Question 7
Multiple Choice
Let P ( x , y ) P ( x , y ) P ( x , y ) denote the point where the terminal side of angle θ \theta θ (in standard position) meets the unit circle. Use the information to evaluate the six trigonometric functions of θ \theta θ . P P P is in Quadrant IV and y = − 3 4 y = - \frac { 3 } { 4 } y = − 4 3 .
A) sin θ = − 7 4 \sin \theta = - \frac { \sqrt { 7 } } { 4 } sin θ = − 4 7 , cos θ = 3 4 \cos \theta = \frac { 3 } { 4 } cos θ = 4 3 , tan θ = − 7 3 \tan \theta = - \frac { \sqrt { 7 } } { 3 } tan θ = − 3 7 , sec θ = 4 3 \sec \theta = \frac { 4 } { 3 } sec θ = 3 4 , csc θ = 4 7 7 \csc \theta = \frac { 4 \sqrt { 7 } } { 7 } csc θ = 7 4 7 , cot θ = 3 7 7 \cot \theta = \frac { 3 \sqrt { 7 } } { 7 } cot θ = 7 3 7 B) sin θ = − 7 4 \sin \theta = - \frac { \sqrt { 7 } } { 4 } sin θ = − 4 7 , cos θ = 3 4 \cos \theta = \frac { 3 } { 4 } cos θ = 4 3 , tan θ = 7 3 \tan \theta = \frac { \sqrt { 7 } } { 3 } tan θ = 3 7 , sec θ = 4 3 \sec \theta = \frac { 4 } { 3 } sec θ = 3 4 , csc θ = − 4 7 7 \csc \theta = - \frac { 4 \sqrt { 7 } } { 7 } csc θ = − 7 4 7 , cot θ = 3 7 7 \cot \theta = \frac { 3 \sqrt { 7 } } { 7 } cot θ = 7 3 7 C) sin θ = − 3 4 \sin \theta = - \frac { 3 } { 4 } sin θ = − 4 3 , cos θ = 7 4 \cos \theta = \frac { \sqrt { 7 } } { 4 } cos θ = 4 7 , tan θ = − 3 7 7 \tan \theta = - \frac { 3 \sqrt { 7 } } { 7 } tan θ = − 7 3 7 , sec θ = 4 7 7 \sec \theta = \frac { 4 \sqrt { 7 } } { 7 } sec θ = 7 4 7 , csc θ = − 4 3 \csc \theta = - \frac { 4 } { 3 } csc θ = − 3 4 , cot θ = − 7 3 \cot \theta = - \frac { \sqrt { 7 } } { 3 } cot θ = − 3 7 D) sin θ = 7 4 cos θ = − 3 4 tan θ = − 7 3 \sin \theta = \frac { \sqrt { 7 } } { 4 } \quad \cos \theta = - \frac { 3 } { 4 } \quad \tan \theta = - \frac { \sqrt { 7 } } { 3 } sin θ = 4 7 cos θ = − 4 3 tan θ = − 3 7 sec θ = 4 3 \sec \theta = \frac { 4 } { 3 } sec θ = 3 4 , csc θ = 7 7 \csc \theta = \frac { \sqrt { 7 } } { 7 } csc θ = 7 7 , cot θ = − 7 7 \cot \theta = - \frac { \sqrt { 7 } } { 7 } cot θ = − 7 7 E) sin θ = 3 4 \sin \theta = \frac { 3 } { 4 } sin θ = 4 3 , cos θ = − 7 4 \cos \theta = - \frac { \sqrt { 7 } } { 4 } cos θ = − 4 7 , tan θ = 3 7 7 \tan \theta = \frac { 3 \sqrt { 7 } } { 7 } tan θ = 7 3 7 , sec θ = − 4 7 7 \sec \theta = - \frac { 4 \sqrt { 7 } } { 7 } sec θ = − 7 4 7 , csc θ = 4 3 \csc \theta = \frac { 4 } { 3 } csc θ = 3 4 , cot θ = 7 3 \cot \theta = \frac { \sqrt { 7 } } { 3 } cot θ = 3 7
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