Find the point (x, y) on the unit circle that corresponds to the real number t. $t = \frac { 19 \pi } { 3 }$
A) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$ .
B) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$ .
C) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right)$ .
D) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { 1 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right)$ .
E) $t = \frac { 19 \pi } { 3 }$ 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.   $t = \frac { 19 \pi } { 3 }$  
A) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$ .
B) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$ .
C) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right)$ .
D) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { 1 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right)$ .
E) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( - \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$ .

Quizplus · Accepted Answer

The Answer of Find the point (x, y) on the unit circle that corresponds to the real number t.   $t = \frac { 19 \pi } { 3 }$  
A) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right)$ .
B) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$ .
C) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right)$ .
D) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( \frac { 1 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right)$ .
E) $t = \frac { 19 \pi } { 3 }$ corresponds to the point $\left( - \frac { \sqrt { 3 } } { 2 } , - \frac { 1 } { 2 } \right)$ .

Find the Point (X, Y) on the Unit Circle That t=19π3t = \frac { 19 \pi } { 3 }t=319π​

Find the Point (X, Y) on the Unit Circle That $t = \frac { 19 \pi } { 3 }$