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Exercise: Sum and Difference Formulas · Q21

Q.Prove that cos⁡7x−cos⁡5xsin⁡7x+sin⁡5x=−tan⁡x\dfrac{\cos 7x - \cos 5x}{\sin 7x + \sin 5x} = -\tan x.

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Numerator, by cos⁡X−cos⁡Y=−2sin⁡X+Y2sin⁡X−Y2\cos X-\cos Y=-2\sin\dfrac{X+Y}{2}\sin\dfrac{X-Y}{2} with X=7x,Y=5xX=7x,Y=5x:

cos⁡7x−cos⁡5x=−2sin⁡6xsin⁡x.\cos7x-\cos5x = -2\sin6x\sin x.

Denominator, by sin⁡X+sin⁡Y=2sin⁡X+Y2cos⁡X−Y2\sin X+\sin Y=2\sin\dfrac{X+Y}{2}\cos\dfrac{X-Y}{2}:

sin⁡7x+sin⁡5x=2sin⁡6xcos⁡x.\sin7x+\sin5x = 2\sin6x\cos x.

So …

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