Evaluate (if possible)the sine,cosine,and tangent of the real number.

t = \frac { 9 \pi } { 4 }

t = \frac { 9 \pi } { 4 }

corresponds to the point

( x , y ) = \left( \frac { \sqrt { 2 } } { 2 } , - \frac { \sqrt { 2 } } { 2 } \right)

\begin{array} { l } \sin \left( \frac { 9 \pi } { 4 } \right) = - \frac { \sqrt { 2 } } { 2 } \\\cos \left( \frac { 9 \pi } { 4 } \right) = \frac { \sqrt { 2 } } { 2 } \\\tan \left( \frac { 9 \pi } { 4 } \right) = - 1\end{array}

t = \frac { 9 \pi } { 4 }

corresponds to the point

( x , y ) = \left( \frac { \sqrt { 2 } } { 2 } , \frac { \sqrt { 2 } } { 2 } \right)

\begin{array} { l } \sin \left( \frac { 9 \pi } { 4 } \right) = \frac { \sqrt { 2 } } { 2 } \\\cos \left( \frac { 9 \pi } { 4 } \right) = \frac { \sqrt { 2 } } { 2 } \\\tan \left( \frac { 9 \pi } { 4 } \right) = 1\end{array}

t = \frac { 9 \pi } { 4 }

corresponds to the point

( x , y ) = \left( - \frac { \sqrt { 2 } } { 2 } , - \frac { \sqrt { 2 } } { 2 } \right)

\begin{array} { l } \sin \left( \frac { 9 \pi } { 4 } \right) = - \frac { \sqrt { 2 } } { 2 } \\\cos \left( \frac { 9 \pi } { 4 } \right) = - \frac { \sqrt { 2 } } { 2 } \\\tan \left( \frac { 9 \pi } { 4 } \right) = 1\end{array}

t = \frac { 9 \pi } { 4 }

corresponds to the point

( x , y ) = \left( - \frac { \sqrt { 2 } } { 2 } , \frac { \sqrt { 2 } } { 2 } \right)

\begin{array} { l } \sin \left( \frac { 9 \pi } { 4 } \right) = \frac { \sqrt { 2 } } { 2 } \\\cos \left( \frac { 9 \pi } { 4 } \right) = - \frac { \sqrt { 2 } } { 2 } \\\tan \left( \frac { 9 \pi } { 4 } \right) = - 1\end{array}

E)Not possible

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

Multiple Choice