How do you find A, B and C, given that A sin ( B x + C ) = cos ( cos^(-1) sin x + sin^(-1) cos x ) + sin (cos^(-1) sin x + sin^(-1) cos x )?

Aug 2, 2016

$\sqrt{2} \sin \left(2 x - \frac{\pi}{4}\right)$

Explanation:

Use $\sin x = \cos \left(\frac{\pi}{2} - x\right) , \cos x = \sin \left(\frac{\pi}{2} - x\right)$

${\cos}^{- 1} \cos a = a , {\sin}^{- 1} \sin b = b$ and

$\sin 2 A - \sin 2 B = 2 \cos \left(A + B\right) \sin \left(A - B\right)$.

Let $u = {\cos}^{- 1} \sin x + {\sin}^{- 1} \cos x$

$= {\cos}^{- 1} \cos \left(\frac{\pi}{2} - x\right) + {\sin}^{- 1} \sin \left(\frac{\pi}{2} - x\right)$

$= \left(\frac{\pi}{2} - x\right) + \left(\frac{\pi}{2} - x\right)$

$= \pi - 2 x$

So, the given equation is

$A \sin \left(B x + C\right)$

$= \cos \left(\pi - 2 x\right) + \sin \left(\pi - 2 x\right)$

$= - \cos 2 x + \sin 2 x$

$= \sin 2 x - \sin \left(\frac{\pi}{2} - 2 x\right)$

$= 2 \cos \left(\frac{\pi}{4}\right) \sin \left(2 x - \frac{\pi}{4}\right)$

$= \sqrt{2} \sin \left(2 x - \frac{\pi}{4}\right)$