Prove that cos2 2x – cos2 6x = sin 4x sin 8x

Asked by Pragya Singh | 1 year ago |  113

##### Solution :-

We know that,

cosA+cosB=2cos$$( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})$$

cosA-cosB=-2sin $$( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})$$

L.H.S. =cos22x-cos26x

=(cos2x+cos6x)(cos2x-6x)

$$[ 2cos(\dfrac{2x+6x}{2})cos( \dfrac{2x-6x}{2})]$$

$$[ - 2sin(\dfrac{2x+6x}{2})sin( \dfrac{2x-6x}{2})]$$

We get

= [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]

It can be written as

= [2 cos 4x cos 2x] [–2 sin 4(–sin 2x)]

So we get

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= RHS

Answered by Abhisek | 1 year ago

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