# How do you find the value of sin^(-1)(sin (cos^(-1) (sin (pi/12))))?

Jul 17, 2016

${\sin}^{-} 1 \left(\sin \left({\cos}^{-} 1 \left(\sin \left(\frac{\pi}{12}\right)\right)\right)\right) = 5 \frac{\pi}{12}$.

#### Explanation:

We will use the following Rules :

$\left(R 1\right) : \sin \theta = \cos \left(\frac{\pi}{2} - \theta\right)$

$\left(R 2\right) : {\cos}^{-} 1 \left(\cos \theta\right) = \theta , \theta \in \left[o , \pi\right]$

$\left(R 3\right) : {\sin}^{-} 1 \left(\sin \theta\right) = \theta , \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

Let us note that, by

$\left(R 1\right) , \sin \left(\frac{\pi}{12}\right) = \cos \left(\frac{\pi}{2} - \frac{\pi}{12}\right) = \cos \left(5 \frac{\pi}{12}\right)$

So, ${\cos}^{-} 1 \left(\sin \left(\frac{\pi}{12}\right)\right) = {\cos}^{-} 1 \left(\cos \left(5 \frac{\pi}{12}\right)\right)$, where,

$5 \frac{\pi}{12} \in \left[0 , \pi\right]$, so, using $\left(R 2\right)$, we get,

${\cos}^{-} 1 \left(\cos \left(5 \frac{\pi}{12}\right)\right) = 5 \frac{\pi}{12}$

Again, as $5 \frac{\pi}{12} \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, by $\left(R 3\right)$, we have,

${\sin}^{-} 1 \left(\sin \left(5 \frac{\pi}{12}\right)\right) = 5 \frac{\pi}{12}$

Finally, ${\sin}^{-} 1 \left(\sin \left({\cos}^{-} 1 \left(\sin \left(\frac{\pi}{12}\right)\right)\right)\right) = 5 \frac{\pi}{12}$.

Hope, this will be of Help! Enjoy Maths.!

Jul 17, 2016

$\frac{5 \pi}{12} = {75}^{o}$

#### Explanation:

Use $\sin a = \cos \left(\frac{\pi}{2} - a\right) \mathmr{and} ,$

if $y = f \left(x\right) , x = {f}^{- 1} y , f {f}^{- 1} y = y \mathmr{and} {f}^{- 1} y f \left(x\right) = x$.

The given expression is

${\sin}^{- 1} \sin {\cos}^{- 1} \sin \left(\frac{\pi}{12}\right)$

$= {\sin}^{- 1} \sin \left({\cos}^{- 1} \cos \left(\frac{\pi}{2} - \frac{\pi}{12}\right)\right)$

$= {\sin}^{- 1} \sin \left(\frac{5 \pi}{12}\right)$

$= \frac{5 \pi}{12}$