# How do you find sin^-1(sin ((5pi)/6))?

Mar 27, 2018

${\sin}^{-} 1 \left(\sin \frac{5 \pi}{6}\right) = \frac{\pi}{6}$

#### Explanation:

$\sin \left(\frac{5 \pi}{6}\right) = \frac{1}{2}$, so we're really being asked for ${\sin}^{-} 1 \left(\frac{1}{2}\right) .$

Let

$y = {\sin}^{-} 1 \left(\frac{1}{2}\right)$

Then, from the definition of arcsine,

$\frac{1}{2} = \sin \left(y\right)$

( $y = {\sin}^{-} 1 \left(a\right) \Leftrightarrow a = \sin \left(y\right)$ )

In other words, we want to know where the sine function returns $\frac{1}{2.}$

Sine is equal to $\frac{1}{2}$ for $\frac{\pi}{6} + 2 n \pi , \frac{5 \pi}{6} + 2 n \pi ,$ where $n$ is any integer. But really, since the domain of arcsine is $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right] , \frac{\pi}{6}$ is the only valid answer.

Mar 27, 2018

$\frac{\pi}{6}$

#### Explanation:

Hint:
We know that ,
color(red)(sin^-1(sintheta)=theta, where, color(red)(theta in [-pi/2,pi/2]
Even though,
color(blue)(sin^-1(sin((5pi)/6))!=((5pi)/6), because, color(blue)((5pi)/6 !in [-pi/2,pi/2]
So, we reduce $\theta = \frac{5 \pi}{6.}$ But how ? By using Addition and Subtraction Formulas for sine and cosine .Now enjoy the answer.

Here,

$\frac{5 \pi}{6} = \pi - \frac{\pi}{6}$

(sin^-1(sin((5pi)/6))=sin^-1(sin(pi-pi/6))

$= {\sin}^{-} 1 \left(\sin \left(\frac{\pi}{6}\right)\right) \ldots . \to$, Apply color(red)([sin(pi-theta)=sintheta ]

$= \frac{\pi}{6.} \ldots . . \to$, where, $\frac{\pi}{6} \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$