If sin (π cos x) = cos (π sin x), then sin 2 x =
Options
- \[\pm \frac{3}{4}\]
- \[\pm \frac{4}{3}\]
- \[\pm \frac{1}{3}\]
none of these
Solution
\[ \pm \frac{1}{3}\]
\[\sin \left( \pi \cos x \right) = \cos \left( \pi \sin x \right)\]
\[\text{ As we know that }\sin x = - \cos \left( \frac{\pi}{2} + x \right)\]
\[ \Rightarrow - \cos \left( \frac{\pi}{2} + \pi \cos x \right) = \cos \left( \pi \sin x \right)\]
\[ \Rightarrow \frac{- \pi}{2} - \pi \cos x = \pi \sin x\]
\[ \Rightarrow \pi \sin x - \pi \cos x = \frac{\pi}{2}\]
\[ \Rightarrow \sin x - \cos x = \frac{1}{2}\]
Squaring both sides we get,
\[ \sin^2 x + \cos^2 x - 2 \sin x \cos x = \frac{1}{4}\]
\[ \Rightarrow 1 - \sin 2x = \frac{1}{4}\]
\[ \Rightarrow \sin 2x = \frac{1}{3}\]
\[\text{ And we know that }\sin x = \cos \left( \frac{\pi}{2} - x \right)\]
\[ \Rightarrow \cos \left( \frac{\pi}{2} - \pi \cos x \right) = \cos \left( \pi \sin x \right)\]
\[ \Rightarrow \frac{\pi}{2} - \pi \cos x = \pi \sin x\]
\[ \Rightarrow \pi \sin x + \pi \cos x = \frac{\pi}{2}\]
\[ \Rightarrow \sin x + \cos x = \frac{1}{2}\]
Squaring both sides we get,
\[ \Rightarrow \sin {}^2 x + \cos {}^2 x + 2\sin x \cos x = \frac{1}{4}\]
\[ \Rightarrow 1 + \sin 2x = \frac{1}{4}\]
\[ \Rightarrow \sin 2x = \frac{1}{3}\]
\[\text{Therefore, }\sin 2x = \pm \frac{1}{3}\]
