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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove : $\cot 4x(\sin 5x+\sin 3x)=\cot x(\sin 5x-\sin 3x)$

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Toolbox:
  • $\sin A+\sin B=2\sin\large\frac{A+B}{2}$$.\cos\large\frac{A-B}{2}$
  • $2\cos A\sin B=\sin(A+B)-\sin (A-B)$
L.H.S
$\cot 4x(\sin 5x+\sin 3x)$
$\Rightarrow \cot 4x\times 2\sin \large\frac{5x+3x}{2}.$$\cos \large\frac{5x-3x}{2}$
$\Rightarrow \cot x\times 2\sin 4x.\cos x$
$\Rightarrow \large\frac{\cos 4x}{\sin 4x}$$\times 2\sin 4x.\cos x$
$\Rightarrow 2\cos 4x\cos x\times \large\frac{\sin x}{\sin x}$
$\Rightarrow \large\frac{\cos x}{\sin x}$$2\cos 4x\sin x$
$\Rightarrow \cot x(\sin (4x+x)-\sin (4x-x))$
$\Rightarrow \cot x(\sin 5x-\sin 3x)$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 
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