Question

Find the range of y, if ${{\sin }^{-1}}x=y$.
(A) $0\le y\le \pi$
(B) $-\dfrac{\pi }{2}\le y\le \dfrac{\pi }{2}$
(C) $0 < y < \pi$
(D) $-\dfrac{\pi }{2} < y < \dfrac{\pi }{2}$

Answer 1

In this question, we have to find the range of y.. We know the domain of sine inverse function which is $\left[ -1,1 \right]$ . Put the extremum value in the function ${{\sin }^{-1}}x=y$ and get the extremum values of y. We know that $\sin \left( \dfrac{\pi }{2} \right)=1$ and $\sin \left( -\dfrac{\pi }{2} \right)=-1$ . Also, we know the property, ${{\sin }^{-1}}\left( \sin x \right)=x$ . Now, solve it further and get the extremum values of y are the range.

Complete step-by-step answer:
According to the question, it is given that

& {{\sin }^{-1}}x=y \\\ & y={{\sin }^{-1}}x \\\ \end{aligned}$$ We know the domain of the sine inverse function which is $$\left[ -1,1 \right]$$ . Here, the extremum values of the domain of inverse sine function are -1 which is minimum and 1 which is maximum. Putting the extremum values in the function $${{\sin }^{-1}}x=y$$ , we get the maximum and minimum values of the range of the function $${{\sin }^{-1}}x=y$$ . Putting $$x=-1$$ in the function $${{\sin }^{-1}}x=y$$ , we get $${{\sin }^{-1}}(-1)=y$$ …………………(1) We know that, $$\sin \left( -\dfrac{\pi }{2} \right)=-1$$ ……………(2) From equation (1) and (2), we get $${{\sin }^{-1}}\left( \sin \left( -\dfrac{\pi }{2} \right) \right)=y$$ ……………(3) We know the property, $${{\sin }^{-1}}\left( \sin x \right)=x$$ . Using this property in equation (3), we get $$\begin{aligned} & {{\sin }^{-1}}\left( \sin \left( -\dfrac{\pi }{2} \right) \right)=y \\\ & \Rightarrow -\dfrac{\pi }{2}=y \\\ \end{aligned}$$ Putting $$x=1$$ in the function $${{\sin }^{-1}}x=y$$ , we get $${{\sin }^{-1}}(1)=y$$ …………………(1) We know that, $$\sin \left( \dfrac{\pi }{2} \right)=1$$ ……………(2) From equation (1) and (2), we get $${{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{2} \right) \right)=y$$ ……………(3) We know the property, $${{\sin }^{-1}}\left( \sin x \right)=x$$ . Using this property in equation (3), we get $$\begin{aligned} & {{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{2} \right) \right)=y \\\ & \Rightarrow \dfrac{\pi }{2}=y \\\ \end{aligned}$$ Now, we have got the extremum values of y which are $$-\dfrac{\pi }{2}$$ and $$\dfrac{\pi }{2}$$ . The minimum value of the function $${{\sin }^{-1}}x=y$$ is $$-\dfrac{\pi }{2}$$ and its maximum value is $$\dfrac{\pi }{2}$$ . So, $$-\dfrac{\pi }{2}\le y\le \dfrac{\pi }{2}$$ . Hence, option (B) is the correct one. **Note:** In this question, one can think of converting the sine inverse into cosine inverse which leads to complexity and increases the chance of mistake. So, never convert into other trigonometric functions if the question asks for range.