# How do you find the exact value for sin (cos^-1 (-1/4))?

Apr 15, 2015

${\cos}^{-} 1 \left(- \frac{1}{4}\right)$ is a $t$ in $\left[0 , \pi\right]$ with $\cos t = - \frac{1}{4}$

Since $t$ is in quadrant 1 or 2, $\sin t$ is positive.

The problem becomes: $\cos t = - \frac{1}{4}$ and $\sin t > 0$. Find $\sin t$.

Either draw a picture of $t$ on the coordinate system,
or draw a right triangle with acute angle $t$, adjacent side 1 and hypotenuse 4
or use ${\sin}^{2} t + {\cos}^{2} = 1$ to get

$\sin t = \frac{\sqrt{15}}{4}$