MTH 1120, College Trigonometry—Exam 1
Sept. 28, 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. Let P be the wrapping function, and let t be a real number for which P (t) is a second quadrant
point which lies on the line whose equation is 12x + 5y = 0. Give exact values for cos t, sin t,
tan t, cos(t + π/2), and tan(t − π).
Solution: The coordinates (x, y) of the point P (t) must satisfy both of the equations x2 +y 2 = 1
and 12x+5y = 0. The latter equation implies that y = −12x/5. Substituting for y in the equation
of the unit circle, we arrive at x2 + (−12x/5)2 = 1, or 25x2 + 144x2 = 25. Thus, x2 = 25/169,
and x = ±5/13. But we are given that P (t) lies in the second quadrant, so we will take the
minus sign, giving x = −5/13. Then y = −12x/5 = 12/13, and P (t) = (−5/13, 12/13). Thus,
cos t = −5/13, sin t = 12/13, and tan t = −12/5. The point P (t + π/2) lies on the unit circle
a quarter of a counterclockwise revolution from P (t), so its coordinates are (−12/13, −5/3),
making cos(t + π/2) = −12/13. Finally, P (t − π) lies on the unit circle diametrically opposite
from P (t), so its coordinates are (5/13, −12/13). Therefore, tan(t − π) = −12/5.
2. Find the exact value of s in the given interval that has the given circular function value. Give
your reasoning explicitly.
1
π
and ≤ s ≤ π.
2
2
√
5π
(b) tan s = 3 and
≤ s ≤ 4π.
2
√
2 3
(c) sec s = 2, csc s = −
, and − 3π ≤ s ≤ −π
3
(a) sin s =
Solution:
(a) From π/2 ≤ s ≤ π, we see that P (s) lies in the second quadrant. Because sin s = 1/2, we
know that s has reference number s∗ = π/6. Hence s = π − π/6 = 5π/6.
(b) From 5π/2 ≤ s ≤√4π, we see that P (s) lies in the second, the third, or the fourth quadrant.
Because tan s = 3 > 0, the reference number s∗ = π/3 and P (s) must be in the third
quadrant. Therefore, s = 3π + π/3 = 10π/3.
(c) We are given sec s = 2, and this means that the reference number s∗ must be π/3. Because
sec s > 0 while csc s < 0, the point P (s) must lie in the fourth quadrant. Hence s =
−2π − π/3 = −7π/3.
3. Give a sinusoidal function whose graph appears Figure 1. (The labeled points are, respectively,
the maximum and minimum points on the graph in the interval shown.) Be sure to give the
reasoning that supports your conclusions.
y
4
2
(0, 2)
-2
2
4
6
8
x
-2
(2 !, "4)
-4
Figure 1: A Graph
Solution: The vertical distance from the minimum to the maximum is 6, and the horizontal
distance from the maximum to the minimum is 2π. So the graph is that of a cosine function
of period 4π, of amplitude 3, shifted
x one unit downward. Hence, an equation for the graph is
x
y + 1 = 3 cos
, or y = 3 cos
− 1.
2
2
4. In a right triangle, 4ABC, labeled conventionally, it is given that ∠A = 36◦ and a = 42.8.
Solve the triangle, giving your answers correct to at least four digits to the right of the decimal
points.
Solution: In a right triangle, we have sin A = a/c, or c = a/ sin A. Hence c = 42.8/ sin 36◦ ∼
72.81570919. Similarly, b = a/ tan 36◦ ∼ 58.9091462. Finally, ∠B = 90◦ − ∠A = 90◦ − 36◦ =
54◦ .
5. Consider the triangle (not drawn to scale) shown in Figure 2. Solve the triangle completely
C
a
b
A
c
B
Figure 2: A Triangle
given that a = 4, ∠C = 50◦ , and ∠A = 35◦ . Give each of your answers correct to at least
four digits to the right of the decimal point.
Solution: The sum of the angles in any triangle must be 180◦ , so ∠B = 180◦ − 50◦ − 35◦ = 95◦ .
By the Law of Sines,
sin 35◦
sin 95◦
sin 50◦
=
=
.
4
b
c
Hence b = (4 sin 95◦ )/(sin 35◦ ) ∼ 6.947249817, and c = (4 sin 50◦ )/(sin 35◦ ) ∼ 5.342230919.
6. In each part below, show how to use the Fundamental Identities to transform the first expression into the second.
(a)
(b)
sin α
1 − cos α
1 + cos α
ii.
sin α
1
1
i.
−
tan β + sec β tan β − sec β
i.
ii. 2 sec β
Solution:
(a)
sin α
sin α(1 + cos α)
sin α(1 + cos α)
=
=
1 − cos α
(1 − cos α)(1 + cos α)
1 − cos2 α
sin
α(1 + cos α)
1 + cos α
=
.
=
sin α
sin
α · sin α
(b)
1
1
(tan β − sec β) − (tan β + sec β)
−
=
tan β + sec β tan β − sec β
(tan β + sec β)(tan β − sec β)
tan
β − sec β − tan
β − sec β
=
2
2
tan β − sec β
−2 sec β
=
tan2 β − (1 + tan2 β)
−2 sec β
= XX2
= 2 sec β.
X2X
tan X
β − 1 −X
tan
β
X
X