We know that
sinA+sinB=2sin\( ( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})\)
L.H.S. =sinx+sin3x+sin5x+sin7x
=(sinx+sin5x)+(sin3x+sin7x)
Using the formula and simplifying,
\(2sin ( \dfrac{x+5x}{2}).cos( \dfrac{x-5x}{2})\)
\( 2sin ( \dfrac{3x+7x}{2})cos( \dfrac{3x-7x}{2})\)
=2cos2x[sin3x+sin5x]
\( 2cos2x [2sin (\dfrac{3x+5x}{2}).cos( \dfrac{3x-5x}{2})]\)
=2cos2x[2sin4x.cos(-x)]
Therefore we have,
L.H.S=4cos2xsin4xcosx
R.H.S
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