# How do you find the exact value of sin[arccos(-2/3)+arcsin(1/4)]?

Jul 26, 2016

$\sin \left(\arccos \left(- \frac{2}{3}\right) + \arcsin \left(\frac{1}{4}\right)\right) = \frac{5 \sqrt{3}}{12} - \frac{1}{6}$

#### Explanation:

Let $\alpha = \arccos \left(- \frac{2}{3}\right)$ and $\beta = \arcsin \left(\frac{1}{4}\right)$

$\alpha$ is in Q2, so $\sin \left(\alpha\right) > 0$ and we find:

$\sin \left(\alpha\right) = \sqrt{1 - {\cos}^{2} \left(\alpha\right)} = \sqrt{1 - {\left(- \frac{2}{3}\right)}^{2}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$

$\beta$ is in Q1, so $\cos \left(\beta\right) > 0$ and we find:

$\cos \left(\beta\right) = \sqrt{1 - {\sin}^{2} \left(\beta\right)} = \sqrt{1 - {\left(\frac{1}{4}\right)}^{2}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$

Then:

$\sin \left(\alpha + \beta\right)$

$= \sin \alpha \cos \beta + \sin \beta \cos \alpha$

$= \left(\frac{\sqrt{5}}{3}\right) \left(\frac{\sqrt{15}}{4}\right) + \left(\frac{1}{4}\right) \left(- \frac{2}{3}\right)$

$= \frac{5 \sqrt{3}}{12} - \frac{1}{6}$