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Prove $\large\frac{\sin x-\sin 3x}{\sin^2x-\cos^2x}$$=2\sin x$

1 Answer

Toolbox:
  • $\sin x-\sin y=2\cos\large\frac{x+y}{2}$$\sin \large\frac{x-y}{2}$
  • $\cos 2x=\cos^2x-\sin^2x$
$\large\frac{\sin x-\sin 3x}{\sin^2x-\cos^2x}$
$\Rightarrow \large\frac{-(\sin 3x-\sin x)}{-(\cos^2x-\sin^2x)}$
$\Rightarrow \large\frac{2\cos \Large\frac{3x+x}{2}.\sin \Large\frac{3x-x}{2}}{\cos 2x}$
$\Rightarrow \large\frac{2\cos 2x.\sin x}{\cos 2x}$
$\Rightarrow 2\sin x$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 
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