# How do you simplify sin(sin^-1(1/2)+cos^-1(3/5))?

Jun 18, 2016

$\sin \left({\sin}^{- 1} \left(\frac{1}{2}\right) + {\cos}^{- 1} \left(\frac{3}{5}\right)\right) = \textcolor{g r e e n}{1.3}$

#### Explanation:

${\sin}^{- 1}$ is by definition an angle in $\left[0 , \pi\right)$
within this interval $\sin \left({\sin}^{- 1} \left(\theta\right)\right) = \theta$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow \sin \left({\sin}^{- 1} \left(\frac{1}{2}\right)\right) = \frac{1}{2}$

${\cos}^{- 1}$ is by definition an angle in $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right]$
within this range a $\cos = \frac{y}{r}$ of $\frac{3}{5}$ implies a standard triangle with
$\textcolor{w h i t e}{\text{XXX}} \frac{x}{r} = \sin = \frac{4}{5}$
So ${\cos}^{- 1} \left(\frac{3}{5}\right) = {\sin}^{- 1} \left(\frac{4}{5}\right)$
and, again, since this is in QI,
$\textcolor{w h i t e}{\text{XXX}} \sin \left({\cos}^{- 1} \left(\frac{3}{5}\right)\right) = \sin \left({\sin}^{- 1} \left(\frac{4}{5}\right)\right) = \frac{4}{5}$

Therefore
$\sin \left({\sin}^{- 1} \left(\frac{1}{2}\right) + {\cos}^{- 1} \left(\frac{3}{5}\right)\right)$
$\textcolor{w h i t e}{\text{XXX}} = \frac{1}{2} + \frac{4}{5}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{13}{10} \left(= 1 \frac{3}{10} = 1.3\right)$