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Evaluate $\sin 3x+\sin 2x-\sin x=4\sin x\cos\large\frac{x}{2}$$\cos\large\frac{3x}{2}$

1 Answer

Toolbox:
  • $\sin 3x=2\sin \large\frac{3x}{2}$$\cos \large\frac{3x}{2}$
  • $\sin x-\sin y=2\cos \large\frac{x+y}{2}$$.\sin \large\frac{x-y}{2}$
L.H.S
$\sin 3x+(\sin 2x-\sin x)$
$\Rightarrow 2\sin \large\frac{3x}{2}$$.\cos \large\frac{3x}{2}$$+2\cos \large\frac{3x}{2}$$\sin \large\frac{x}{2}$
$\Rightarrow 2\cos \large\frac{3x}{2}$$[\sin \large\frac{3x}{2}$$+\sin \large\frac{x}{2}]$
$\Rightarrow 2\cos \large\frac{3x}{2}$$2\sin \large\frac{(3x/2)+(x/2)}{2}$$\cos \large\frac{(3x/2)-(x/2)}{2}$
$\Rightarrow 4\cos \large\frac{3x}{2}$$\sin x\cos\large\frac{x}{2}$
$\Rightarrow 4\sin x\cos\large\frac{x}{2}$$\cos \large\frac{3x}{2}$=R.H.S
Hence proved.
answered Apr 18, 2014 by sreemathi.v
 

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