FINAL EXAM SAMPLE - SOLUTIONS
2.2.28. Given cos t = 12 and sin t < 0, find all trignomoetric
functions of t.
√
Solution: We have sin t = ± 1 − cos2 t. Given the fact that
sin t < 0, we have
s
r
r
√
µ ¶2
3
3
1
1
sin t = − 1 −
=− 1− =−
=−
.
2
4
4
2
3.1.8. Verify the identity
tan2 α − sin2 α = tan2 α sin2 α.
Solution: We compute both sides in terms of sin α and cos α:
sin2 α
sin2 α cos2 α
sin2 α
− sin2 α =
−
=
cos2 α
cos2 α
cos2 α
2
2
2
2
2
sin α − sin α cos α
sin α(1 − cos α)
sin4 α
=
=
=
;
2
2
cos α
cos α
cos2 α
sin2 α
sin4 α
RHS =
· sin2 α =
= LHS.
cos2 α
cos2 α
LHS =
The other trigonometric functions are
√
tan t =
√
3
1
cot t = − √ = −
;
3
3
√
2 3
1
2
csc t =
= −√ = −
.
sin t
3
3
√
− 23
sin t
= − 3;
=
1
cos t
2
sec t =
1
= 2;
cos t
7
3.3.20. If α and β are acute angles, such that tan α = − 24
and
3
cot β = 4 , for a second quadrant angle α and a third quadrant
angle β, find:
2.4.4. Find the exact values of the six trigonometric functions
of θ if θ is in standard position
and´ the point P (−1, 2) is all
³
π
solutions of the equation cos x −
= −1.
3
Solution: First we compute the distance to the origin, given
x = −1, y = 2:
q
p
√
r = x2 + y 2 = (−1)2 + 22 = 5.
sin(α − β) = sin α cos β − cos α sin β;
cos(α − β) = cos α cos β + sin α sin β.
We need the values of sin α, cos α, sin β, and√cos β.
To find sin α and cos α we use sec α = ± 1 + tan2 α. Since
α is in the second quadrant, we have
s
r
r
µ
¶
625
25
7 2
49
sec α = − 1 + −
=−
=− .
=− 1+
24
576
576
24
¢
¡
7
·
This gives cos α = − 24
, so we get sin α = tan α · cos α = − 24
25
¡ 24 ¢
7
− 25 = 25 .
To find the trigonometric
p functions sin β and cos β, we start
with the identity csc β = ± 1 + cot2 β. Since β is in the third
quadrant, we have
s
r
µ ¶2
3
9
csc β = − 1 −
=
=− 1+
4
16
r
25
5
=−
=− .
16
4
Then we get sin β = csc1 β = − 54 , and cos β = sin β cot β =
¡ 4¢ 3
− 5 · 3 = − 35 . Having found the desired values, we go back to
the forumlas for sin(α+β), cos(α+β), sin(α−β) and cos(α−β),
and we compute:
¢ ¡ 24 ¢ ¡ 4 ¢
¡
7
21
96
75
sin(α + β) = 25
· − 5 = − 125
· − 53 + − 25
+ 125
= 125
= 35 ;
¢
¡ 24 ¢ ¡ 3 ¢
¡
7
72
28
cos(α + β) = − 25 · − 5 − 25
· − 45 = 125
+ 125
= 100
= 45 ;
125
¡ 3 ¢ ¡ 24 ¢ ¡ 4 ¢
7
21
96
117
sin(α − β) = 25 · − 5 − − 25 · − 5 = − 125 + 125 = 125 ;
¢
¡
¡ 24 ¢ ¡ 3 ¢
7
72
28
46
· − 5 + 25
· − 45 = 125
− 125
= 125
.
cos(α − β) = − 25
2.5.4(b) & 2.5.15(b). Find the reference angle θR for θ = 2π
,
3
.
and use it to find sec 2π
3
is is the second
Solution: The reference angle is π3 . Since 2π
3
quadrant, we have
π
3
(f ) tan(α − β).
sin(α + β) = sin α cos β + cos α sin β;
Now we compute the trigonometric functions:
√
opp
10
adj
1
sin θ =
=
;
cos θ =
= ;
hyp
3
hyp
3
√
√
√
opp
10
10
1
adj
tan θ =
==
=
= 10; cot θ =
= √
;
adj
1
opp
10
10
√
3 10
hyp
3
hyp
3
=
sec θ =
= = 3;
csc θ =
= √
.
adj
1
opp
10
10
= − sec
(c) tan(α + β);
(e) cos(α − β);
cos(α + β) = cos α cos β − sin α sin β;
θ
3
2.4.22. Find the values of
1
the six trigonometric functions
of the angle θ
Solution: First we find the opposite side, using Pythogoras
Theorem.
q
p
√
opp = (hyp)2 − (adj)2 = 32 − 12 = 10.
2π
3
(b) cos(α + β);
(d) sin(α − β);
Solution: We use the formulas
Now we compute the trigonometric functions:
√
√
y
5
2 5
2
x
−1
sin θ = = √ =
;
cos θ = = √ = −
;
r
5
r
5
5
5
y
2
x
−1
1
tan θ = =
= −2;
cot θ = =
=− ;
x
−1
y
2
2
√
√
√
r
5
5
r
sec θ = =
csc θ = =
= − 5;
.
x
−1
y
2
sec
(a) sin(α + β);
= −2.
1
¡
¢¤
¡
¢
£
12
;
3.6.22. Compute (a) sin 12 cos−1 − 53 ; (b) cos 12 sin−1 13
¢
¡1
40
−1
.
(c) tan 2 tan
9
¡
¢
Solution: (a) Put α = cos−1 − 53 , so that what we need to
compute is
r
1 − cos α
sin(α/2) = ±
.
2
The angle α has cos α = − 35 , and satisfies 0 ≤ α ≤ π. This
gives 0 ≤ α/2 ≤ π/2, which means that sin(α/2) ≥ 0, so we get
s
s
s
¢
¡
r
√
8
1 − − 53
1 + 35
2 5
4
2
5
sin(α/2) = ±
=
=
=
= √ =
.
2
2
2
5
5
5
3.4.6. Find the exact values of sin(θ/2), cos(θ/2), and tan(θ/2),
given that csc θ = − 35 and −90◦ < θ < 0◦ .
1
Solution: First of all, sin θ =
= − 53 . Second, using
csc θ
p
cos θ = ± 1 − sin2 θ, combined with the fact that θ is is quadrant IV, we get
q
q
¢2 q
¡
9
16
cos θ = 1 − − 53 = 1 − 25
=
= 45 .
25
Thirdly, we use the equalities
r
1 − cos θ
,
sin(θ/2) = ±
2
r
cos(θ/2) = ±
1 + cos θ
,
2
12
,
13
(b) Put β = sin−1
so that what we need to compute is
r
1 + cos β
cos(β/2) = ±
.
2
together with −45◦ < θ/2 < 0◦ , which gives the fact that θ/2
is in quadrant IV. So we have
s
s
r
√
1
1 − 45
1
10
1
5
sin(θ/2) = −
=−
=−
= −√
;
=−
2
2
10
10
10
s
s
r
√
9
1 + 45
9
3 10
3
5
cos(θ/2) =
=
=
=
= √
;
2
2
10
10
10
√
√
10
− 10
sin(θ/2)
1
10
10
√
tan(θ/2) =
=− .
=−
=
· √
3 10
cos(θ/2)
10
3
3 10
The angle β has sin β = 12
, and satisfies −π/2 ≤ β ≤ π/2
13
(quadrants I or IV). In fact, since sin β > 0, we have 0 ≤ β ≤
π/2, so we get
q
q
¡ ¢2 q
25
5
cos β = 1 − 12
= 1 − 144
=
= 13
.
13
169
169
Since we have 0 ≤ β/2 ≤ π/4, we get cos(β/2) > 0, which means
that
s
s
r
√
5
18
1 + 13
3 13
9
3
13
cos(β/2) = ±
=
=
=
= √
.
2
2
13
13
13
10
3.5.34. Find all soultions of the equation
sin 5x − sin x = 2 cos 3x.
40
,
9
(c) Put γ = tan−1
Solution: We factor the left hand side using the formula
µ
¶
µ
¶
u−v
u+v
sin u − sin v = 2 sin
cos
,
2
2
so that what we need to compute is
tan(γ/2) =
1 − cos γ
.
sin γ
, and satisfies −π/2 < γ < π/2
The angle γ has tan γ = 40
9
(quadrants I or IV). This gives sec γ > 0, so we get
q
q
p
¡ ¢2 q
1681
sec γ = 1 + tan2 γ = 1 + 40
= 1 + 1600
=
= 41
.
9
81
81
9
with u = 5x and v = x. The equation becomes
2 sin 2x cos 3x = 2 cos 3x.
If we move all terms to the left, the equation becomes
This gives cos γ =
We now get
2 sin 2x cos 3x − 2 cos 3x = 0,
and it can be factored as:
tan(γ/2) =
2 cos 3x(sin 2x − 1) = 0.
9
41
and sin γ = tan γ · cos γ =
9
41
40
41
1−
=
32
41
40
41
=
40
9
·
9
41
=
40
.
41
32
4
32 41
·
=
= .
41 40
40
5
This leads to two equations:
b = 32.32◦ , a = 263.6, and c = 574.3, solve the
4.1.8. Given A
triangle 4ABC.
Solution: Using the Law of Sines:
cos 3x = 0 and sin 2x = 1.
To solve the first equation cos 3x = 0, we make the substitution
u = 3x and we get two families: u = π2 +2nπ and u = 3π
+2nπ,
2
n integer. Going back to the substitution, the first family gives
3x = π2 + 2nπ, which gives
x=
π
2nπ
+
, n integer,
6
3
and the second family gives 3x =
3π
2
a
b
c
=
=
,
sin A
sin B
sin C
we have
(1)
This gives
+ 2nπ, which gives
π
2nπ
x= +
, n integer.
2
3
574.3 · sin 32.32◦
' 1.16.
263.3
This shows that the problem has no solution, because sin C must
always be less than 1.
Comments. This is an SSA problem. It is possible to have two
solutions (see Exercise 4.1.12 - Sample Exam II), one solution
b = 25◦ ), or no solution at all (as
(try the exercise above with A
in the above exercise).
sin C =
(2)
To solve the second equation sin 2x = 1 we make the substitution
v = 2x and we get one family: u = π2 + 2nπ. Going back to the
substitution, this family gives 2x = π2 + 2nπ, which gives
x=
π
+ nπ, n integer.
4
263.3
574.3
,
=
sin 32.32◦
sin C
(3)
Conclusion: The given equation has all three families (1), (2),
and (3) as solutions.
2
4.2.8. Given a = 10, b = 15, and c = 12, solve the triangle
4ABC.
Solution: Use the Law of Cosine:
µ 2
µ 2
¶
2
2¶
15 + 122 − 102
b = cos−1 b + c − a
A
= cos−1
=
2bc
2 · 15 · 12
¶
µ
269
= cos−1
' 41.65◦ ;
360
µ 2
µ 2
¶
2
2¶
10 + 152 − 122
b = cos−1 a + c − b
B
= cos−1
=
2ac
2 · 10 · 15
µ
¶
181
= cos−1
' 52.89◦ .
300
6.3.12. Find vertices, foci an asymptotes for the hyperbola
(y − 1)2
(x − 3)2
−
= 1.
25
4
Solution: Since the “plus fraction carries x,” matches the form
(x − h)2
(y − k)2
−
= 1,
2
a
b2
with a = 5, b = 2, h = 3, and k = 1. We have
p
√
√
c = a2 + b2 = 25 + 4 = 29.
Given the form (6) we know that the vertices are V (h ± a, k) =
V (3 ± 5, 1). The two vertices then √
are V1 (−2, 1) and V2 (8, 1).
The foci√are F (h ± c, k) = √
F (3 ± 29, 1). The two foci are
F1 (3 − 29, 1) and F2 (3 + 29, 1). The asymptotes for the
hyperbola (6) are
b ' 180◦ − 41.65◦ − 52.89◦ = 86.46◦ .
Then C
6.1.8. Find the focus, vertex and directrix for the parabola:
x−h
y−k
x−h
y−k
=
and
=−
,
a
b
a
b
(y + 1)2 = −12(x + 2).
Solution: Since the equation has no x2 term, the parabola has
horizontal symmetry axis, and the equation is of the form
that is:
(y − k)2 = 4p(x − h),
and the vertex is V (h, k), the focus is F (h + p, k), and the directrix has equation x = h − p. In our case we get h = −2, k = −1,
and 4p = −12, which gives p = −3. So the vertex is V (−2, −1),
the focus is F (−2 − 3, −1) = F (−5, −1), and the directrix is:
x = −2 − (−3), that is, x = 1.
x−3
y−1
x−3
y−1
=
and
=−
.
5
2
5
2
6.3.20. Find the equation of the hyperbola with vertices V1 (0, 2),
V2 (2, 2), and foci F1 (−2, 2) and F2 (4, 2).
Solution: Notice that the focal axis is horizontal, so the equation of the hyperbola will be of the form (6). We know that the
center C(h, k) is the midpoint of the ¡line segment
¢ that joins the
2+2
vertices, so in our case the center is 0+2
= (1, 2), which
,
2
2
means that h = 1 and k = 2. For a hyperbols with equation (6)
we know that the vertices are V (h ± a, k), so we get h − a = 0
and h + a = 2, which gives a = 1. Finally, we know that the
foci are F (h ± c, k), so we get h − c = −2 and h + c = 4, which
gives c = 3. This gives
6.2.10. Find the foci and vertices of the ellipse
(x + 2)2
(y − 3)2
+
= 1.
25
4
Solution: Since the largest denominator is 25, the equation
matches the form
(x − h)2
(y − k)2
+
= 1,
a2
b2
(6)
b2 = c2 − a2 = 32 − 12 = 9 − 1 = 8,
(4)
so the equation of the hyperbola is
with a = 5, b = 2, h = −2, and k = 3. We have
p
√
√
c = a2 − b2 = 25 − 4 = 21.
(x − 1)2 −
(y − 2)2
= 1.
8
Given the form (4) we know that the vertices are V (h ± a, k) =
V (−2 ± 5, 3). The two vertices then are
√ V1 (−7, 3) and V2 (3, 3).
The foci √
are F (h ± c, k) = F (−2
√ ± 21, 3). The two foci are
F1 (−2 − 21, 3) and F2 (−2 + 21, 3).
6.5.35(a) & 6.5.41(b). Change
6.2.20. Find the equation of the ellipse with vertices V (0, ±7)
and foci F (0, ±2).
Solution: Notice that the focal axis is vertical, so the equation
of the ellipse will be of the form
Solution: (a) Here r = 3 and θ = π/4, so the the rectangular
coordinates (x, y) are
√
3 2
x = r cos θ = 3 · cos(π/4) =
,
2
√
3 2
y = r sin θ = 3 · sin(π/4) =
.
2
(y − k)2
(x − h)2
+
= 1.
2
a
b2
(a) (3, π/4) from polar to rectangular coordinates;
(b) (−1, 1) from rectangular to polar coordinates (r, θ) with
r > 0 and 0 ≤ θ < 2π.
(5)
We know that the center C(h, k) is the midpoint of the line
segment that joins the vertices, so in our case the center is (0, 0),
which means that h = 0 and k = 0. For an ellipse with equation
(5) we know that the vertices are V (h, k ± a), so we get a = 7.
Finally, we know that the foci are F (h, k ± c), so we get c = 2.
This gives
(b) Here x = −1 and y = 1. Then
q
p
√
r = x2 + y 2 = (−1)2 + 12 = 2.
The angle θ is then given by
√
√
x
2
2
−1
y
1
cos θ = = √ = −
and sin θ = = √ =
.
r
2
r
2
2
2
b2 = a2 − c2 = 72 − 22 = 49 − 4 = 45,
so the equation of the ellipse is
This gives θ = π −
x2
y2
+
= 1.
45
49
3
π
4
=
3π
.
4
6.5.56. Find the polar equation of the curve
(x + 2)2 + (y − 3)2 = 13.
Solution: We compute the left hand side:
LHS = x2 + 4x + 4 + y 2 − 6y + 9 = (x2 + y 2 ) + 4x − 6y + 13 =
= r2 + 4r cos θ − 6r sin θ + 13.
The equation is now
r 2 + 4r cos θ − 6r sin θ + 13 = 13,
which is the same as
r 2 + 4r cos θ − 6r sin θ = 0.
(7)
Comment. One can actually factor r, and eliminate it (without
loosing the case r = 0) so the equation (7) can be written as
r + 4 cos θ − 6 sin θ = 0,
or
r = −4 cos θ + 6 sin θ.
6.5.68. Find the equation of
r(3 cos θ − 4 sin θ) = 12
in rectangular coordinates x, y.
Solution: The left hand side is
r(3 cos θ − 4 sin θ) = 3r cos θ − 4r sin θ = 3x − 4y,
so the equation is
3x − 4y = 12.
6.6.6. Find the focus (or foci), the vertex (or vertices) and the
directrix for the conic given by the polar equation
r=
3
.
2 − 2 sin θ
Solution: The polar equations are of one of the following types
de
;
1 + e cos θ
de
(iii)r =
;
1 + e sin θ
(i)r =
de
;
1 − e cos θ
de
(iv)r =
.
1 − e sin θ
(ii)r =
The right hand side of our equation can be re-written as
3
RHS =
3
2
=
,
2(1 − sin θ)
1 − sin θ
so the given equation is of the form (iv) with d =
This is a parabola, with
3
2
and e = 1.
• focus F (0, 0);
• horizontal directrix, d units below the focus;
• vertex on the y-axis, below the origin, having θ = −π/2.
In our case the directrix is y = − 32 , and the focus with polar
coordinates θ = − π2 and
r=
3
3
¢ = .
¡
4
2 − 2 sin − π2
So the vertex is V (0, − 43 ).
Remark. The vertex can also be found using the fact that it is
placed half-way between the focus and the directrix.
4