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Evaluate (If Possible)the Sine,cosine,and Tangent of the Real Number t=π2t = - \frac { \pi } { 2 }

Question 16

Multiple Choice

Evaluate (if possible) the sine,cosine,and tangent of the real number.​ t=π2t = - \frac { \pi } { 2 }


A) t=π2t = - \frac { \pi } { 2 } corresponds to the point (x,y) =(0,0) ( x , y ) = ( 0,0 ) . sin(π2) =0cos(π2) =0tan(π2) =0\begin{array} { l } \sin \left( - \frac { \pi } { 2 } \right) = 0 \\\cos \left( - \frac { \pi } { 2 } \right) = 0 \\\tan \left( - \frac { \pi } { 2 } \right) = 0\end{array}
B) t=π2t = - \frac { \pi } { 2 } corresponds to the point (x,y) =(1,0) ( x , y ) = ( - 1,0 ) . sin(π2) =1cos(π2) =0tan(π2) =0\begin{array} { l } \sin \left( - \frac { \pi } { 2 } \right) = - 1 \\\cos \left( - \frac { \pi } { 2 } \right) = 0 \\\tan \left( - \frac { \pi } { 2 } \right) = 0\end{array}
C) t=π2t = - \frac { \pi } { 2 } corresponds to the point (x,y) =(0,1) ( x , y ) = ( 0 , - 1 ) . sin(π2) =1cos(π2) =0tan(π2)  is undefined \begin{array} { l } \sin \left( - \frac { \pi } { 2 } \right) = - 1 \\\cos \left( - \frac { \pi } { 2 } \right) = 0 \\\tan \left( - \frac { \pi } { 2 } \right) \text { is undefined }\end{array}
D) t=π2t = - \frac { \pi } { 2 } corresponds to the point (x,y) =(1,0) ( x , y ) = ( - 1,0 ) . sin(π2) =0cos(π2) =1tan(π2) =0\begin{array} { l } \sin \left( - \frac { \pi } { 2 } \right) = 0 \\\cos \left( - \frac { \pi } { 2 } \right) = - 1 \\\tan \left( - \frac { \pi } { 2 } \right) = 0\end{array}
E) Not possible

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