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Evaluate (If Possible)the Sine,cosine,and Tangent of the Real Number t=25π3t = \frac { 25 \pi } { 3 }

Question 1

Multiple Choice

Evaluate (if possible) the sine,cosine,and tangent of the real number.​ t=25π3t = \frac { 25 \pi } { 3 }


A) t=25π3t = \frac { 25 \pi } { 3 } corresponds to the point (x,y) =(12,32) ( x , y ) = \left( \frac { 1 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right) . sin(25π3) =12cos(25π3) =32tan(25π3) =3\begin{array} { l } \sin \left( \frac { 25 \pi } { 3 } \right) = - \frac { 1 } { 2 } \\\cos \left( \frac { 25 \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 2 } \\\tan \left( \frac { 25 \pi } { 3 } \right) = - \sqrt { 3 }\end{array}
B) t=25π3t = \frac { 25 \pi } { 3 } corresponds to the point (x,y) =(12,32) ( x , y ) = \left( \frac { 1 } { 2 } , \frac { \sqrt { 3 } } { 2 } \right) . sin(25π3) =32cos(25π3) =12tan(25π3) =3\begin{array} { l } \sin \left( \frac { 25 \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 2 } \\\cos \left( \frac { 25 \pi } { 3 } \right) = \frac { 1 } { 2 } \\\tan \left( \frac { 25 \pi } { 3 } \right) = \sqrt { 3 }\end{array}
C) t=25π3t = \frac { 25 \pi } { 3 } corresponds to the point (x,y) =(12,32) ( x , y ) = \left( - \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right) . sin(25π3) =12cos(25π3) =32tan(25π3) =33\begin{array} { l } \sin \left( \frac { 25 \pi } { 3 } \right) = - \frac { 1 } { 2 } \\\\\cos \left( \frac { 25 \pi } { 3 } \right) = - \frac { \sqrt { 3 } } { 2 } \\\\\tan \left( \frac { 25 \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 3 }\end{array}
D) t=25π3t = \frac { 25 \pi } { 3 } corresponds to the point (x,y) =(12,32) ( x , y ) = \left( \frac { 1 } { 2 } , - \frac { \sqrt { 3 } } { 2 } \right) . sin(25π3) =12cos(25π3) =32tan(25π3) =33\begin{array} { l } \sin \left( \frac { 25 \pi } { 3 } \right) = - \frac { 1 } { 2 } \\\\\cos \left( \frac { 25 \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 2 } \\\\\tan \left( \frac { 25 \pi } { 3 } \right) = - \frac { \sqrt { 3 } } { 3 }\end{array}
E) Not possible

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