Question

1. use the unit circle and reference angles to find the exact value for each trig function.

a) sin(3π/4) =
b) cos - 120° =
c) tan 135° =
d) cot(3π/2) =
e) csc 330° =
f) sec(π/6) =

Explanation:

Step1: Find reference angles

For \( \sin\frac{3\pi}{4} \), the reference angle \( \theta'=\pi - \frac{3\pi}{4}=\frac{\pi}{4} \). In the second - quadrant, \( \sin\theta>0 \), so \( \sin\frac{3\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2} \).
For \( \cos(- 120^{\circ}) \), since \( \cos(-\alpha)=\cos\alpha \), \( \cos(-120^{\circ})=\cos120^{\circ} \). The reference angle \( \theta' = 180^{\circ}-120^{\circ}=60^{\circ} \). In the second - quadrant, \( \cos\theta<0 \), so \( \cos120^{\circ}=-\cos60^{\circ}=-\frac{1}{2} \).
For \( \tan135^{\circ} \), the reference angle \( \theta'=180^{\circ}-135^{\circ}=45^{\circ} \). In the second - quadrant, \( \tan\theta<0 \), so \( \tan135^{\circ}=-\tan45^{\circ}=-1 \).
For \( \cot\frac{3\pi}{2} \), \( \cot\theta=\frac{\cos\theta}{\sin\theta} \). \( \cos\frac{3\pi}{2}=0 \), \( \sin\frac{3\pi}{2}=-1 \), so \( \cot\frac{3\pi}{2}=\frac{0}{-1}=0 \).
For \( \csc330^{\circ} \), \( \csc\theta=\frac{1}{\sin\theta} \). The reference angle \( \theta' = 360^{\circ}-330^{\circ}=30^{\circ} \). In the fourth - quadrant, \( \sin\theta<0 \), \( \sin330^{\circ}=-\sin30^{\circ}=-\frac{1}{2} \), so \( \csc330^{\circ}=\frac{1}{-\frac{1}{2}}=-2 \).
For \( \sec\frac{\pi}{6} \), \( \sec\theta=\frac{1}{\cos\theta} \). \( \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \), so \( \sec\frac{\pi}{6}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3} \).

Answer:

a) \( \frac{\sqrt{2}}{2} \)
b) \( -\frac{1}{2} \)
c) \( -1 \)
d) \( 0 \)
e) \( -2 \)
f) \( \frac{2\sqrt{3}}{3} \)