Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Asked by Pragya Singh | 1 year ago |  105

##### Solution :-

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

Answered by Abhisek | 1 year ago

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