# How do you evaluate sin^-1(sin((5pi)/3))?

Mar 18, 2016

$= \frac{5 \pi}{3}$

#### Explanation:

Given ${\sin}^{-} 1 \left(\sin \left(\frac{5 \pi}{3}\right)\right)$
We know that
${\sin}^{-} 1 \left(\sin \left(x\right)\right) = \cancel{{\sin}^{-} 1} \left(\cancel{\sin} \left(x\right)\right) = \left(x\right)$
or $= x$

Following this we obtain
${\sin}^{-} 1 \left(\sin \left(\frac{5 \pi}{3}\right)\right) = \frac{5 \pi}{3}$

Mar 18, 2016

$\frac{4 \pi}{3} , \frac{5 \pi}{3}$

#### Explanation:

Trig unit circle and trig table give -->
$\sin \left(\frac{5 \pi}{3}\right) = \sin \left(- \frac{\pi}{3} + \frac{6 \pi}{3}\right) = \sin \left(- \frac{\pi}{3} + 2 \pi\right) =$
$= - \sin \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2.}$
Next, find $\arcsin \left(- \frac{\sqrt{3}}{2}\right)$
$x = - \frac{\sqrt{3}}{2}$ --> 2 solutions -->
arc $x = \frac{4 \pi}{3}$ and arc $x = - \frac{\pi}{3}$ , or $x = \frac{5 \pi}{3}$ (co-terminal)

Check.
$x = \frac{4 \pi}{3}$ --> $\sin x = - \sin \frac{\pi}{3} = - \frac{\sqrt{3}}{2} = \sin \left(\frac{5 \pi}{3}\right)$. OK
$x = \frac{5 \pi}{3}$ --> $\sin x = - \sin \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2} = \sin \left(\frac{5 \pi}{3}\right) .$ OK