MTH 1120, College Trigonometry—Exam 2
Nov. 9, 2011
Instructions: Work the following problems; give your reasoning as appropriate; show your supporting calculations. Do not give decimal approximations unless the nature of a problem requires
them. Write your solutions on your own paper; your paper is due at 3:15 pm. Complete solutions
to the exam problems will be available from the course web-site later this evening.
1. If sin θ =
9
, and θ is in the second quadrant, give exact values for
41
(a) cos θ
(b) tan θ
(c) csc θ
(d) sin 2θ
θ
(e) tan
2
Solution:
(a) cos2 θ = 1 − sin2 θ = 1 − 81/1681 = 1600/1681 = (40/41)2 . It is given that θ lies in the
second quadrant, where the cosine function is negative. Hence, cos θ = −40/41.
(b) tan θ = sin θ/ cos θ = (9/41)/(−40/41) = −9/40.
(c) csc θ = 1/ sin θ = 41/9.
(d) sin 2θ = 2 sin θ cos θ = 2(9/41)(−40/41) = −720/1681.
(e) tan(θ/2) = (sin θ)/(1 + cos θ) = (9/41)/[1 + (−40/41)] = (9/41)/(1/41) = 9.
2. Evaluate the following expressions exactly.
15
(a) sin arcsin
17
12
(b) sin arccos
13
15
(c) tan sin−1
17
h
x i
(d) tan 2 cos−1
3
Solution:
(a) sin[arcsin(15/17)] = 15/17.
(b) sin2 θ = 1 − cos2 θ, so sin2 [arccos(12/13)] = 1 − (12/13)2 = 25/169 = (5/13)2 . We may
choose the + sign because arccos(12/13) lies in the first quadrant, where the sine function
is positive. Hence, sin[arccos(12/13)] = 5/13.
(c) Let us consider a right triangle 4ABC, conventionally labeled, where
√ a = 15 and√c = 17.
Then ∠A = sin−1 (15/17), and, by the Pythagorean Theorem, b = 172 − 152 = 64 = 8.
Hence, tan[sin−1 (15/17)] = a/b = 15/8.
(d) We begin with a right triangle 4ABC, conventionally labeled, and with
√ b = x, c =
−1 (x/3), and, by the Pythagorean Theorem, a =
3. Then ∠A = cos
9 − x2 . Thus
√
tan[cos−1 (x/3)] = 9 − x2 /x. But tan(2θ) = 2 tan θ/(1 − tan2 θ), so
√
√
h
i
2( 9 − x2 /x)
2x 9 − x2
−1 x
√
tan 2 cos
=
.
=
3
2x2 − 9
1 − ( 9 − x2 /x)2
3. Solve 4ABC, labeled conventionally, if ∠A = 48◦ , b = 12, and c = 20. Give your solutions
correct to five digits to the right of the decimal point.
Solution: By the Law of Cosines,
a2 = b2 + c2 − 2bc cos ∠A, or
a2 = 144 + 400 − 480 cos 48◦ ∼ 222.8173089, so that
a ∼ 14.92706632087.
Again by the Law of Cosines,
a2 + b2 − c2
222.817308947748 + 144 − 400
∼
∼ −0.09262450487, so that
2ab
2 · 14.92706632087 · 12
∠C ∼ 95.31461103571◦ .
cos ∠C =
Finally,
∠B = 180◦ − ∠A − ∠B ∼ 36.68538964291.
Answers correct to five digits to the right of the decimal are therefore 14.92707 for a, 95.31461◦
for ∠C, and 36.68539◦ for ∠B.
4. Find all solutions in the interval [0, 2π) for each of the following equations. Do not give
decimal approximations.
3
4
(b) 2 sin2 x + cos x − 1 = 0
(a) sin2 θ =
Solution:
√
(a) √
If sin2 θ = 3/4, then sin θ = ± 3/2. The only solutions for the equation sin θ =
3/2 in [0, 2π) are θ = π/3 and θ = 2π/3.The only solutions for the equation sin θ =
√
− 3/2 in [0, 2π) are θ = 4π/3 and θ = 5π/3. Consequently, the solutions we seek are
π 2π 4π
5π
, , , and
.
3 3 3
3
(b) We replace sin2 x with 1 − cos2 x in the equation and rearrange to obtain the quadratic
equation 2 cos2 x − cos x − 1 = 0. The Quadratic Formula now gives
p
1 ± 1 − (4)(2)(−1)
1±3
1
cos x =
=
= 1 or − .
2·2
4
2
If cos x = 1 and 0 ≤ x < 2π, then x = 0. If cos x = −1/2 and 0 ≤ x < 2π, then x = 2π/3
4π
2π
.
or x = 4π/3. The solutions we desire are therefore 0, , and
3
3
5. Find all solutions in [0, 2π) for the equation tan2 θ − 0.7 tan θ − 4.08 = 0. Give your solutions
correct to at least five digits to the right of the decimal point.
Solution: By the Quadratic Formula,
p
0.7 ± (0.7)2 − 4 · (−4.08)
0.7 ± 4.1
tan x =
=
.
2
2
Thus, tan x = 2.4 or tan x = −1.7. The solutions to the equation tan x = 2.4 that lie in
[0, 2π) are arctan 2.4 ∼ 1.17600520710 and π + arctan 2.4 ∼ 4.31759786068. The solutions
to the equation tan x = −1.7 that lie in [0, 2π) are π + arctan(−1.7) ∼ 2.10252039405 and
2π + arctan(−1.7) ∼ 5.24411304764. Solutions correct to five digits are therefore 1.17600 (or
1.17601), 2.10252, 4.31760, and 5.24411.
6. Transform the equation x2 − y 2 = x into polar coordinates. Give your solution in the form
r = f (θ).
Solution: We substitute x = r cos θ and y = r sin θ into the equation x2 − y 2 = x:
x2 − y 2 = x;
r2 cos2 θ − r2 sin2 θ = r cos θ;
r(cos2 θ − sin2 θ) = cos θ;
cos θ
r=
.
2
cos θ − sin2 θ
This has the desired form. It may—but need not—be written as r =
cos θ
.
cos 2θ