# If Sin (π Cos X) = Cos (π Sin X), Then Sin 2 X = - Mathematics

MCQ

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,
$\Rightarrow \sin {}^2 x + \cos {}^2 x + 2\sin x \cos x = \frac{1}{4}$
$\Rightarrow 1 + \sin 2x = \frac{1}{4}$
$\Rightarrow \sin 2x = \frac{1}{3}$
$\text{Therefore, }\sin 2x = \pm \frac{1}{3}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 7 Values of Trigonometric function at sum or difference of angles
Q 14 | Page 28