# If tan(theta) = 3/4 and sin(theta) < 0, how do you find cos(theta)?

May 5, 2017

$- \frac{4}{5}$

#### Explanation:

Since $\tan \left(\theta\right) = \frac{3}{4} > 0$ and $\sin \left(\theta\right) < 0$, $\theta$ is in quadrant 3 (since $\tan \left(\theta\right) > 0$ iff $\theta$ is in quadrant 1 or 3 and $\sin \left(\theta\right) < 0$ iff $\theta$ is in quadrant 3 or 4).

$\cos \left(\theta\right)$ must then be negative.

Remember that ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$. Divide both sides by ${\cos}^{2} \left(\theta\right)$ to get ${\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) + 1 = \frac{1}{\cos} ^ 2 \left(\theta\right)$. Since $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$, this is simply ${\tan}^{2} \left(\theta\right) + 1 = \frac{1}{\cos} ^ 2 \left(\theta\right)$. Thus, ${\cos}^{2} \left(\theta\right) = \frac{1}{{\tan}^{2} \left(\theta\right) + 1}$.

Since $\tan \left(\theta\right) = \frac{3}{4}$, ${\cos}^{2} \left(\theta\right) = \frac{1}{{\left(\frac{3}{4}\right)}^{2} + 1} = \frac{16}{25}$. Since $\cos \left(\theta\right)$ is negative, $\cos \left(\theta\right) = - \sqrt{\frac{16}{25}} = - \frac{4}{5}$.