Question

What is the value of $si{n}^{-1}\left(sinx\right))$ in the interval $[-\frac{23}{2}\pi ,\frac{21\pi }{2}]$

• A. -11 $\pi$ - x
  
• B. -11 $\pi$ + x
  
• C. $x$
• D. $-x$

Answer 1

The correct option is A -11 $\pi$ - x


We are concerned with the function $f\left(x\right)=si{n}^{-1}\left(sinx\right)$. Lets analyse the function and get the values for different intervals. We know,
$si{n}^{-1\left(sin\right(x\left)\right)}=\{\begin{array}{c}-2n\pi +x,xϵ[2n\pi -\frac{\pi }{2},(2n\pi +\frac{\pi }{2}\left)\right]\\ (2n+1)\pi -x,xϵ\left[\right(2n+1)\pi -\frac{\pi }{2},(2n+1)\pi +\frac{\pi }{2}]\end{array}$
We are given the interval $[\frac{-23}{2}\pi ,\frac{21\pi }{2}]$
Which is equal to the interval $\left[\right(2(-6)+1)\pi -\frac{\pi }{2},(2(-6)+1)+1)\pi +\frac{\pi }{2}]$ which gives n = -6.
$\therefore$ value of $si{n}^{-1}\left(sinx\right)$ will be $(2n+1)\pi -x=(-11\pi -x)$