Find the reference number and the terminal point $P ( x , y )$ determined by $t = - \frac { 11 \pi } { 6 }$
A) $- \frac { \pi } { 6 }$, $P ( x , y ) = \left( \frac { \sqrt { j } } { 2 } , - \frac { 1 } { 2 } \right)$

B) $\pi$, $P ( x , y ) = ( - 1,0 )$

C) $\frac { \pi } { 6 }$, $P ( x , y ) = \left( \frac { \sqrt { 1 } } { 2 } , \frac { 1 } { 2 } \right)$

D) $\frac { \pi } { 3 }$, $P ( x , y ) = \left( \frac { 1 } { 2 } , \frac { \sqrt { 0 } } { 2 } \right)$

E) $\frac { 11 \pi } { 6 }$, $P ( x , y ) = \left( - \frac { \sqrt { x } } { 2 } , - \frac { 1 } { 2 } \right)$

Question

Find the reference number and the terminal point $P ( x , y )$ determined by $t = - \frac { 11 \pi } { 6 }$
A) $- \frac { \pi } { 6 }$, $P ( x , y ) = \left( \frac { \sqrt { j } } { 2 } , - \frac { 1 } { 2 } \right)$

B) $\pi$, $P ( x , y ) = ( - 1,0 )$

C) $\frac { \pi } { 6 }$, $P ( x , y ) = \left( \frac { \sqrt { 1 } } { 2 } , \frac { 1 } { 2 } \right)$

D) $\frac { \pi } { 3 }$, $P ( x , y ) = \left( \frac { 1 } { 2 } , \frac { \sqrt { 0 } } { 2 } \right)$

E) $\frac { 11 \pi } { 6 }$, $P ( x , y ) = \left( - \frac { \sqrt { x } } { 2 } , - \frac { 1 } { 2 } \right)$

Quizplus · Accepted Answer

The terminal point determined by $t=-\frac{11\pi}{6}$ on the unit circle is $(\cos(-\frac{11\pi}{6}),\sin(-\frac{11\pi}{6}))=(\cos(\frac{\pi}{6}),-\sin(\frac{\pi}{6}))=(\frac{\sqrt{3}}{2},\frac{1}{2})$. Therefore, the reference number is $\frac{\pi}{6}$ and the terminal point is $P(\frac{\sqrt{3}}{2},\frac{1}{2})$. The correct answer choice is C.

Find the Reference Number and the Terminal Point P(x,y)P ( x , y )P(x,y)

Find the Reference Number and the Terminal Point $P ( x , y )$