Question

If sin (π cos x) = cos (π sin x), then sin 2 x =

(a) $\pm \frac{3}{4}$

(b) $\pm \frac{4}{3}$

(c) $\pm \frac{1}{3}$

(d) none of these

Answer 1

$$\begin{gathered} \left(\mathrm{c}\right)\pm \frac{1}{3} \\ \sin \left(\mathrm{\pi }\cos x\right)=\cos \left(\mathrm{\pi }\sin x\right) \\ \mathrm{As}\mathrm{we}\mathrm{know}\mathrm{that}\sin x=-\cos \left(\frac{\mathrm{\pi }}{2}+x\right) \\ \Rightarrow -\cos \left(\frac{\mathrm{\pi }}{2}+\mathrm{\pi }\cos x\right)=\cos \left(\mathrm{\pi }\sin x\right) \\ \Rightarrow \frac{-\mathrm{\pi }}{2}-\mathrm{\pi }\cos x=\mathrm{\pi }\sin x \\ \Rightarrow \mathrm{\pi }\sin x-\mathrm{\pi }\cos x=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \sin x-\cos x=\frac{1}{2} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}, \\ {\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x=\frac{1}{4} \\ \Rightarrow 1-\sin 2x=\frac{1}{4} \\ \Rightarrow \sin 2x=\frac{1}{3} \\ \mathrm{And}\mathrm{we}\mathrm{know}\mathrm{that}\sin x=\cos \left(\frac{\mathrm{\pi }}{2}-x\right) \\ \Rightarrow \cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{\pi }\cos x\right)=\cos \left(\mathrm{\pi }\sin x\right) \\ \Rightarrow \frac{\mathrm{\pi }}{2}-\mathrm{\pi }\cos x=\mathrm{\pi }\sin x \\ \Rightarrow \mathrm{\pi }\sin x+\mathrm{\pi }\cos x=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \sin x+\cos x=\frac{1}{2} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}, \\ \Rightarrow \sin {}^{2}x+\cos {}^2\mathrm{x}+2\sin x\cos x=\frac{1}{4} \\ \Rightarrow 1+\sin 2x=\frac{1}{4} \\ \Rightarrow \sin 2x=\frac{1}{3} \\ \mathrm{Therefore},\sin 2x=\pm \frac{1}{3} \end{gathered}$$