Question
Prove that: $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^2 x \sin 4 x$
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Answer
We have L.H.S. = sin 2x + 2 sin 4x + sin 6x
= [ sin 4x + sin 2x ] + [sin 6x + sin 4x]
$ = 2\sin \left( {\frac{{4x + 2x}}{2}} \right)\cos \left( {\frac{{4x - 2x}}{2}} \right)$$ + 2\sin \left( {\frac{{6x + 4}}{2}} \right)\cos \left( {\frac{{6x - 4x}}{2}} \right)$
$\left[ {\because \sin C + \sin D = } \right.$$\left. { = 2\sin \left( {\frac{{C - D}}{2}} \right)\cos \left( {\frac{{C - D}}{2}} \right)} \right]$
= 2 sin 2x cos x + 2 sin 5x cos x
= 2 cos x [sin 3x + sin 5x]
$ = 2\cos x\left[ {2\sin \left( {\frac{{5x + 3x}}{2}} \right) \cdot \cos \left( {\frac{{5x - 3x}}{2}} \right)} \right]$
$ = 2\cos x\left[ {2\sin 4x \cdot \cos x} \right]= 4cos^2x\sin4x= R.H.S.$
= [ sin 4x + sin 2x ] + [sin 6x + sin 4x]
$ = 2\sin \left( {\frac{{4x + 2x}}{2}} \right)\cos \left( {\frac{{4x - 2x}}{2}} \right)$$ + 2\sin \left( {\frac{{6x + 4}}{2}} \right)\cos \left( {\frac{{6x - 4x}}{2}} \right)$
$\left[ {\because \sin C + \sin D = } \right.$$\left. { = 2\sin \left( {\frac{{C - D}}{2}} \right)\cos \left( {\frac{{C - D}}{2}} \right)} \right]$
= 2 sin 2x cos x + 2 sin 5x cos x
= 2 cos x [sin 3x + sin 5x]
$ = 2\cos x\left[ {2\sin \left( {\frac{{5x + 3x}}{2}} \right) \cdot \cos \left( {\frac{{5x - 3x}}{2}} \right)} \right]$
$ = 2\cos x\left[ {2\sin 4x \cdot \cos x} \right]= 4cos^2x\sin4x= R.H.S.$
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