find the terminal point on the unit circle determined by -\\frac{11\\pi}{6} radians. use exact values, not…

Explanation:

Step1: Recall the unit - circle definitions

For a point ((x,y)) on the unit circle (x = \cos\theta) and (y=\sin\theta), where (\theta) is the angle in radians. Here (\theta=-\frac{11\pi}{6}).

Step2: Use the trigonometric identities

We know that (\cos(-\alpha)=\cos\alpha) and (\sin(-\alpha)=-\sin\alpha). So (\cos(-\frac{11\pi}{6})=\cos(\frac{11\pi}{6})) and (\sin(-\frac{11\pi}{6})=-\sin(\frac{11\pi}{6})). Also, (\frac{11\pi}{6}=2\pi-\frac{\pi}{6}). And (\cos(2\pi - \alpha)=\cos\alpha), (\sin(2\pi-\alpha)=-\sin\alpha). So (\cos(\frac{11\pi}{6})=\cos(2\pi - \frac{\pi}{6})=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}) and (\sin(\frac{11\pi}{6})=\sin(2\pi - \frac{\pi}{6})=-\sin(\frac{\pi}{6})=-\frac{1}{2}). Then (\sin(-\frac{11\pi}{6})=-(-\frac{1}{2})=\frac{1}{2})

Answer:

((\frac{\sqrt{3}}{2},\frac{1}{2}))