# How do you simplify the expression sin(arcsin(3/5)+arctan(-2))?

Mar 6, 2018

$\sin \left(\arcsin \left(\frac{3}{5}\right) + \arctan \left(- 2\right)\right) = - \frac{1}{\sqrt{5}}$

#### Explanation:

Let us note that the range of $\arcsin x$ and $\arctan x$ is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

Now let $\arcsin \left(\frac{3}{5}\right) = A$ and $\arctan \left(- 2\right) = B$,

then $\sin A = \frac{3}{5}$ and $\tan B = - 2$

therefore $\cos A = \sqrt{1 - {\left(\frac{3}{5}\right)}^{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

and $\sec B = \sqrt{1 + {\left(- 2\right)}^{2}} = \sqrt{5}$ i,e. $\cos B = \frac{1}{\sqrt{5}}$

and as $\sin A = \tan B \cos B = - \frac{2}{\sqrt{5}}$

we have $\sin \left(\arcsin \left(\frac{3}{5}\right) + \arctan \left(- 2\right)\right)$

= $\sin \left(A + B\right)$

= $\sin A \cos B + \cos A \sin B$

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

= $\frac{3 - 8}{5 \sqrt{5}}$

= $- \frac{5}{5 \sqrt{5}}$

= $- \frac{1}{\sqrt{5}}$