# How do you evaluate sin^-1(sin((11pi)/10))?

Mar 29, 2018

Evalute the inner bracket first. See below.

#### Explanation:

sin(11*pi/10) = sin((10+1)pi/10 =sin(pi + pi/10)

Now use the identity:

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

I leave the nitty-gritty substitution for you to solve.

Mar 29, 2018

${\sin}^{-} 1 \left(\sin \left(\frac{11 \pi}{10}\right)\right) = - \frac{\pi}{10}$

#### Explanation:

Note:

color(red)((1)sin(pi+theta)=-sintheta

color(red)((2)sin^-1(-x)=-sin^-1x

color(red)((3)sin^-1(sintheta)=theta,where,theta in[-pi/2,pi/2]

WE have,

${\sin}^{-} 1 \left(\sin \left(\frac{11 \pi}{10}\right)\right) = {\sin}^{-} 1 \left(\sin \left(\frac{10 \pi + \pi}{10}\right)\right)$

$= {\sin}^{-} 1 \left(\sin \left(\pi + \frac{\pi}{10}\right)\right) \ldots \ldots \ldots \to A p p l y \left(1\right)$

$= {\sin}^{-} 1 \left(- \sin \left(\frac{\pi}{10}\right)\right) \ldots \ldots \ldots . . \to A p p l y \left(2\right)$

$= - {\sin}^{-} 1 \left(\sin \left(\frac{\pi}{10}\right)\right) \ldots \ldots \ldots . \to A p p l y \left(3\right)$

$= - \frac{\pi}{10} \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

Hence,

${\sin}^{-} 1 \left(\sin \left(\frac{11 \pi}{10}\right)\right) = - \frac{\pi}{10}$