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

Solution

Using the identity:

\[ \cos\theta=\sin\left(\frac{\pi}{2}-\theta\right) \]

Therefore,

\[ \sin(\pi\cos x) = \sin\left(\frac{\pi}{2}-\pi\sin x\right) \]

For

\[ \sin A=\sin B \]

either

\[ A=B+2n\pi \]

or

\[ A=\pi-B+2n\pi \]

Taking the principal relation,

\[ \pi\cos x = \frac{\pi}{2}-\pi\sin x \]

Dividing by \(\pi\),

\[ \cos x+\sin x=\frac{1}{2} \]

Now square both sides:

\[ (\sin x+\cos x)^2=\left(\frac{1}{2}\right)^2 \]

\[ \sin^2x+\cos^2x+2\sin x\cos x=\frac{1}{4} \]

Using

\[ \sin^2x+\cos^2x=1 \]

we get

\[ 1+\sin2x=\frac{1}{4} \]

Therefore,

\[ \sin2x=-\frac{3}{4} \]

Similarly, from the second condition,

\[ \pi\cos x = \pi-\left(\frac{\pi}{2}-\pi\sin x\right) \]

\[ \cos x-\sin x=\frac{1}{2} \]

Squaring again,

\[ 1-\sin2x=\frac{1}{4} \]

Hence,

\[ \sin2x=\frac{3}{4} \]

Therefore,

\[ \boxed{ \sin2x=\pm\frac{3}{4} } \]

Final Answer

\[ \boxed{ \sin2x=\pm\frac{3}{4} } \]

Correct Option: (a)