
M cos 10◦ = cos 10◦ sin 10◦ sin 50◦ sin 70◦
1
= sin 20◦ sin 50◦ sin 70◦
2
1
= cos 70◦ sin 50◦ sin 70◦
2
1
= sin 140◦ sin 50◦
4
1
= sin 40◦ sin 50◦
4
1
= cos 50◦ sin 50◦
4
1
= sin 100◦
8
1
= sin 80◦
8
1
= cos 10◦
8
Since M cos 10◦ =
M is equal to 18 .
1
8
cos 10◦ , we have that the value of the original expression
11. We begin by computing sin 20◦ cos 20◦ cos 40◦ cos 80◦ .
1
sin 40◦ cos 40◦ cos 80◦
2
1
= sin 80◦ cos 80◦
4
1
= sin 160◦
8
1
= sin 20◦
8
sin 20◦ cos 20◦ cos 40◦ cos 80◦ =
This implies that cos 20◦ cos 40◦ cos 80◦ = 1/8.
12. We begin by computing cos π/10 sin π/10 sin π/5.
cos
π
π
π
1
π
π
sin
cos = sin cos
10
10
5
2
5
5
2π
1
= sin
4
5
1
π
= cos
4
10
This implies that sin π/10 cos π/5 = 1/4.
90
Page 148
1.
cos 3α = cos (2α + α)
= cos 2α cos α − sin 2α sin α
= 2 cos2 α − 1 cos α − 2 sin α cos α sin α
= 2 cos3 α − cos α − 2 sin2 α cos α
= 2 cos3 α − cos α − 2 1 − cos2 α cos α
= 4 cos3 α − 3 cos α
2.
sin 3α = 3 sin α − 4 sin3 α
3
3
3
=3
−4
5
5
27
9
= −4·
5
125
117
=
125
If sin α = 3/5, then cos α = ±4/5. Therefore,
cos 3α = 4 cos3 α − 3 cos α
3
4
4
=4 ±
−3 ±
5
5
256 12
=±
∓
125
5
44
=∓
125
3. If cos α = 4/5, then sin α = ±3/5. Therefore,
sin 3α = 3 sin α − 4 sin3 α
3
3
3
−4 ±
=3 ±
5
5
9
27
=± ∓4·
5
125
117
=±
125
91
cos 3α = 4 cos3 α − 3 cos α
3
4
4
=4
−3
5
5
256 12
=
−
125
5
44
=−
125
4. (a)
cos 4α = 2 cos2 2α − 1
2
= 2 2 cos2 α − 1 − 1
= 2 4 cos4 α − 4 cos2 α + 1 − 1
= 8 cos4 α − 8 cos2 α + 1
(b)
cos 4α = 1 − 2 sin2 2α
= 1 − 8 sin2 α cos2 α
= 1 − 8 sin2 α 1 − sin2 α
= 1 − 8 sin2 α + 8 sin4 α
5.
sin 3α cos α − cos 3α sin α = 3 sin α − 4 sin3 α cos α − 4 cos3 α − 3 cos α sin α
= 3 sin α cos α − 4 sin3 α cos α − 4 cos3 α sin α + 3 sin α cos α
= 2 sin α cos α 3 − 2 sin2 α − 2 cos2 α
= 2 sin α cos α (3 − 2)
= sin 2α
Alternatively, applying the sine difference formula,
sin 3α cos α − cos 3α sin α = sin (3α − α) = sin 2α
6.
sin 3α cos 3α
3 sin α − 4 sin3 α 4 cos3 α − 3 cos α
−
=
−
sin α
cos α
sin α
cos
α
= 3 − 4 sin2 α − 4 cos2 α − 3
= 6 − 4 sin2 α − 4 cos2 α
=2
92
Alternatively, using the result from the previous exercise,
sin 3α cos 3α
sin 3α cos α − cos 3α sin α
−
=
sin α
cos α
sin α cos α
sin 2α
=
sin α cos α
2 sin α cos α
=
sin α cos α
=2
7. (a) Applying the result of Exercise 9 from Section 3 of this chapter,
4 sin α sin (60◦ + α) sin (60◦ − α) = 4 sin α sin2 60◦ cos2 α − cos2 60◦ sin2 α
1
3
cos2 α − sin2 α
= 4 sin α
4
4
= 3 sin α cos2 α − sin3 α
= 3 sin α 1 − sin2 α − sin3 α
= 3 sin α − 4 sin3 α
= sin 3α
(b) Applying the result of Exercise 8 from Section 3 of this chapter,
4 cos α cos (60◦ + α) cos (60◦ − α) = 4 cos α cos2 60◦ cos2 α − sin2 60◦ sin2 α
3
1
cos2 α − sin2 α
= 4 cos α
4
4
= cos3 α − 3 sin2 α cos α
= cos3 α − 3 1 − cos2 α cos α
= 4 cos3 α − 3 cos α
8.
sin 4α = 2 sin 2α cos 2α
= 4 sin α cos α 2 cos2 α − 1
= 8 sin α cos3 α − 4 sin α cos α
sin 4α
=⇒
= 8 cos3 α − 4 cos α
sin α
9. In the penultimate step, we apply the following identity:
3
(A − B) = A3 − 3A2 B + 3AB 2 − B 3 ,
taking A = cos2 α and B = sin2 α.
93
sin 3α sin3 α + cos 3α cos3 α = 3 sin α − 4 sin3 α sin3 α + 4 cos3 α − 3 cos α cos3 α
= 3 sin4 α − 4 sin6 α + 4 cos6 α − 3 cos4 α
= 3 sin4 α − 4 sin4 α 1 − cos2 α + 4 cos4 α 1 − sin2 α − 3 cos4 α
= sin4 α 4 cos2 α − 1 + cos4 α 1 − 4 sin2 α
= sin4 α 4 cos2 α − sin2 α − cos2 α + cos4 α sin2 α + cos2 α − 4 sin2 α
= 3 sin4 α cos2 α − sin6 α + cos6 α − 3 cos4 α sin2 α
3
= cos2 α − sin2 α
= cos3 2α
Page 150
1.
r
α
1+1
cos α = 1 =⇒ cos = ±
= ±1
2
2
cos α is equal to 1 when α is an integer multiple of 2π. If α = 4nπ for some
integer n (i.e., an even integer multiple of 2π), then cos α/2 = cos 2nπ = 1.
Otherwise, if α = (4n + 2) π for some integer n (i.e., an odd integer multiple
of 2π), then cos α/2 = cos (2n + 1) π = −1.
For a particular example of each case, we may set n = 0.
α = 4 (0) π = 0:
cos 0 = 1, cos
0
= cos 0 = 1
2
α = (4 (0) + 2) π = 2π:
cos 2π = 1, cos
2π
= cos π = −1
2
2. (a) We take the positive square root because 60◦ /2 = 30◦ lies in the first
quadrant.
r
60◦
1 + cos 60◦
cos
=
2
2
r
1 + 1/2
=
2
r
3
=
4
√
3
=
2
94
(b) We take the positive square root because 120◦ /2 = 60◦ lies in the first
quadrant.
r
120◦
1 + cos 120◦
cos
=
2
2
r
1 − 1/2
=
2
r
1
=
4
1
=
2
(c) We take the negative square root because 240◦ /2 = 120◦ lies in the
second quadrant.
r
240◦
1 + cos 240◦
cos
=−
2
2
r
1 − 1/2
=−
2
r
1
=−
4
1
=−
2
3.
95
α
Quadrant α?
α/2
Quadrant α/2?
780◦
I
390◦
I
1020◦
IV
510◦
II
√
− 3/2
1140◦
I
570◦
III
√
− 3/2
1380◦
IV
690◦
IV
−60◦
IV
−30◦
IV
−300◦
I
−150◦
III
√
− 3/2
−420◦
IV
−210◦
II
√
− 3/2
−660◦
I
−330◦
I
−780◦
IV
−390◦
IV
4.
r
◦
1 − cos 30◦
2
s
√
1 − 3/2
=
2
s
√
2− 3
=
4
p
√
2− 3
=
2
sin 15 =
r
◦
1 + cos 30◦
2
s
√
1 + 3/2
=
2
s
√
2+ 3
=
4
p
√
2+ 3
=
2
cos 15 =
96
cos α/2
√
√
√
√
√
3/2
3/2
3/2
3/2
3/2
5. Because |cos α| ≤ 1, 1 ± cos α ≥ 0, so the expressions under the square roots
in the sine and cosine half-angle formulas will not be negative.
6. The square root sign in the half-angle formula prevents it from being a
rational trigonometric function, so the Principle of Analytic Continuation
does not apply.
7. (a)
α
β
α
γ
β
γ
α
β
γ
α
β
tan tan + tan tan + tan tan = tan tan + tan
tan + tan
2
2
2
2
2
2
2
2
2
2
2
α
β
γ
α+β
α
β
= tan tan + tan tan
1 − tan tan
2
2
2
2
2
2
β
α
β
α
= tan tan + 1 − tan tan
2
2
2
2
=1
Alternatively, we can recall the extended tangent addition formula derived in Exercise 6 of Section 4 of this chapter:
tan (α + β + γ) =
tan α + tan β + tan γ − tan α tan β tan γ
1 − tan α tan β − tan α tan γ − tan β tan γ
Since α/2 + β/2 + γ/2 = π/2, tan (α/2 + β/2 + γ/2) is undefined. This
implies that the denominator of the tangent addition formula is 0 (assuming that none of tan α/2, tan β/2, or tan γ/2 are undefined). Therefore, we can conclude
tan
α
β
α
γ
β
γ
tan + tan tan + tan tan = 1.
2
2
2
2
2
2
(b) Following a similar approach as to the previous part, we begin by noting
γ
that sin α+β
2 = cos 2 .
4 cos
β
γ
α
β
α+β
α
cos cos = 4 cos cos sin
2
2
2
2
2
2
α
β
α
β
α
β
= 4 cos cos
sin cos + cos sin
2
2
2
2
2
2
α
α
β
β
β
α
= 4 sin cos cos2 + 4 sin cos cos2
2
2
2
2
2
2
β
α
= 2 sin α cos2 + 2 sin β cos2
2
2
1 + cos β
1 + cos α
= 2 sin α
+ 2 sin β
2
2
= sin α + sin α cos β + sin β + sin β cos α
= sin α + sin β + sin (α + β)
= sin α + sin β + sin (π − α − β)
= sin α + sin β + sin γ
97
Page 152
1. Because 1 + cos α is non-negative, division by 1 + cos α does not change the
sign of sin α, which means tan (α/2) and sin α/ (1 + cos α) have the same
sign.
2.
tan
r
1 − cos α
1 + cos α
r
1 − cos α 1 − cos α
=±
·
1 + cos α 1 − cos α
s
2
(1 − cos α)
=±
1 − cos2 α
s
2
(1 − cos α)
=±
sin2 α
1 − cos α
=±
sin α
α
=±
2
For acute angles α, we need to take the positive branch of the square root so
that the signs of both sides of the half-angle formula agree. By the Principle
of Analytic Continuation, since the positive branch is correct for all acute
angles and both sides of the formula are rational trigonometric expressions,
it is correct for all angles in general, so we have
tan
α
1 − cos α
=
2
sin α
Page 153
1.
1
1
1
cos (α − β) − cos (α + β) = (cos α cos β + sin α sin β − cos α cos β + sin α sin β)
2
2
2
1
= (2 sin α sin β)
2
= sin α sin β
1
1
1
sin (α + β) + sin (α − β) = (sin α cos β + cos α sin β + sin α cos β − cos α sin β)
2
2
2
1
= (2 sin α cos β)
2
= sin α cos β
98
2.
1
1
cos (75◦ − 15◦ ) − cos (75◦ + 15◦ )
2
2
1
1
= cos 60◦ − cos 90◦
2
2
1 1 1
= · − ·0
2 2 2
1
=
4
sin 75◦ sin 15◦ =
3.
1
1
sin (75◦ + 15◦ ) + sin (75◦ − 15◦ )
2
2
1
1
◦
◦
= sin 90 + sin 60
2
√2
1
1
3
= ·1+ ·
2 √ 2 2
2+ 3
=
4
sin 75◦ cos 15◦ =
4. (a)
1
1
cos (75◦ + 15◦ ) + cos (75◦ − 15◦ )
2
2
1
1
= cos 90◦ + cos 60◦
2
2
1
1 1
= ·0+ ·
2
2 2
1
=
4
cos 75◦ cos 15◦ =
Alternatively, by the cosine subtraction formula, cos 60◦ = cos 75◦ cos 15◦ +
sin 75◦ sin 15◦ so
cos 75◦ cos 15◦ = cos 60◦ − sin 75◦ sin 15◦ =
1 1
1
− =
2 4
4
Alternatively, using the idea of sine and cosine being cofunctions,
cos 75◦ cos 15◦ = sin (90◦ − 75◦ ) sin (90◦ − 15◦ ) = sin 15◦ sin 75◦ =
1
4
For the above two alternative solutions, we apply the result from Exercise 2 above that sin 75◦ sin 15◦ = 1/4.
99
(b)
1
1
sin (15◦ + 75◦ ) + sin (15◦ − 75◦ )
2
2
1
1
◦
◦
= sin 90 − sin 60
2
√2
3
1
1
= ·1− ·
2 √ 2 2
2− 3
=
4
cos 75◦ sin 15◦ =
Alternatively, by the sine addition formula, sin 90◦ = sin 75◦ cos 15◦ +
cos 75◦ sin 15◦ so
√
√
2+ 3
2− 3
cos 75◦ sin 15◦ = sin 90◦ − sin 75◦ cos 15◦ = 1 −
=
4
4
For the above alternative solution,
√we
apply the result from Exercise 3
above that sin 75◦ cos 15◦ = 2 + 3 /4.
5.
2 cos
π
π
π
π
π
π
+ α cos
− α = cos
+ α + − α + cos
+α− +α
4
4
4
4
4
4
π
= cos cos 2α
2
= cos 2α
Alternatively, applying the result of Exercise 8 in Section 3 of this chapter,
π
π
π
π
2 cos
+ α cos
− α = 2 cos2 cos2 α − sin2 sin2 α
4
4
4
4
1
1
2
2
cos α − sin α
=2
2
2
= cos2 α − sin2 α
= cos 2α
6.
sin (α + β) sin (α − β) + sin (β + γ) sin (β − γ) + sin (γ + α) sin (γ − α)
1
1
1
1
1
1
= cos 2β − cos 2α + cos 2γ − cos 2β + cos 2α − cos 2γ
2
2
2
2
2
2
=0
100
7.
sin α sin (β − γ) + sin β sin (γ − α) + sin γ sin (α − β)
1
1
1
1
= cos (α − β + γ) − cos (α + β − γ) + cos (β − γ + α) − cos (β + γ − α)
2
2
2
2
1
1
+ cos (γ − α + β) − cos (γ + α − β)
2
2
=0
Page 155
1. Recall from the previous section that cos (γ + δ) + cos (γ − δ) = 2 cos γ cos δ.
Following Example 54, we let γ = (α + β) /2 and δ = (α − β) /2. Substituting for γ and δ, we arrive at the first formula,
cos α + cos β = 2 cos
α+β
α−β
cos
2
2
Similarly, we recall that cos (γ − δ) − cos (γ + δ) = 2 sin γ sin δ. Performing
the same substitution as above, we obtain
cos β−cos α = 2 sin
α+β
α−β
α+β
α−β
sin
⇐⇒ cos α−cos β = −2 sin
sin
2
2
2
2
2.
cos 70◦ + sin 40◦ = sin 20◦ + sin 40◦
20◦ − 40◦
20◦ + 40◦
cos
= 2 sin
2
2
= 2 sin 30◦ cos (−10◦ )
1
= 2 · cos 10◦
2
= cos 10◦
3.
55◦ + 65◦
55◦ − 65◦
cos
2
2
= 2 cos 60◦ cos (−5◦ )
1
= 2 · cos 5◦
2
= cos 5◦
cos 55◦ + cos 65◦ = 2 cos
101
4.
100◦ + 20◦
100◦ − 20◦
cos
+ cos 140◦
2
2
= 2 cos 60◦ cos 40◦ + cos 140◦
1
= 2 · cos 40◦ + cos 140◦
2
= cos 40◦ + cos 140◦
140◦ + 40◦
140◦ − 40◦
= 2 cos
cos
2
2
= 2 cos 90◦ cos 50◦
cos 20◦ + cos 100◦ + cos 140◦ = 2 cos
=0
5.
sin 78◦ + cos 132◦ = sin 78◦ − cos 48◦
= sin 78◦ − sin 42◦
78◦ + 42◦
78◦ − 42◦
= 2 cos
sin
2
2
= 2 cos 60◦ sin 18◦
1
= 2 · sin 18◦
2
= sin 18◦
6.
cos 15◦ + sin 15◦
sin 75◦ + sin 15◦
=
◦
◦
cos 15 − sin 15
sin 75◦ − sin 15◦
2 sin 45◦ cos 30◦
=
2 cos 45◦ sin 30◦
= tan 45◦ cot 30◦
√
= 3
7. (a)
sin (α + β) = sin (π − α − β) = sin γ
(b)
cos (α + β) = − cos (π − α − β) = − cos γ
102
(c)
sin 2α + sin 2β + sin 2γ = 2 sin (α + β) cos (α − β) + sin 2γ
= 2 sin γ cos (α − β) + 2 sin γ cos γ
= 2 sin γ (cos (α − β) + cos γ)
= 2 sin γ (cos (α − β) − cos (α + β))
= 2 sin γ (2 sin α sin β)
= 4 sin α sin β sin γ
8.
π
2π
4π
π
4π
sin α + sin α +
+ sin α +
= 2 sin α +
cos −
+ sin α +
3
3
3
3
3
π
4π
= sin α +
+ sin α +
3
3
5π
π
= 2 sin α +
cos −
6
2
=0
9. We first note that
sin kα+sin (k + 2)α = 2 sin
(k + 2)α + kα
(k + 2)α − kα
cos
= 2 sin (k + 1)α cos α.
2
2
Therefore,
sin α + 2 sin 3α + sin 5α = sin α + sin 3α + sin 3α + sin 5α
= 2 sin 2α cos α + 2 sin 4α cos α
= 2 cos α (sin 2α + sin 4α)
= 2 cos α (2 sin 3α cos α)
= 4 cos2 α sin 3α.
10.
sin (β − γ) sin (γ − α) sin (α − β)
+
+
sin β sin γ
sin γ sin α
sin α sin β
sin β cos γ
cos β sin γ
sin γ cos α cos γ sin α sin α cos β
cos α cos β
=
−
+
−
+
−
sin β sin γ
sin β sin γ
sin γ sin α
sin γ sin α
sin α sin β
sin α sin β
= cot γ − cot β + cot α − cot γ + cot β − cot α
=0
This result also follows from the identity which was proven in Exercise 7 of
the previous section. This can be seen by rewriting the left-hand side of the
equation as a single fraction with a common denominator and noting that
the numerator of this fraction is zero.
103
11.
2α − β − γ
γ−β
cos
+ sin (β − γ)
2
2
2α − β − γ
β−γ
β−γ
β−γ
= 2 sin
cos
+ 2 sin
cos
2
2
2 2
β−γ
2α − β − γ
β−γ
= 2 cos
sin
+ sin
2
2
2
β−γ
α−γ
α−β
= 2 cos
2 sin
cos
2
2
2
α−γ
β−γ
α−β
sin
cos
= 4 cos
2
2
2
sin (α − β) + sin (α − γ) + sin (β − γ) = 2 sin
12.
sin (α + β + γ) + sin (α − β − γ) + sin (α + β − γ) + sin (α − β + γ)
= 2 sin α cos (β + γ) + sin (α + β − γ) + sin (α − β + γ)
= 2 sin α cos (β + γ) + 2 sin α cos (β − γ)
= 2 sin α (cos (β + γ) + cos (β − γ))
= 2 sin α (2 cos β cos γ)
= 4 sin α cos β cos γ
Page 158
1. Let β = 2γ.
sin2 β + cos2 β = sin2 2γ + cos2 2γ
2 2
2a
1 − a2
=
+
1 + a2
1 + a2
2
4a
1 − 2a2 + a4
=
+
1 + 2a2 + a4
1 + 2a2 + a4
2
4
1 + 2a + a
=
1 + 2a2 + a4
=1
2.
2a
1 − a2
2a
2
= 1 + a2
1−a
1 + a2
sin 2β
=
cos 2β
tan 2β =
104
Page 160
1.
sin α =
2·2·3
12
=
22 + 3 2
13
−5
22 − 32
=
22 + 32
13
These values give the Pythagorean triple 5, 12, 13, provided we take the
absolute value of cos α. However, because cos α is negative, α is not an
acute angle, so it cannot correspond to an angle in a right triangle.
cos α =
2.
sin α =
2·8·5
80
=
82 + 5 2
89
82 − 52
39
=
2
2
8 +5
89
This corresponds to the right triangle with legs of 39 and 80 and a hypotenuse
of 89.
cos α =
3.
2
(2pq) + q 2 − p2
2
= 4p2 q 2 + q 4 − 2q 2 p2 + p4 = q 4 + 2q 2 p2 + p4 = q 2 + p2
The above shows that q 2 + p2 is the hypotenuse.
Page 161 (First)
1.
• sin 20◦ cos 20◦ ≈ 0.3214
• sin 10◦ cos 10◦ ≈ 0.1710
• sin 5◦ cos 5◦ ≈ 0.0868
• sin 1◦ cos 1◦ ≈ 0.0174
• sin 70◦ cos 70◦ ≈ 0.3214
• sin 80◦ cos 80◦ ≈ 0.1710
• sin 85◦ cos 85◦ ≈ 0.0868
• sin 89◦ cos 89◦ ≈ 0.0174
2.
1
sin 30 cos 30 = ·
2
√
2
sin 45◦ cos 45◦ =
2
√
3
◦
◦
sin 60 cos 60 =
2
◦
◦
105
√
√
3
3
=
2
4
√
2
1
·
=
2
2
√
1
3
· =
2
4
2
Page 161 (Second)
1. In the below answers for this question, n represents an arbitrary integer.
(a)
sin x cos x =
(b)
1
1
1
=⇒ sin 2x =
2
2
2
=⇒ sin 2x = 1
π
=⇒ 2x = + 2nπ
2
π
=⇒ x = + nπ
4
√
√
1
3
3
=⇒ sin 2x =
=⇒ sin 2x = 3
sin x cos x =
2
2
2
√
Because 3 > 1, there are no values of x which satisfy the given equation.
√
(c)
√
√
3
3
1
sin x cos x =
=⇒ sin 2x =
4
2
√ 4
3
=⇒ sin 2x =
π 2 π
=⇒ 2x =
±
+ 2nπ
π2 π6 =⇒ x =
±
+ nπ
4
12
2. (c) has no solution because sin x cos x ≤ 0.5 < 0.6.
3. sin x cos x = N has a solution when |N | ≤ 1/2.
sin x cos x = N =⇒
1
sin 2x = N =⇒ sin 2x = 2N
2
2x = arcsin 2N gives one solution to the above equation. Another (not
necessarily distinct) solution is 2x = π − arcsin 2N . All solutions to the
equation can be generated by adding an integer multiple of 2π to either of
the above angles. Therefore, the general solution to sin x cos x = N is
x=
1
1
arcsin 2N + nπ or x = (π − arcsin 2N ) + nπ.
2
2
Using the properties of cofunctions, this solution can be written in a slightly
more compact manner as follows:
π 1
x=
± arccos 2N + nπ
4
2
106
Page 162 (First)
1.
√
1 1
1
3
sin 30 + cos 30 = +
> + =1
2
2
2 2
◦
◦
2.
sin 0 + cos 0 = 1
3.
√
√
2
2 √
π
π
sin + cos =
+
= 2
4
4
2
2
Page 162 (Second)
√
1. Yes, because 1.414 < 2. (Note: sin x + cos x must attain the value 1.414
for some x because of the Intermediate Value Theorem).
√
2. No, because 1.415 > 2.
3. We square both sides of the equation to solve for x.
√
π
π
sin x+cos x = 2 =⇒ 1+sin 2x = 2 =⇒ sin 2x = 1 =⇒ 2x = +2nπ =⇒ x = +nπ
2
4
2
4. Since we know the maximum value of (sin
the minimum
√ x + cos x) is 2, √
value of sin x + cos x cannot be less than − 2. The value of − 2 is attained
when x = 5π
4 .
√
√
√
5π
5π
2
2
sin
−
=− 2
+ cos
=−
4
4
2
2
Page 163 (First)
1. sin x + cos x is maximized when x + π4 = π2 + 2nπ, where n is an integer.
Therefore,
π
x = + 2nπ
4
√
2. The minimum value of sin x + cos x is − 2. This is achieved when
x=−
3π
+ 2nπ.
4
Page 163 (Second)
1. Yes. Because sin α and cos α are both positive, α must be in the first quadrant.
2. The minimum value of 3 sin x + 4 cos x is −5. This occurs when
π
x = −α − + 2nπ.
2
107
√
3. Let α be the positive acute angle such that sin α = 7/ 53.
√
√
7
2
2 sin x + 7 cos x = 53 √ sin x + √ cos x = 53 sin (α + x)
53
53
The√above shows
√ that the maximum and minimum values of 2 sin x + 7 cos x
are 53 and − 53, respectively.
Page 164
1.
cos
sin
π
=
16
v
u
u 1 + cos π
t
8
2
v
u
p
u 1 + 1 2 + √2
t
2
=
2
s
p
√
2+ 2+ 2
=
4
r
q
√
1
=
2+ 2+ 2
2
π
=
16
v
u
u 1 − cos π
t
8
2
v
u
p
u 1 − 1 2 + √2
t
2
=
2
s
p
√
2− 2+ 2
=
4
r
q
√
1
=
2− 2+ 2
2
2.
108
α
cos α
sin α
π
16
π
32
p
√
2+ 2+ 2
q
p
√
1
2+ 2+ 2+ 2
s2 r
q
p
√
1
2+ 2+ 2+ 2+ 2
v2
s
u
r
q
u
p
√
1t
2+ 2+ 2+ 2+ 2+ 2
2
p
√
2− 2+ 2
q
p
√
1
2− 2+ 2+ 2
s2 r
q
p
√
1
2− 2+ 2+ 2+ 2
v2
s
u
r
q
u
p
√
1t
2− 2+ 2+ 2+ 2+ 2
2
π
64
π
128
q
1
r2
Page 166
1.
1√
2 ≈ 0.7071
2
q
√
1
2 + 2 ≈ 0.9239
2
r
q
√
1
2 + 2 + 2 ≈ 0.9809
2
s
r
q
√
1
2 + 2 + 2 + 2 ≈ 0.9952
2
2.
22
r
3
2
s
2
4
q
2−
√
2 ≈ 3.0615
q
√
2 − 2 + 2 ≈ 3.1214
r
2−
q
2+
2+
√
2 ≈ 3.1365
v
s
u
r
u
q
√
t
5
2
2 − 2 + 2 + 2 + 2 ≈ 3.1403
109
1
r2
q
3. (a)
π
cos
=
12
r
1 + cos (π/6)
2
s
√
1 + 3/2
=
2
q
√
1
2+ 3
=
2
≈ 0.9659
(b)
r
1 + cos (π/12)
2
s
p
√
1 + 2 + 3/2
=
2
r
q
√
1
=
2+ 2+ 3
2
≈ 0.9914
π
=
cos
24
(c)
π
cos
=
48
r
1 + cos (π/24)
2
v
q
u
p
√
u
t 1 + 2 + 2 + 3/2
=
2
s
r
q
√
1
=
2+ 2+ 2+ 3
2
≈ 0.9979
(d)
π
cos
=
96
=
r
1 + cos (π/48)
2
v
r
u
q
u
p
u 1 + 2 + 2 + 2 + √3/2
t
2
v
s
u
r
u
q
√
1t
=
2+ 2+ 2+ 2+ 3
2
≈ 0.9995
110
These values approach 1 because for small angles x, cos x gets closer to
1. More formally, the limit of cos x as x approaches zero is 1.
4. (a)
q
√
π
12 sin
= 6 2 − 3 ≈ 3.1058
12
(b)
r
q
√
π
24 sin
= 12 2 − 2 + 3 ≈ 3.1326
24
(c)
s
π
48 sin
= 24 2 −
48
r
q
2+
2+
√
3 ≈ 3.1394
(d)
v
s
u
r
u
q
√
t
π
= 48 2 − 2 + 2 + 2 + 3 ≈ 3.1410
96 sin
96
Page 168
1.
99
√
X
1
1
1
1
√ +√
√ + ... + √
√
√
=
√
1+ 2
2+ 3
99 + 100 k=1 k + k + 1
=
99
X
√
k=1
=
99
X
√
k+
√
√
k+1− k
√
·√
k+1
k+1− k
k+1−
k=1
√
k
√
√ √
√ √ 1 +
3 − 2 + ... +
100 − 99
√
√
= 100 − 1
=
√
1
√
2−
= 10 − 1
=9
111
2.
1 + 3 + . . . + (2n + 1) =
=
n
X
k=0
n
X
2k + 1
2
(k + 1) − k 2
k=0
2
= 1 − 02 + 22 − 12 + . . . + (n + 1)2 − n2
2
= (n + 1) − 02
= (n + 1)
2
= n2 + 2n + 1
3.
(1 − x) P = (1 − x) (1 + x) 1 + x2 1 + x4 1 + x8 1 + x16
= 1 − x2 1 + x2 1 + x4 1 + x8 1 + x16
= 1 − x4 1 + x4 1 + x8 1 + x16
= 1 − x8 1 + x8 1 + x16
= 1 − x16 1 + x16
= 1 − x32
Thus,
P =
1 − x32
= 1 + x + . . . + x31
1−x
4.
sin 20◦ P = sin 20◦ cos 20◦ cos 40◦ cos 80◦
1
= sin 40◦ cos 40◦ cos 80◦
2
1
= sin 80◦ cos 80◦
4
1
= sin 160◦
8
1
= sin 20◦
8
Thus, P = 81 .
112
Page 170
1.
2 sin x (sin x + sin 3x + . . . + sin 99x) = 2 sin x
49
X
sin (2k + 1) x
k=0
=
49
X
2 sin x sin (2k + 1) x
k=0
=
49
X
cos 2kx − cos (2k + 2)x
k=0
= (cos 0 − cos 2x) + (cos 2x − cos 4x) + . . . + (cos 98x − cos 100x)
= 1 − cos 100x
Thus, the original sum, sin x+. . .+sin 99x, is equal to (1 − cos 100x) / (2 sin x).
Notice that this is equivalent to sin2 50x/ sin x, which is what would be found
using the formula at the bottom of page 169.
2.
2 sin
99
99π
πX
π
π
sin x + . . . + sin x +
= 2 sin
sin x + k
8
4
8
4
k=0
99
X
π
π
sin x + k
8
4
k=0
99
X
1 π
1 π
=
cos x + k −
− cos x + k +
2 4
2 4
k=0
π
199π
= cos x −
− cos x +
8
8
π
7π
= cos x −
− cos x +
8
8
π
7π
= cos x −
+ cos π − x −
8
8
π
= 2 cos x −
8
π
π
= 2 cos x cos + 2 sin x sin
8
8
=
2 sin
Dividing by 2 sin π8 , we find that the original sum is equal to cot π8 cos x +
sin x. Notice that this is equivalent to sin (x + 99π/8) / sin (π/8), which is
what would be found using the formula at the bottom of page 169.
An alternative way to evaluate this sum is by noticing
that the terms of this
sum repeat in periods of 8 because sin x + 8π
=
sin
(x).
Furthermore, the
4
113
first 8 terms of the sum total to zero (try verifying this yourself using the
sine addition formulas). Therefore, to evaluate the whole sum, all we need
to evaluate is the sum of the last four terms:
96π
97π
98π
99π
sin x +
+ sin x +
+ sin x +
+ sin x +
.
4
4
4
4
By the periodicity in the sum, we can equivalently evaluate the first four
terms of the sum. Using the sine addition formulas, we find that the sum is
equal to
π
3π
π
+ sin x +
+ sin x +
sin x + sin x +
4
2
4
√
√
√
√
2
2
2
2
= sin x +
sin x +
cos x + cos x −
sin x +
cos x
2
2
2
2
√
= sin x + cos x + 2 cos x.
Combining this
√ result with our initial evaluation of this sum shows that
cot π8 = 1 + 2.
3. We begin by multiplying the sum by 2 sin x.
2 sin x (cos 2x + cos 4x + . . . + cos 2nx) =
=
n
X
k=1
n
X
2 sin x cos 2kx
sin (2k + 1)x − sin (2k − 1)x
k=1
= (sin 3x − sin x) + . . . + (sin (2n + 1)x − sin (2n + 1)x)
= sin (2n + 1)x − sin x
Dividing by 2 sin x gives us the value of the sum as
sin(2n + 1)x 1
− .
2 sin x
2
π
4. We begin by multiplying the sum by 2 sin 2k
.
X
n
π
π
2π
nπ
jπ
π
2 sin
cos + cos
+ . . . + cos
=
2 sin
cos
2k
k
k
k
2k
k
j=1
n
X
π
1
π
1
=
sin
j+
− sin
j−
k
2
k
2
j=1
nπ
π
π
= sin
+
− sin
k
2k
2k
nπ
π
nπ
π
π
= sin
cos
+ cos
sin
− sin
k
2k
k
2k
2k
π
, we find that the value of the sum is
Dividing by 2 sin 2k
1
nπ
π
nπ
sin
cot
+ cos
−1 .
2
k
2k
k
114
5. The heights of the perpendiculars are given by sin kπ
12 , where k ranges from
π
1 to 11. We consider this sum multiplied by 2 sin 24
.
11
X
k=1
11
2 sin
kπ X
π
sin
=
cos
24
12
k=1
π
kπ
−
12
24
− cos
kπ
π
+
12
24
π
23π
− cos
24
24
π
= 2 cos
24
= cos
π
π
Dividing this result by 2 sin 24
gives the value of the sum as cot 24
.
Alternatively, we can use the formula for series of sines with angles in arithmetic progression, setting x = 0, n = 11, α = π/12 to find that the sum of
the altitudes is
sin
11π
11π
π
6π
sin
sin
cos
12
24 =
24 =
24 = cot π .
π
π
π
24
sin
sin
sin
24
24
24
Using
half-angle formulasp
for sine and cosine, we can show that cos π/12 =
p the
√
√
1
1
2 + 3 and sin π/12 = 2 2 − 3. Therefore,
2
π
1 + cos
π
12
cot
=
π
24
sin
12
√
1p
2+ 3
1+
2
=
√
1p
2− 3
2p
√
2+ 2+ 3
= p
√
2− 3
s
√
2
2+ 3
√
=p
√ +
2− 3
2− 3
p
√
√
2 2− 3
√
=
+2+ 3
2− 3
q
√ √ √
=2 2− 3 2+ 3 +2+ 3
q
√
√
=2 2+ 3+2+ 3
r
r !
√
1
3
=2
+
+2+ 3
2
2
√
√
√
= 2 + 6 + 2 + 3.
115
p
p
√
√
√
Let’s
elaborate on how to evaluate 2 + 3. Suppose 2 + 3 = a +
√
b for
numbers a and b. Then, squaring both sides, we get
√
√ some rational
2 + 3 = a + 2 ab + b. Matching the rational and irrational parts, we get
two equations relating a and b: a + b = 2 and 4ab = 3. Solving this system
shows that a and b are 1/2 and 3/2 (the order doesn’t matter because the
equations are symmetric). Thus, we have shown that
r
r
q
√
1
3
2+ 3=
+
.
2
2
Chapter 8: Graphs of Trigonometric Functions
Page 177
1. Since k = 5, Period is 5, and Frequency is
π
5
2. Since k = 14 , Period is 14 , and Frequency is
π
1/4
= 4π
3. Since k = 45 , Period is 45 , and Frequency is
π
4/5
=
5π
4
4. Since k = 45 , Period is 54 , and Frequency is
π
5/4
=
4π
5
5. Period is
2π
3
y
1
0
π
2
x
π
3π
2
2π
2π
3π
4π
−1
6. Period is
2π
1/3
= 6π
y
1
0
1π
−1
7. Period is
2π
3/2
=
4π
3
116
5π
x
6π
y
1
x
2π
0
1π
−1
8. Period is
2π
2/3
= 3π
y
1
x
0
1π
2π
3π
2π
3π
−1
9. Period is
2π
2/3
= 3π
y
1
0
1π
4π
−1
10. Period is
2π
3/2
=
4π
3
y
1
x
2π
0
1π
−1
11. For y = f (3x)
117
5π
x
6π
y
1
x
0
−1
0
1
2
3
4
5
6
1
2
3
4
5
6
−1
For y = f ( x3 )
y
1
x
0
−1
0
−1
Page 179
1. y = 2 sin (x)
y
2
1
0
−1
−2
2. y =
1
2
1π
x
2π
1π
x
2π
sin (x)
y
1
0
−1
3. y = 3 sin (2x)
118
y
3
2
1
0
−1
−2
−3
4. y =
1
2
1π
x
2π
1π
x
2π
1π
x
2π
sin (3x)
y
1
0
−1
5. y = 4 cos (x)
y
4
3
2
1
0
−1
−2
−3
−4
6. For y = 3f (x)
y
3
2
1
0
−1
0
−1
−2
−3
x
1
2
3
4
7. For y = 31 f (x)
119
5
6
y
1
x
0
−1
0
1
2
3
4
5
6
−1
Page 181
1. y = sin x −
π
6
y
1
1π
x
2π
1π
x
2π
1π
x
2π
0
−1
2. y = sin x +
π
6
y
1
0
−1
3. y = 2 sin x −
π
2
y
2
1
0
−1
−2
4. y =
1
2
sin x +
π
2
120
y
1
1π
x
2π
1π
x
2π
1π
x
2π
1π
x
2π
0
−1
5. y = cos x −
π
4
y
1
0
−1
6. y = 3 cos x +
π
3
y
3
2
1
0
−1
−2
−3
7. y = sin (x − 2x)
y
1
0
−1
8. Since the sine wave has shifted right by π3 . The equation is
π
y = sin x −
3
121
9. Since the sine wave has shifted right by 2π
3 . The equation is
2π
y = sin x −
3
10. Since the sine wave has shifted left by π6 . The equation is
π
y = sin x −
6
11. Note: this looks like the sine wave has shifted right by
the sine wave is on its wave down.
π
4,
but at that point,
The actual shift is the x intercept on the far left, calculate its location by
π
3π
4 − π = − 4 as it’s half a period away, that is, if 2π radians is one period,
then π radians is a half period
Therefore, the sine wave has shifted left by 3π
4 . The equation is
3π
y = sin x +
4
Page 183
1. y = sin 12 x −
π
6
y
1
2π
3π
x
4π
2π
3π
x
4π
0
1π
−1
2. y = sin
1
3x
−
π
6
y
1
0
1π
−1
3. y = cos 2 x +
π
3
122
y
1
0
1π
2π
3π
x
4π
−1
4. Since period is π, making k =
2π
π
= 2, making the equation
y = sin 2(x − β)
To solve for β, substitute x = 0, y = −1, as the sine wave passes point
(0, −1), giving
−1 = sin 2(0 − β)
−1 = sin −2β
π
− = −2β
2
π
=β
4
π
β=
4
Therefore the equation is
π
y = sin 2 x −
4
5. Since half period is 2π, its full period is 4πmaking k =
equation
1
y = sin (x − β)
2
To solve for β, substitute x =
giving
5π
6 ,y
2π
4π
= 12 , making the
= 1, as the sine wave passes point ( 5π
6 , 1),
1 5π
1 = sin
−β
2 6
π
1 5π
=
−β
2
2 6
5π
π=
−β
6
π
= −β
6
π
β=−
6
Therefore the equation is
y = sin
1
π
x−
2
6
123
Page 186
1. Answer A) equal to sin x since
sin(x + 2π) = sin(x + 2π − 2π)
= sin(x)
2. Answer C) equal to − sin x since
sin(x + 3π) = sin(x + 3π − 2π)
= sin(x + π)
= − sin(x)
3. Answer B) equal to cos x since
9
9
sin x + π = sin x + π − 4π
2
2
π
= sin x +
2
= cos(x)
4. Answer D) equal to − cos x since
π
sin x −
= − cos(x)
2
5. Answer B) equal to cos x since
3
3
sin x − π = sin − π + 2π
2
2
π
= sin x +
2
= cos(x)
6. Answer D) equal to − cos x since
19
19
sin x + π = sin x + π − 10π
2
2
π
= sin x −
2
= − cos(x)
124
7. Answer D) equal to − cos x since
19
19
− sin x − π = − sin x + π + 10π
2
2
π
= − sin x +
2
= − cos(x)
8. Answer B) equal to cos x since
157
157
sin x +
π = sin x +
π − 78π
2
2
π
= sin x +
2
= cos(x)
9. Answer D) equal to − cos x since
157
157
sin x −
π = sin x −
π + 78π
2
2
π
= sin x −
2
= − cos(x)
10. Note: there is no question 10 in textbook (printing error?)
11. Using definition that p is called a half-period of function f is f (x + p) =
−f (x), for all values of x which f (x) and f (x + p) are defined
Since cos(x) has period of 2π radians, therefore its half-period is π radians,
or we can compare diagrams
y
y = cos(x)
1
0
1π
2π
3π
−1
x
4π
y = cos(x + π)
Which can be seen that cos(x + π) is the same as − cos(x), therefore by
definition, cos x has a half-period of π radians
125
For y = tan x
sin(x + π)
cos(x + π)
− sin x
=
− cos x
= tan x
tan(x + π) =
So π is period of y = tan x
For y = cot x
cos(x + π)
sin(x + π)
− cos x
=
− sin x
= cot x
cot(x + π) =
So π is period of y = cot x
12. Using definition that q is called a half-period of function f is f (x + q) =
−f (x), for all values of x which f (x) and f (x + q) are defined
f (x + 2q) = f (x + q + q)
= −f (x + q)
= − (−f (x))
= f (x)
Therefore if q is half period of some function f , then 2q is a period of f
13. (a)
π(4n + 1)
cos (x + kπ/2) = cos x +
2
= cos (x + 2nπ + π/2)
= cos(x + π/2)
= − sin x
(b)
π(4n + 2)
cos (x + kπ/2) = cos x +
2
= cos (x + 2nπ + π)
= cos(x + π)
= − cos x
126
(c)
π(4n + 3)
cos (x + kπ/2) = cos x +
2
3π
= cos x + 2nπ +
2
3π
= cos(x +
)
2
π
= cos(x − )
2
= sin x
(d)
π(4n)
cos (x + kπ/2) = cos x +
2
= cos (x + 2nπ)
= cos x
14. (a) As the equation has a negative coefficient, we need to shift this by a
half-period of π radians
y = −2 sin x
= 2 sin(x + π)
Comparing coefficients, we get a = 2 and k = 1 and β = π
(b) As the equation has a negative coefficient, we need to shift this by a
half-period of π radians
y = 2 sin(x − π/3)
= 2 sin(x − π/3 + π)
= 2 sin(x + 2π/3)
Comparing coefficients, we get a = 2 and k = 1 and β = −2π/3
(c) As the equation has a negative coefficient, we need to shift this by a
half-period of π radians
y = −2 sin(x + π/4)
= 2 sin(x + π/4 + π)
= 2 sin(x + 5π/4)
Comparing coefficients, we get a = 2 and k = 1 and β = −5π/4
127
(d) As the equation is a cosine function, we need to re-express as a sine
function
3 cos x = 3 sin(x + π/2)
Comparing coefficients, we get a = 3 and k = 1 and β = π/2
(e) As the equation is a cosine function, we need to re-express as a sine
function
y = 3 cos(x − π/6)
= 3 sin(x − π/6 + π/2)
= 3 sin(x + π/3)
Comparing coefficients, we get a = 3 and k = 1 and β = −π/3
(f) As the equation is a cosine function, we need to re-express as a sine
function. Then re-express as a postive amplitude
y = −3 cos(x − π/6)
= −3 sin(x − π/6 + π/2)
= 3 sin(x − π/6 + π/2 + π)
= 3 sin(x + 4π/3)
Comparing coefficients, we get a = 3 and k = 1 and β = −4π/3
15.
cos(x − π/5)
y
1
0
1π
2π
3π
−1
π
2
3π
2
16. We need to shift to the right by
We need to shift to the left by
128
x
4π
17. To express k as an odd number, let k = 2n + 1 for any integer n
π(2n + 1)
tan (x + kπ/2) = tan x +
2
= tan (x + 2nπ + π/2)
= tan (x + π/2)
sin(x + π/2)
cos(x + π/2)
− cos x
=
sin x
= − cot x
=
To express k as an even number, let k = 2n for any integer n
π(2n)
tan (x + kπ/2) = tan x +
2
= tan (x + 2nπ)
= tan x
Page 188
1. a) y = tan x −
π
6
y
3
2
1
0
−1
−2
−3
2π
3
b) y = 3 tan (x)
y
3
2
1
0
π
2
−1
−2
−3
c) y = cot x +
π
4
2π
3
1π
1π
3π
2
129
x
2π
x
2π
y
3
2
1
0
−1
−2
−3
3π
4
7π
4
1π
x
2π
2. Not really, although there is a trigonometric identity where
π
tan x = cot
−x
2
This equation has a negative x inside. Therefore it won’t be in form of
y = cot(x + ϕ) (note the positive x inside)
Page 189
1. (a) For y = y1 + y2
y
4
3
2
1
0
−1
−2
−3
−4
x
2π
1π
(b) For y = y1 + y2
130
y
4
3
2
1
0
−1
−2
−3
−4
1π
x
2π
1π
x
2π
(c) For y = y2 + y3
y
4
3
2
1
0
−1
−2
−3
−4
2. Graph 1a appear to be a sinusoidal function
Page 191
1. For y = 2 sin x+3 cos x, compare with standard form y = A sin kx+B cos kx,
we get values of A = 2, B = 3, and k = 1
To convert to the other standard form y = a sin k(x + ϕ), we calculate its
amplitude by
p
a = A2 + B 2
p
= 22 + 3 2
√
= 13
131
While its phase ϕ = α/k is calculated by
cos α = √
A
+ B2
A2
2
cos α = √
13
√
k=1
α = arccos(2/ 13)Since
√
ϕ = arccos(2/ 13)
√
√
Therefore the equation is y = 2 sin (x + ϕ) where ϕ = arccos 2 13
√
√
Since amplitude is 2, then its maximum value is 2
√
√
2. Since amplitude is 13, then its maximum value is 13
3. For y = sin x + cos x, compare with standard form y = A sin kx + B cos kx,
we get values of A = 1, B = 1, and k = 1
To convert to the other standard form y = a sin k(x + ϕ), we calculate its
amplitude a by
p
a = A2 + B 2
p
= 12 + 12
√
= 2
While its phase ϕ = α/k is calculated by
cos α = √
A
+ B2
A2
1
cos α = √
2
α = π/4
Since k = 1
ϕ = π/4
√
Therefore the equation is y = 2sin (x + π/4)
√
√
Since amplitude is 2, then its maximum value is 2
4. For y = sin x − cos x, compare with standard form y = A sin kx + B cos kx,
we get values of A = 1, B = 1, and k = 1
To convert to the other standard form y = a sin k(x − β), we calculate its
amplitude by
p
a = A2 + B 2
p
= 12 + 12
√
= 2
132
While its phase β = α/k is calculated by
cos α = √
A
+ B2
A2
1
cos α = √
2
α = π/4
Since k = 1
β = π/4
√
Therefore the equation is y = 2 sin (x − π/4)
√
√
Since amplitude is 2, then its maximum value is 2
5. For y = 4 sin x+3 cos x, compare with standard form y = A sin kx+B cos kx,
we get values of A = 4, B = 3, and k = 1
To convert to the other standard form y = a sin k(x + ϕ), we calculate its
amplitude by
p
a = A2 + B 2
p
= 42 + 32
√
= 25
=5
While its phase ϕ = α/k is calculated by
cos α = √
cos α =
A
+ B2
A2
4
5
4
α = arccos( )
5
Since k = 1
4
ϕ = arccos( )
5
Therefore the equation is y = 5 sin (x + ϕ) where ϕ = arccos(4/5)
Since amplitude is 5, then its maximum value is 5
6. For y = sin 2x+3 cos 2x, compare with standard form y = A sin kx+B cos kx,
we get values of A = 1, B = 3, and k = 2
133
To convert to the other standard form y = a sin k(x + ϕ), we calculate its
amplitude by
p
a = A2 + B 2
p
= 12 + 32
√
= 10
While its phase ϕ = α/k is calculated by
cos α = √
A
+ B2
A2
1
cos α = √
10
√
α = arccos(1/ 10)
Since k = 2
1
1
ϕ = arccos( √ )
2
10
Therefore the equation is y =
√
10 sin (x + ϕ) where ϕ =
1
2
arccos( √110 )
7. For y = sin(x − π/4), compare with standard form y = a sin k(x − β), we get
values of a = 1, β = π/4, and k = 1
To convert to the other standard form y = A sin kx − B cos kx, we need to
solve for A and B using the definition A = a cos kβ and B = a sin kβ
A = 1 cos(π/4)
1
=√
2
B = 1 sin(π/4)
1
=√
2
Therefore the equation is y =
√1
2
sin x −
√1
2i
cos x
8. For y = 4 sin(x + π/6), compare with standard form y = a sin k(x + ϕ), we
get values of a = 4, ϕ = π/6, and k = 2
To convert to the other standard form y = A sin kx + B cos kx, we need to
134
solve for A and B using the definition A = a cos kβ and B = a sin kβ
A = 4 cos(2π/6)
= 4 × 1/2
=2
B = 4 sin(2π/6)
√
= 4 × 3/2
√
=2 3
√
Therefore the equation is y = 2 sin 2x + 2 3 cos 2x
Page 194
1. (a)
y = 2 sin(x + π/6) + cos(x + π/6)
y = 2 (sin x cos π/6 + cos x sin π/6) + (cos x cos π/6 − sin x sin π/6)
!
!
√
√
3
1
3
1
sin x + cos x +
cos x − sin x
y=2
2
2
2
2
√
√
1
3
cos x − sin x
y = 3 sin x + cos x +
2
2
√ !
√
1
3
y=
3−
cos x
sin x + 1 +
2
2
(b)
y = 2 sin 2(x + π/4) − cos 2(x + π/4)
y = 2 sin(2x + π/2) − cos(2x + π/2)
y = 2 (sin 2x cos π/2 + cos 2x sin π/2) + (cos 2x cos π/2 − sin 2x sin π/2)
y = 2 (sin 2x × 0 + cos 2x × 1) + (cos 2x × 0 − sin 2x × 1)
y = 2 cos 2x − sin 2x
y = − sin 2x + 2 cos 2x
2. (a)
y1 + y1 = 2 sin x + sin(x − π/4)
= 2 sin x + (sin x cos π/4 − cos x sin π/4)
1
1
= 2 sin x + √ sin x − √ cos x
2
2
1
1
= 2+ √
sin x − √ cos x
2
2
(b) The functions have different frequencies, which would need advanced
mathematics topics such as Fourier Series
135
Page 196
1.
y = −x + sin x
y
5
4
3
2
1
0
−1
−2
−3
−4
−5
1π
x
2π
2.
y = x2 + sin x
y
2
1
x
- π2
π
2
3.
y = x2 + cos x
Note this is an even function since f (−x) = f (x)
y
2
1
x
π
2
- π2
136
4.
y = x3 + sin x
y
5
4
3
2
1
0
−1
−2
−3
−4
−5
- π2
π
2
5.
y = x2 +
x
1
sin x
10
y
2
1
x
- π2
π
2
6.
y = cos x +
2
1
−2π
−1π
−1
−2
1
sin 20x
10
y
1π
7.
y = 2 sin x +
137
x
2π
1
sin 20x
10
y
2
1
−2π
−1π
−1
−2
1π
x
2π
Page 197
1. In y = sin 2x + sin 3x. The period of sin 2x has form of m(2π/2), while the
period of sin 3x has form of n(2π/3).
To be a period of both parts
m×
2π
2π
=n×
2
3
3m = 2n
Using m = 2 and n = 3. the period of each part will be 2π radians
2. In y = sin 3x + sin 6x. The period of sin 3x has form of m(2π/3), while the
period of sin 6x has form of n(2π/6).
To be a period of both parts
m×
2π
2π
=n×
3
6
6m = 3n
2m = 1n
Using m = 1 and n = 2. the period of each part will be 2π/3 radians
3. In y = sin 4x + sin 6x. The period of sin 4x has form of m(2π/4), while the
period of sin 6x has form of n(2π/6).
To be a period of both parts
m×
2π
2π
=n×
4
6
6m = 4n
3m = 2n
Using m = 2 and n = 3. the period of each part will be π radians
4.
√
√
√
√
5. In y = sin 2x + sin 3 √
2x. The period of sin √
2x has form of m(2π/ 2),
while the period of sin 3 2x has form of n(2π/3 2).
138
To be a period of both parts
2π
2π
m× √ =n× √
2
3 2
3m = 1n
√
√
Using m = 1 and n = 3. the period of each part will be 2π/ 2 = 2π
radians
Page 199
1.
y = sin x −
1
1
sin 2x + sin 3x
2
3
y
1
−2π
−1π
1π
x
2π
−1
2.
y = sin x −
1
1
1
1
sin 2x + sin 3x − sin 4x + sin 5x
2
3
4
5
y
1
−2π
−1π
1π
x
2π
−1
3. The limit will be a sawtooth wave
y
1
−2π
−1π
1π
−1
139
x
2π
4.
y = cos x −
1
1
1
cos 3x +
cos 5x −
cos 7x
9
25
49
y
1
−2π
−1π
t
2π
1π
−1
Therefore, the limiting function is the triangular function
y
1
−2π
−1π
x
2π
1π
−1
Page 200
1. The graph will be
y
1
t
1
2
3
4
5
6
2. The graph will be
y
1
t
1
2
3
4
5
−1
3. The graph will be
140
6
y
8
8
16
24
32
t
4. The graph will be
40 y
4
6
12
18
24
t
Tide is rising fastest at 3rd, 4th, 15th, and 16th hour
Tide is receding fastest at 9th, 10th, 21st, and 22nd hour
Tide is rising slowest at 1st, 5th, 13th and 17th hour
Tide is receding slowest at 7th, 11th, 19th and 23rd hour
Page 203
1. Substituing x values
• sin π = 0
• sin 2π = 0
• sin 3π = 0
• sin 4π = 0
2. For y = sin 4πx, the value of k is k = 4π, therefore the period is
period is 21
3. Since period is
2π
k ,
if period is 3
2π
k
2π
k=
3
3=
141
2π
4π ,
so the
Therefore the equation is y = sin 2π
3 x
4. Since period is
2π
k ,
if period is 2
2π
k
2π
k=
2
2=
Therefore the equation is y = sin πx
5. Since period is
2π
k ,
if period is n
2π
k
2π
k=
n
n=
Therefore the equation is y = sin 2π
n x
Page 205
1. The period is estimated to be 12 months by distance between peaks at t = 8
and t = 20. Therefore the value for k is k = 2π/20
The amplitude is estimated to be mid-way between peak (15) and trough
(9.5), which is (15 − 9.5)/2 = 2.75
π
Therefore the equation is y = 2.75 sin 10
(x − π)
2. If (a) is in northern hemisphere, (b) and (c) would be in northern hemisphere too as the peaks occur at similar month. (d) would be in southern
hemispehere as the wave is shifted by half-period
3. For all graphs, the ’average’ number of daylight hours occur at month 4,
month 10, month 16, and month 22
Chapter 9: Inverse Functions and Trigonometric
Equations
Page 213
1. (a) arcsin 0.5 = π/6
(b) arccos 0.5 = π/3
(c) arctan 0.5 = π/4
√
(d) arcsin(− 3/2) = −π/3
√
(e) arccos(− 3/2) = 5π/6
√
(f) arctan(− 3) = −π/3
142
(g) Note: not valid as the domain of arcsin x is −1 ≤ x ≤ 1
2. (a)
sin(arcsin 0.5)
= sin(π/6)
= −0.5
(b)
cos(arccos 0.5)
= cos(π/3)
= 0.5
(c)
tan(arctan(−1))
= tan(π/4)
= −π/4
(d)
arcsin(sin(π/3))
√
= arcsin(− 3/2)
= π/3
(e)
arccos(cos 11π/6)
√
= arccos(− 3/2)
= π/6
3. Let θ = arccos b which mean cos θ = b. If we assume angle θ is acute, then
we can draw a right angle triangle
1
√
1 − b2
θ
b
Giving sin θ =
√
1 − b2 when 0 ≤ θ ≤ π/2
For π/2 ≤ θπ, sin θ = sin(π − θ), making the sign positive
For π ≤ θ3π/2, sin θ = − sin(θ − π), making the sign negative
For 3π/2 ≤ θ2π, sin θ = − sin(2θ − θ), making the sign negative
143
4. Let θ = arcsin b which mean sin θ = b. If we assume angle θ is acute, then
we can draw a right angle triangle
1
θ
√
b
1 − b2
Giving tan θ =
√ b
1−b2
when 0 ≤ θ ≤ π/2
For π/2 ≤ θπ, tan θ = − tan(π − θ), making the sign negative
For π ≤ θ3π/2, tan θ = tan(θ − π), making the sign positive
For 3π/2 ≤ θ2π, tan θ = − tan(2θ − θ), making the sign negative
5. Let θ = arctan b which mean tan θ = b. If we assume angle θ is acute, then
we can draw a right angle triangle
√
1 + b2
b
θ
1
Giving cos θ =
√ 1
1+b2
when 0 ≤ θ ≤ π/2
For π/2 ≤ θπ, cos θ = − cos(π − θ), making the sign negative
For π ≤ θ3π/2, cos θ = − cos(θ − π), making the sign negative
For 3π/2 ≤ θ2π, cos θ = cos(2θ − θ), making the sign positive
6. Since angle α is acute, then we can draw a right angle triangle
c
β
a
α
b
Note that sin α = a/b and cos β = a/b, and β = π/2 − α
sin α = cos β
arccos(sin α) = β
arccos(sin α) = π/2 − α
For π/2 ≤ α ≤ π, the expression sin α becomes sin(π − α), therefore making
arccos(sin α) = arccos(sin(π − α))
= π/2 − (π/2 − α)
=α
144
For π ≤ α ≤ 3π/2, the expression sin α becomes − sin(α − π), therefore
making
arccos(sin α) = arccos(− sin(α − π))
= arccos(sin(−α + π))
= π/2 − (−α + π)
= α − π/2
For 3π/2 ≤ α ≤ 2π, the expression sin α becomes − sin(2π − α), therefore
making
arccos(sin α) = arccos(− sin(2π − α))
= arccos(sin(−2π + α))
= arccos(sin(α))
= π/2 − α
7. Note that the angles for parts (a) to (e) are less than π/2, so no change in
angle is needed, unlike (f)
(a) arcsin(sin π/11) = π/11
(b) arcsin(sin 2π/11) = 2π/11
(c) arcsin(sin 3π/11) = 3π/11
(d) arcsin(sin 4π/11) = 4π/11
(e) arcsin(sin 5π/11) = 5π/11
(f)
arcsin(sin 6π/11)
= arcsin(sin(π − 6π/11))
= arcsin(sin 5π/11)
= 5π/11
8. Graph of y = cos(arccos x)
145
y
1
x
−1
1
9. Graph of y = arccos(cos x)
y
1π
−2π
−1π
1π
−1π
10. Letting α = arcsin 3/5 and β = arcsin 5/13. Then drawing the triangles
5
3
α
4
13
5
β
12
146
x
2π
sin(α + β) = sin α cos β + cos α sin β
= 3/5 × 12/13 + 4/5 × 5/13
= 56/65
11. Letting α = arctan a and β = arctan b. Therefore we have tan α = a amd
tan β = b
tan α + tan β
1 − tan α tan β
a+b
=
1 − ab
tan(α + β) =
12. Let tan α = 1/3 and tan β = 1/2 and tan γ = 1
tan α + tan β
1 − tan α tan β
1/3 + 1/2
tan(α + β) =
1 − 1/3 × 1/2
tan(α + β) = 1
tan(α + β) =
But Since both tan(α + β) = 1 and tan γ = 1. Therefore α + β = γ
Note: I think the author meant Question 11 instead of ’Problem 8’
13. Using plane geometry, draw another row of squares underneath the diagram
and draw lines AH and HC.
Note that the lengths of AH and HC are the same, making triangle AHC
an isosceles triangle, therefore ̸ CAH = ̸ ACH
Since triangle AHG and HEC is congruent with triangle BCD, ̸ BAH = β,
and ̸ HCE = β
Since triangle F CA is congruent with triangle ACD, ̸ F CA = α
Therefore
̸
F CA + ̸ ACH + ̸ HCE = 90◦
α + ̸ ACH + β = 90◦
̸
ACH = 90◦ − α − β
Since ̸ γ = 45◦ , ̸ CAH = α + β and ̸ CAH = ̸ ACH
α + β = 90◦ − α − β
2(α + β) = 90◦
α + β = 45◦
α+β =γ
147
E
C
α
A
γ
β
B
D
G
H
Note: I think the author meant Question 12 instead of ’Problem 9’
Page 220
1. From the graph, sin x > 1/2 when π/6 ≤ x ≤ 5π/6 and 13π/6 ≤ x ≤ 17π/6
2. Using the general equation πk + (−1)k (α) for any integer k
For sin x = −1/2, the value for α is arcsin(−1/2) = −π/6, so the solution is
πk + (−1)k (−π/6)
3. Using the general equation 2nπ ± (α) for any integer n
√
√
For cos x = 2/2, the value for α is arccos( 2/2) = π/4, so the solution is
2πn ± π/4
4. Using the general equation nπ + (α) for any integer n
For tan x = 1, the value for α is arctan(1) = π/4, so the solution is πn + π/4
5. Using the general equation πk + (−1)k (α) for any integer k
For sin x = −1, the value for α is arcsin(−1) = −π/2, so the solution is
πk + (−1)k (−π/2)
Page 221
1. Using the general equation πk + (−1)k (α) for any integer k
For sin x = sin π/5, the value for α is π/5, so the solution is πk + (−1)k (π/5)
2. Using the general equation πk + (−1)k (α) for any integer k
For sin x = sin π/2, the value for α is π/2, so the solution is πk + (−1)k (π/2)
148
y
1
cos α
0
α
1π 2π − α 2π
3π
4π
5π
x
6π
−1
3.
Since period of cos x is 2π, the solutions are 2πn + α and 2πn − α for any
integer n
4. Using the general equation 2nπ ± (α) for any integer n
For cos x = cos π/5, the value for α is π/5, so the solution is 2πn ± π/5
5. Since period of tan x is π, then if α is a solution, then the general equation
for solutions are α + πn for any integer n
y
3
2
1
x
0
π
α
3π
1πα + π
2π
2
2
−1
−2
−3
6. Using the general equation nπ + (α) for any integer n
For tan x = tan π/5, the value for α is π/5, so the solution is πn + π/5
7. Using the general equation πk + (−1)k (α) for any integer k
sin x = − sin α
sin x = sin(−α)
x = πk + (−1)k (−α)
8. Using the general equation 2nπ ± (α) for any integer n
cos x = − cos α
cos x = cos(π − α)
x = 2nπ ± (π − α)
9. Using the general equation πk + (−1)k (α) for any integer k
sin x = sin α
sin x = sin(α)
x = πk + (−1)k (α)
149
Page 225
1. Using the general equation πk + (−1)k (α) for any integer k
sin 2x = 1
2x = πk + (−1)k × arcsin 1
2x = πk + (−1)k × π/2
1
x=
πk + (−1)k × π/2
2
2. Using the general equation πk + (−1)k (α) for any integer k
sin x/2 = 1/2
x/2 = πk + (−1)k × arcsin 1/2
x/2 = πk + (−1)k × π/6
x = 2 πk + (−1)k × π/6
3.
cos x = sin 2x
cos x = 2 sin x cos x
0 = 2 sin x cos x − cos x
0 = cos x (2 sin x − 1)
Therefore there are two set of solutions. Using general solutions 2nπ ± (α)
cos x = 0
x = 2nπ ± (arccos 0)
x = 2nπ ± (π/2)
and πk + (−1)k (α)
2 sin x − 1 = 0
sin x = 1/2
x = πk + (−1)k (arcsin 1/2)
x = πk + (−1)k (π/6)
Note: I think the question is supposed to be cos x = sin2 x?
150
4.
sin x = sin 3x
sin x = sin(2x + x)
sin x = sin 2x cos x + cos 2x sin x
sin x = 2 sin x cos x cos x + (cos2 x − sin2 x) sin x
sin x = 2 sin x(1 − sin2 x) + (1 − sin2 x − sin2 x) sin x
sin x = 2 sin x − 2 sin3 x + sin x − 2 sin3 x
0 = −4 sin3 x + 2 sin x
0 = −2 sin x(2 sin2 −1)
Therefore there are multiple sets of solutions to solve and using the general
equation πk + (−1)k (α)
sin x = 0
x = πk + (−1)k (arcsin 0)
x = πk
The other sets of solutions are
2 sin2 x − 1 = 0
sin2 x = 1/2
√
sin x = ±1/ 2
√
For sin x = 1/ 2
√
x = πk + (−1)k (arcsin 1/ 2)
x = πk + (−1)k × π/4
√
For sin x = −1/ 2
√
x = πk + (−1)k (arcsin −1/ 2)
x = πk + (−1)k × −π/4
Note: I think the questions is supposed to be sin x = sin3 x?
151
5.
cos x = sin 4x
cos x = 2 sin 2x cos 2x
cos x = 2 × 2 sin x cos x × (cos2 x − sin2 x)
cos x = 4 sin x cos x × (1 − sin2 x − sin2 x)
0 = 4 sin x cos x × (1 − 2 sin2 x) − cos x
0 = cos x 4 sin x(1 − 2 sin2 x) − 1
0 = cos x −8 sin3 x + 4 sin x − 1
Therefore there are multiple sets of solutions to solve and using the general
equation πk + (−1)k (α) and 2nπ ± (α)
cos x = 0
x = 2nπ ± arccos 0
x = 2nπ ± π/2
Solving −8 sin3 x + 4 sin x − 1 = 0 requires using cubic functions, which is
advanced level and is beyond this book’s level
Note: I think the question is supposed to be cos x = sin 4x?
6.
26 sin2 x + cos2 x = 10
26 sin2 x + 1 − sin2 x − 10 = 0
25 sin2 x − 9 = 0
(5 sin x + 3)(5 sin x − 3) = 0
Therefore there are multiple sets of solutions to solve and using the general
equation πk + (−1)k (α)
For 5 sin x + 3 = 0
5 sin x + 3 = 0
sin x = −3/5
x = πk + (−1)k (arcsin −3/5)
For 5 sin x − 3 = 0
5 sin x − 3 = 0
sin x = 3/5
x = πk + (−1)k (arcsin 3/5)
152
7.
cos2 x − cos x = sin2 x
cos2 x − cos x = 1 − cos2 x
2 cos2 x − cos x − 1 = 0
(2 cos x + 1)(cos x − 1) = 0
Therefore there are multiple sets of solutions to solve and using the general
equation 2πn ± (α)
For 2 cos x + 1 = 0
2 cos x + 1 = 0
cos x = −1/2
x = 2πn ± arccos(−1/2)
x = 2πn ± 2π/3
For 2 cos x − 1 = 0
2 cos x − 1 = 0
cos x = 1/2
x = 2πn ± arccos(1/2)
x = 2πn ± π/3
8.
3 tan2 x = 12
tan2 x = 4
tan2 x − 4 = 0
(tan x + 2)(tan x − 2) = 0
Therefore there are multiple sets of solutions to solve and using the general
equation πk + (α)
For tan x + 2 = 0
tan x + 2 = 0
tan x = −2
x = πk + arctan −2
For tan x − 2 = 0
tan x − 2 = 0
tan x = 2
x = πk + arctan 2
153
9.
cos 2x = 2 sin2 x
cos2 x − sin2 x = 2 sin2 x
1 − sin2 x − sin2 x = 2 sin2 x
1 − 4 sin2 x = 0
(1 − 2 sin x)(1 + 2 sin x) = 0
Therefore there are multiple sets of solutions to solve and using the general
equation πk + (−1)k (α)
For 1 − 2 sin x = 0
1 − 2 sin x = 0
sin x = 1/2
x = πk + (−1)k arcsin 1/2
x = πk + (−1)k π/6
For 1 + 2 sin x = 0
1 + 2 sin x = 0
sin x = −1/2
x = πk + (−1)k arcsin −1/2
x = πk + (−1)k (−π/6)
10.
tan2 x = cot x
tan3 x = 1
tan3 x − 1 = 0
(tan x − 1)(tan2 x + tan x + 1) = 0
Therefore there are multiple sets of solutions to solve and using the general
equation πk + α
For tan x − 1 = 0
tan x − 1 = 0
tan x =1
x = πk + arctan 1
x = πk + π/4
154
For tan2 x + tan x + 1 = 0, apply quadratic formula
p
−1 ± 12 − 4(1)(1)
tan x =
2(1)
√
−1 ± −3
tan x =
2
Which does not have any solutiosn as the number inside the square root is
negative
11.
5
= 7 tan x + 3
cos2 x
5 sec2 x = 7 tan x + 3
5(tan2 x + 1) = 7 tan x + 3
5 tan2 x + 5 = 7 tan x + 3
5 tan2 x − 7 tan x + 2 = 0
Apply quadratic formula
tan x =
tan x =
7±
p
49 − 4(5)(2)
2(5)
7±3
10
Therefore there are multiple sets of solutions to solve and using the general
equation πk + α
For tan x =
7+3
10
tan x = 1
x = πk + arctan 1
x = πk + π/4
For tan x =
7−3
10
tan x = 2/5
x = πk + arctan 2/5
12.
√
√
3 tan2 x + 1 = (1 +
( 3 tan x + 1)(tan x + 1) = 0
155
√
3) tan x
Therefore there are multiple sets of solutions to solve and using the general
equation πk + α
√
For 3 tan x + 1 = 0
√
3 tan x + 1 = 0
√
3 tan x = −1
√
tan x = −1/ 3
√
x = πn + arctan −1/ 3
x = πn + (−π/6)
For tan x + 1 = 0
tan x + 1 = 0
tan x = −1
x = πn + arctan −1
x = πn + (−π/4)
13. For Solution 1, re-express the equations such that
with n
x = π/6 + 2πn/3
with n+1
= π/6 + 2π(n + 1)/3
= π/6 + 2πn/3 + 2π/3
= 5π/6 + 2πn/3
with n+2
= π/6 + 2π(n + 2)/3
= π/6 + 2πn/3 + 4π/3
= 9π/6 + 2πn/3
= 3π/2 + 2πn/3
Which corresponds to each line of Solution 2, provided that the number n for
Solution 1 is multiples of 3
Page 227
1. If α = 1.6 then
x = α − tan α
≈ 35.8325
156
Since near x = π/2, the tangent will be very flat and only reaches the x-axis
after travelling a long distance
2. If α = π/4 then
x = α − tan α
≈ −0.2146
Since the tangent is intersecting x axis to the left
Note, the smaller the α, the closer to it intersecting x-axis at x = 0
If α = 0.1 then
x = α − tan α
≈ −0.000334
3. To solve for location of R where x = 0 and noting x ≈ tan x for small x
value
x = α − tan α
0 = α − tan α
tan α = α
α=0
Therefore R will be at origin when we take a tangent at x = 0
4. if x = π/2, then the tangent will be horizontal, therefore not intersecting
with x-axis at all
If x is a bit smaller than π/2, then the tangent will intersect with x-axis to
the left (far away)
If x is a bit bigger than π/2, then the tangent will intersect with x-axis to
the right (far away)
Page 229
1. Using A ≈
h
sin h/2 ,
where h = π/n
Using n = 4, so h = π/4 gives A ≈
π/4
sin π/8 ,
Using n = 8, so h = π/8 gives A ≈
π/8
sin π/16 ,
which is 2.05234
which is 2.01291
2. Since area between 0 ≤ x ≤ π is approximately 2, threfore the area under
curve y = sin x from x = 0 to x = π/2 would be around 1
157
3.
h sin x1 + h sin x2 + h sin x3 + · · · + h sin xm
π
2π
nπ
= h sin + sin
+ · · · + sin
n
n
2π
=h
π
n π
sin n+1
2 2n sin 2 2n
π
sin 2n
=h
π
sin n+1
π
2 2n
sin
π
sin 2n
4
=h
π
sin n+1
n 4 1
√
π
sin 2n
2
if n is very large
sin π/4 1
√
sin π/2n 2
h
≈
2 sin h/2
≈h
letting n = 8, giving h = π/8, gives approximate area of
158
π/8
2 sin π/16
≈ 1.00645