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

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

Explanation

Solution

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

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

& x=\sin \left( -\alpha \right) \\\ & \Rightarrow -\alpha ={{\sin }^{-1}}x \\\ & \Rightarrow \alpha =-{{\sin }^{-1}}x \\\ \end{aligned}$$ We equal the values of $\alpha $ to get $$\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 $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$.