# Homework 6 Solutions ```Math 1613 - Trigonometry
Homework #6 - 2007.09.27
Due Date - 2007.10.04
Solutions
1. Find the reference angle θR if θ has the given measure. Also, state which
quadrant each of the original angles lies in.
a)
143π
4
3
So notice that 143π
4 = 35π + 4 π. The angle 35π gets us to the negative x-axis, and
rotating another 43 π gives that the angle lies in the fourth quadrant and θR = 14 π.
b) 840◦
Here we have 840◦ = 720◦ + 120◦. So our angle lies in the second quadrant with
θR = 180◦ − 120◦ = 60◦ = π3 .
c) − 43π
3
First, − 43π
= −14π − 31 π. This gives an angle in the fourth quadrant with
3
θR = π3 π.
2. Find the exact value.
a) cos − 43π
3
π 1
43π
cos −
= cos
= .
3
3
2
b) csc 143π
4
so we have
π
√
143π
csc
= − csc
= − 2.
4
4
1
c) sec (840◦)
Cosine is negative in the second quadrant, so
π ◦
sec (840 ) = − sec
= −2.
3
3. For this problem, please refer to the following figure.
10
y
θ
x
a) If sin(θ) = 12 , what is y?
Since sin(θ) = 12 , we look at the picture and see that
1
y
sin(θ) = =
→ y = 5.
2 10
b) If cos(θ) = 15 , what is sin(θ)? (Solve this by the Pythagorean Theorem and
the definition of sin(θ)).
If cos(θ) = 15 , then similar to part a), we have that x = 2. Next, we need to find
y, which can be√done by√the Pythagorean Theorem.
So√102 = 42 + y 2 and solving
√
for y gives y = 96 = 4 6. So finally, sin(θ) = 4106 = 2 5 6 .
c) Redo part b) but this time, use the fact that sin2 (θ) + cos2(θ) = 1.
Since we know that cos(θ) = 15 , we get
s
r
√
2 r
1
1
24 2 6
sin(θ) = 1 −
= 1−
=
=
.
5
25
25
5
2
d) If tan(θ) = 3, what are x and y?
Since tan(θ) = 3, we have that 3 =
Pythagorean Theorem, we get that
y
x
which means that x = 13 y. Using the
√
1 2
2
x + y = 100 → y + y = 100 → y = 3 10.
9
2
2
Since we now know y, we can solve for x in x2 + y 2 = 100. This gives x =
Notice that xy = 3.
3
√
10.
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