Download Homework 5 Solutions

Math 1613 - Trigonometry
Homework #5 - 2007.09.18
Due Date - 2007.09.27
Solutions
1. Verify the following identities.
a)
cot(−t) cos(t) + sin(−t) = − csc(t)
cos(t)
cos(t) − sin(t)
sin(t)
1
=−
cos2 (t) + sin2 (t)
sin(t)
1
=−
sin(t)
= − csc(t).
cot(−t) cos(t) + sin(−t) = −
b)
cos(−x)
1 + sin(−x)
=
1 − sin(−x)
cos(x)
cos(−x)
cos(x)
=
1 − sin(−x) 1 + sin(x)
cos(x)
1 − sin(x)
=
·
1 + sin(x) 1 − sin(x)
cos(x)(1 − sin(x))
=
1 − sin2 (x)
cos(x)(1 − sin(x))
=
cos2 (x)
1 − sin(x)
=
cos(x)
1 + sin(−x)
=
.
cos(x)
1
c)
log(tan(−t)) = log(cos(t)) − log(sin(t))
The left side does not equal the right side!
2. Complete the following statements.
a) as t → π + , sin(t) →
as t → π + , sin(t) → 0.
b) as t → π − , cos(t) →
as t → π − , cos(t) → −1.
c) as t → π + , csc(t) →
as t → π + , csc(t) → −∞.
d) as t → − π2
−
, tan(t) →
π −
as t → −
, tan(t) → ∞.
2
2
3. Find the exact values of the six trigonometric functions of θ if θ is in standard
position and P is on the terminal side.
a) P (−3, 7)
First, we note that r =
√
58.
3
7
7
cos(θ) = − √ , sin(θ) = √ , tan(θ) = −
3
√58
√ 58
3
58
58
, csc(θ) =
, cot(θ) = −
sec(θ) = −
3
7
7
b) P (−7, −3)
First, we note that r =
√
58.
7
3
3
cos(θ) = − √ , sin(θ) = − √ , tan(θ) =
7
√ 58
√58
58
58
7
sec(θ) = −
, csc(θ) = −
, cot(θ) =
7
3
3
4. (Problem 36 page 136) The highest advertising sign in the world is a large
letter I situated at the top of the 73-story First Interstate World Center in Los
Angeles. At a distance of 200 feet from a point directly below the sign, the angle
between the ground and the top of the sign is 78.87◦ . Approximate the height of
the top of the sign.
To solve this, we use the fact that the distance of 200 feet forms the base of
a triangle, and the distance to the top of the sign forms the height of the right
triangle (call the height y). The angle opposite is 78.87◦ , which gives
y
tan (78.87◦ ) =
.
200
This gives y ≈ 1016.590716 feet.
3
5. (Problem 38 page 136) The phases of the moon can be described using the
phase angle θ, determined by the sun, the moon, and the earth, as shown in the
figure at the bottom of page 136. Because the moon orbits the earth, θ changes
during the course of a month. The area of the region A of the moon, which appears illuminated to an observer on the earth, is given by A = 21 πR2 (1 + cos(θ)),
where R = 1080 miles is the radius of the moon. Approximate A for the following
positions of the moon.
a) θ = 0◦ (full moon)
1
A = πR2 (1 + cos(0))
2
1
= πR2 (1 + 1)
2
= πR2
= 1166400π mi2
≈ 3664353.672 mi2 .
b) θ = 180◦ (new moon)
1
A = πR2 (1 + cos(π))
2
1 2
= πR (1 − 1)
2
= 0 mi2 .
4
c) θ = 90◦ (first quarter)
π 1 2
A = πR (1 + cos
)
2
2
1
= πR2 (1 + 0)
2
1 2
= πR
2
= 583200π mi2
≈ 1832176.836 mi2 .
5