Question

How do you prove ${{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x$?

Answer 1

Hint: We first assume the angle form as ${{\sin }^{-1}}\left( -x \right)=\alpha $ to find the value of $x$. Then we convert the sign of the expression from negative to positive using $\sin \left( -\alpha \right)=-\sin \alpha $. We take the inverse form to equate it with $\alpha $ and prove the given.

Complete step by step answer:
The given equalities are angles of inverse trigonometric functions.
We assume ${{\sin }^{-1}}\left( -x \right)=\alpha $.
From trigonometric form we get
$\begin{align}
  & {{\sin }^{-1}}\left( -x \right)=\alpha \\
 & \Rightarrow -x=\sin \alpha \\
 & \Rightarrow x=-\sin \alpha \\
\end{align}$
Now we want to change the sign of the trigonometric form using its angle.
We get $-\sin \alpha =\sin \left( -\alpha \right)$. Therefore, $x=\sin \left( -\alpha \right)$.
We again take the inverse trigonometric form and get
\[\begin{align}
  & x=\sin \left( -\alpha \right) \\
 & \Rightarrow -\alpha ={{\sin }^{-1}}x \\
 & \Rightarrow \alpha =-{{\sin }^{-1}}x \\
\end{align}\]
We equal the values of $\alpha $ to get \[\alpha ={{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x\].
Thus proved ${{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x$.

Note: We need to remember the conditions for the inverse formula. We are taking the inverse using the general solution. The values of angles are in the primary range of $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$. 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 $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$.