# How do you find the exact value of Sin(arcsin(3/5)+(arctan-2))?

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

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

## First solution

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

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

Let $A = \arcsin \left(\frac{3}{5}\right)$ and $B = \arctan \left(\frac{\textcolor{b l u e}{- 2}}{1}\right)$

then $\sin A = \frac{3}{5}$ and

computed using Pythagorean relation ${c}^{2} = {a}^{2} + {b}^{2}$

$\cos A = \frac{4}{5}$

also

$\sin B = \frac{- 2}{\sqrt{5}}$

$\cos B = \frac{1}{\sqrt{5}}$

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

$\sin \left(A + B\right) = \frac{3}{5} \cdot \frac{1}{\sqrt{5}} + \frac{4}{5} \cdot \frac{- 2}{\sqrt{5}} = - \frac{5}{5 \sqrt{5}}$

$\sin \left(A + B\right) = - \frac{1}{\sqrt{5}} = - \frac{\sqrt{5}}{5}$
$\textcolor{g r e e n}{\text{The 4th quadrant angle}} = A + B = - {63.4349}^{\circ}$

## second solution

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

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

Let $A = \arcsin \left(\frac{3}{5}\right)$ and $B = \arctan \left(\frac{2}{\textcolor{b l u e}{- 1}}\right)$

then $\sin A = \frac{3}{5}$ and

computed using Pythagorean relation ${c}^{2} = {a}^{2} + {b}^{2}$

$\cos A = \frac{4}{5}$

also

$\sin B = \frac{2}{\sqrt{5}}$

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

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

$\sin \left(A + B\right) = \frac{3}{5} \cdot \frac{- 1}{\sqrt{5}} + \frac{4}{5} \cdot \frac{2}{\sqrt{5}} = \frac{5}{5 \sqrt{5}}$

$\sin \left(A + B\right) = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

$\textcolor{g r e e n}{\text{The 2nd quadrant angle}} = A + B = {116.565}^{\circ}$
Have a nice day !!! from the Philippines