Question

find the terminal point on the unit circle determined by -\frac{11\pi}{6} radians. use exact values, not decimal approximations. (x,y)=(\square,\square)

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:

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