Find the exact value of the indicated trigonometric function of $\theta$
-$\cos \theta = \frac { 2 } { 5 } , \quad \tan \theta

Question

Find the exact value of the indicated trigonometric function of $\theta$
-$\cos \theta = \frac { 2 } { 5 } , \quad \tan \theta < 0$
Find $\sin \theta$.
A) $- \frac { 3 } { 2 }$
B) $- \frac { \sqrt { 5 } } { 2 }$
C) $- \sqrt { 5 }$
D) $- \frac { \sqrt { 21 } } { 5 }$

Quizplus · Accepted Answer

Given $\cos \theta = \frac{2}{5}$ and $\tan \theta < 0$, this implies $\theta$ is in the second or fourth quadrant where sine is positive in the second quadrant and negative in the fourth quadrant. Since $\tan \theta < 0$, and tangent is the ratio of sine over cosine, for cosine positive and tangent negative, $\theta$ must be in the fourth quadrant where sine is negative. Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we find $\sin \theta = -\sqrt{1 - \cos^2 \theta} = -\sqrt{1 - \left(\frac{2}{5}\right)^2} = -\sqrt{1 - \frac{4}{25}} = -\sqrt{\frac{21}{25}} = -\frac{\sqrt{21}}{5}$.

Find the Exact Value of the Indicated Trigonometric Function Of \theta