Question

$5\sin x-12\cos x=-13\sin 3x$, if $\varphi ={\sin }^{-1}\left(\frac{12}{13}\right)$, then

• A. $\sin (2x-\frac{\varphi }{2})=0$
• B. $\sin (x+\frac{\varphi }{2})=0$
• C. $\cos (x+\frac{\varphi }{2})=0$
• D. $\cos (2x+\frac{\varphi }{2})=0$

Answer 1

Answer: (A), (C)

The correct options are
A $\sin (2x-\frac{\varphi }{2})=0$
C $\cos (x+\frac{\varphi }{2})=0$

$\sin \varphi =\frac{12}{13}=\frac{P}{H}$

$\cos \varphi =\frac{B}{H}=\frac{\sqrt{{13}^2-{12}^2}}{13}=\frac{5}{13}$

$5\sin x-12\cos x=-13\sin 3x$

$\frac{12}{13}\cos x-\frac{5}{13}\sin x=\sin 3x$

$\sin \varphi \cos x-\cos \varphi \sin x=\sin 3x$

$\sin (\varphi -x)=\sin 3x$

$\sin (\varphi -x)-\sin 3x=0$

$2\cos \left(\frac{\varphi -x+3x}{2}\right)\sin \left(\frac{\varphi -x-3x}{2}\right)=0$

$2\cos (\frac{\varphi }{2}+x).\sin (\frac{\varphi }{2}-2x)=0$

$\cos (\frac{\varphi }{2}+x)=0$

$\sin (\frac{\varphi }{2}-2x)=0$

$\sin (-\theta )=-\sin \theta$

$\Rightarrow \sin (2x-\frac{\varphi }{2})=0$