# How do you evaluate Sin[Arccos(-1/5)]?

Jun 29, 2015

$\sin \left(\arccos \left(- \frac{1}{5}\right)\right) = \frac{2 \sqrt{6}}{5}$

#### Explanation:

Let $\alpha = \arccos \left(- \frac{1}{5}\right)$

By definition of $\arccos$, we have $0 \le \alpha \le \pi$

More specifically, since $\cos \left(\alpha\right) = - \frac{1}{5} < 0$,

we have $\frac{\pi}{2} < \alpha \le \pi$

Since $\alpha$ is in this quadrant, $\sin \left(\alpha\right) \ge 0$

Now ${\cos}^{2} \left(\alpha\right) + {\sin}^{2} \left(\alpha\right) = 1$

So:

$\sin \left(\alpha\right) = \sqrt{1 - {\cos}^{2} \left(\alpha\right)}$

(must be positive square root since $\sin \left(\alpha\right) \ge 0$)

=sqrt(1-(-1/5)^2))

$= \sqrt{1 - \frac{1}{25}}$

$= \sqrt{\frac{24}{25}}$

$= \sqrt{6 \cdot \frac{4}{25}}$

$= \sqrt{6} \sqrt{\frac{4}{25}}$

$= \frac{2 \sqrt{6}}{5}$