7) use the unit circle and reference angles to find the exact value for each trig function.\na) sin(3π/4) =…

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} )