Find the point $( x , y )$ on the unit circle corresponding to the real number $t = - \frac { 7 \pi } { 6 }$.
A) $\left( \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right)$
B) $\left( - \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$
C) $\left( \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$
D) $\left( - \frac { 1 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right)$
E) $\left( - \frac { \sqrt { 2 } } { 2 } , - \frac { \sqrt { 2 } } { 2 } \right)$

Question

Quizplus · Accepted Answer

Starting at the positive $x$-axis, we go clockwise by $\frac{7\pi}{6}$ radians. This takes us to the fourth quadrant, which means the $y$-coordinate must be negative. We can find the exact values of $\sin \frac{7\pi}{6}$ and $\cos \frac{7\pi}{6}$ using the fact that they are equal to $\sin \left(\frac{7\pi}{6}-2\piight)$ and $\cos \left(\frac{7\pi}{6}-2\piight)$, respectively, since the sine and cosine functions have period $2\pi$. We have $\cos \frac{7\pi}{6} = \cos (2\pi - \frac{5\pi}{6}) = \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ and $\sin \frac{7\pi}{6} = \sin (2\pi - \frac{5\pi}{6}) = -\sin \frac{5\pi}{6} = -\frac{1}{2}$. Therefore, the point is $\boxed{	extbf{(B) } \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}ight)}$.

Find the Point (x,y)( x , y )(x,y) On the Unit Circle Corresponding to the Real Number

Find the Point $( x , y )$ On the Unit Circle Corresponding to the Real Number