Find the point (x, y) on the unit circle that corresponds to the real number t. $t = \frac { 5 \pi } { 6 }$
A) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right)$ .
B) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( - \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$ .
C) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( - \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$ .
D) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$ .
E) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$ .

Question

Find the point (x, y) on the unit circle that corresponds to the real number t.   $t = \frac { 5 \pi } { 6 }$  
A) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right)$ .
B) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( - \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$ .
C) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( - \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$ .
D) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$ .
E) $t = \frac { 5 \pi } { 6 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$ .

Quizplus · Accepted Answer

The point $\left( - \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$ correctly corresponds to $t = \frac { 5 \pi } { 6 }$ on the unit circle, which is in the second quadrant where sine is positive and cosine is negative.

Find the Point (X, Y) on the Unit Circle That t=5π6t = \frac { 5 \pi } { 6 }t=65π​

Find the Point (X, Y) on the Unit Circle That $t = \frac { 5 \pi } { 6 }$