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Prove that $\sin 4A=4\sin A\cos^3A-4\cos A\sin^3A$

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  • $\sin 2\theta=2\sin \theta\cos\theta$
  • $\cos 2\theta=\cos^2\theta-\sin^2\theta$
L.H.S
$\sin 4A=\sin 2(2A)$
$\Rightarrow 2\sin 2A.\cos 2A$
$\Rightarrow 2[2\sin A\cos A(\cos ^2A-\sin^2A)]$
$\Rightarrow 4\sin A\cos A(\cos ^2A-\sin^2A)$
$\Rightarrow 4\sin A\cos^3A-4\cos A\sin^3A$=R.H.S
Hence proved
answered Apr 18, 2014 by sreemathi.v
 
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