Question

If ${\sin }^{-1}x+{\sin }^{-1}y=\frac{\mathrm{\pi }}{3}$ and ${\cos }^{-1}x-{\cos }^{-1}y=\frac{\mathrm{\pi }}{6}$, find the values of x and y.

Answer 1

$$\begin{gathered} {\cos }^{-1}x-{\cos }^{-1}y=\frac{\mathrm{\pi }}{6} \\ \Rightarrow \frac{\mathrm{\pi }}{2}-{\sin }^{-1}x-\frac{\mathrm{\pi }}{2}+{\sin }^{-1}y=\frac{\mathrm{\pi }}{6}\left[\because {\cos }^{-1}x=\frac{\mathrm{\pi }}{2}-{\sin }^{-1}x\right] \\ \Rightarrow -\left({\sin }^{-1}x-{\sin }^{-1}y\right)=\frac{\mathrm{\pi }}{6} \\ \Rightarrow {\sin }^{-1}x-{\sin }^{-1}y=-\frac{\mathrm{\pi }}{6} \end{gathered}$$
Solving ${\sin }^{-1}x+{\sin }^{-1}y=\frac{\mathrm{\pi }}{3}$ and ${\sin }^{-1}x-{\sin }^{-1}y=-\frac{\mathrm{\pi }}{6}$, we will get
$$\begin{gathered} 2{\sin }^{-1}x=\frac{\mathrm{\pi }}{6} \\ \Rightarrow {\sin }^{-1}x=\frac{\mathrm{\pi }}{12} \\ \Rightarrow x=\sin \frac{\mathrm{\pi }}{12}=\frac{\sqrt{3}-1}{2\sqrt{2}} \\ \mathrm{and} \\ {\sin }^{-1}y=\frac{\mathrm{\pi }}{3}-{\sin }^{-1}x \\ \Rightarrow {\sin }^{-1}y=\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{12} \\ \Rightarrow {\sin }^{-1}y=\frac{\mathrm{\pi }}{4} \\ \Rightarrow y=\sin \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}} \end{gathered}$$