Question

On what interval is the identity ${{\sin }^{-1}}\left( \sin x \right)=x$ valid?

Answer 1

Hint : We first define the difference between principal value range and general solution. The range of $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ satisfies the equation ${{\sin }^{-1}}\left( \sin x \right)=x$. We use an example to find the validity of the identity relation of ${{\sin }^{-1}}\left( \sin x \right)=x$.

Complete step-by-step answer :
The expression of ${{\sin }^{-1}}\left( \sin x \right)=x$ expresses the particular solution of the inverse trigonometric function.
We can have many general solutions for the value of $x\in \mathbb{R}$.
Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty$. In that case we have to use the formula $x=n\pi +{{\left( -1 \right)}^{n}}a$ for $\sin \left( x \right)=\sin a$ where $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$.
${{\sin }^{-1}}\left( \sin x \right)=x$ satisfies only when it lies in the principal value range of $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$.
For given problem ${{\sin }^{-1}}\left( 1 \right)=x$, the general solution will be $x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{2}$. Here $n\in \mathbb{Z}$.
But for ${{\sin }^{-1}}\left( \sin \dfrac{\pi }{6} \right)=\dfrac{\pi }{6}$ satisfies because of the principal value range.

Note : We can always use the concept of period for the inverse trigonometric function. We need to be careful about the shift from $y=\sin x$ to $y=\sin \left( x+\pi \right)$. The addition or subtraction of the constant decides the direction of the shift along with the solution. If the value is positive then the graph shifts left and if the value is negative then it shifts right.