College Trigonometry, 2011a

746 Foundations of Trigonometry
Solution.
1. According to Theorem 10.6, sec (60 ◦ 1
)=
cos(60 ◦ ) . Hence, sec (60◦ )= 1
(1/2) =2.
2. Since sin ( ) √
7π
4 = − 2
2 , csc ( )
7π
4 =
1
sin( 7π 4 ) = 1
− √ = − √ 2
2/2 2
= − √ 2.
3. Since θ = 3 radians is not one of the ‘common angles’ from Section 10.2, we resort to the
calculator for a decimal approximation. Ensuring that the calculator is in radian mode, we
find cot(3) = cos(3)
sin(3) ≈−7.015.
4. If θ is coterminal with 3π 2 , then cos(θ) =cos( ) (
3π
2 = 0 and sin(θ) =sin 3π
)
2 = −1. Attempting
to compute tan(θ) = sin(θ)
−1
cos(θ)
results in
0
, so tan(θ) is undefined.
5. We are given that csc(θ) = 1
sin(θ) = −√ 5sosin(θ) =− √ 1
√
5
= − 5
5 . AswesawinSection10.2,
we can use the Pythagorean Identity, cos 2 (θ)+sin 2 (θ) = 1, to find cos(θ) by knowing sin(θ).
Substituting, we get cos 2 (θ)+
(
−
θ is a Quadrant IV angle, cos(θ) > 0, so cos(θ) = 2√ 5
5 .
√
5
5
) 2
= 1, which gives cos 2 (θ) = 4 5 , or cos(θ) =± 2√ 5
5 .Since
6. If tan(θ) =3,then sin(θ)
cos(θ)
= 3. Be careful - this does NOT mean we can take sin(θ) =3and
cos(θ) = 1. Instead, from sin(θ)
cos(θ)
=3weget: sin(θ) = 3 cos(θ). To relate cos(θ) and sin(θ), we
once again employ the Pythagorean Identity, cos 2 (θ)+sin 2 (θ) = 1. Solving sin(θ) = 3 cos(θ)
for cos(θ), we find cos(θ) = 1 3
sin(θ). Substituting this into the Pythagorean Identity, we find
sin 2 (θ)+ ( 1
3 sin(θ)) 2 = 1. Solving, we get sin 2 (θ) = 9
10 so sin(θ) =± 3√ 10
10 .Sinceπ