# How do you evaluate sin(arccos (1/3)) ?

Jun 10, 2016

$\sin \left(\arccos \left(\frac{1}{3}\right)\right) = \pm \frac{2 \sqrt{2}}{3}$

#### Explanation:

Let $\arccos \left(\frac{1}{3}\right) = \theta$. This means

$\cos \theta = \frac{1}{3}$ and hence

$\sin \theta = \sqrt{1 - {\left(\frac{1}{3}\right)}^{2}} = \pm \sqrt{1 - \frac{1}{9}} = \pm \sqrt{\frac{8}{9}} = \pm \frac{2 \sqrt{2}}{3}$

We are using both plus and minus as if $\cos \theta$ is in first quadrant, $\sin \theta$ could be positive and if $\cos \theta$ is in fourth quadrant, $\sin \theta$ could be negaitive.

As such $\sin \left(\arccos \left(\frac{1}{3}\right)\right) = \sin \theta = \pm \frac{2 \sqrt{2}}{3}$