Question

If $5\sin x-12\cos x=-13\sin 3x$ and $\sin \varphi =\frac{12}{13}$, 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

The correct option is A $\sin (2x-\frac{\varphi }{2})=0$

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

$\Rightarrow \cos \theta \sin x-\sin \theta \cos x=-\sin 3x$

$\Rightarrow \sin (x-\theta )=-\sin 3x$

$\Rightarrow \sin (x-\theta )=-\sin \left(3x\right)$

$\Rightarrow \sin (x-\theta )=\sin (-3x)$

$\Rightarrow x-\theta =-3x$

$\Rightarrow 4x-\theta =0$

$\Rightarrow 2x-\frac{\theta }{2}=0$

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