Find the point $( x , y )$ on the unit circle corresponding to the real number $t = - \frac { \pi } { 6 }$.
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

The point on the unit circle corresponding to $t = -\frac{\pi}{6}$ can be found using the sine and cosine values for $\frac{\pi}{6}$. The cosine gives the x-coordinate, and the sine gives the y-coordinate. For $\frac{\pi}{6}$, $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ and $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. Since $t$ is negative, it indicates a rotation in the opposite direction, making the y-coordinate negative. Thus, the point is $\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$.

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