Prove that $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$
Prove that $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$
$L.H.S.$ $=\sin 2 x+2 \sin 4 x+\sin 6 x$
$=[\sin 2 x+\sin 6 x]+2 \sin 4 x$
$=\left[2 \sin \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]+2 \sin 4 x$
$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=2 \sin 4 x \cos (-2 x)+2 \sin 4 x$
$=2 \sin 4 x \cos 2 x+2 \sin 4 x$
$=2 \sin 4 x(\cos 2 x+1)$
$=2 \sin 4 x\left(2 \cos ^{2} x-1+1\right)$
$=2 \sin 4 x\left(2 \cos ^{2} x\right)$
$=4 \cos ^{2} x \sin 4 x$
$=R.H .S.$
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- [JEE MAIN 2020]
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