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Question -

Prove that:┬аcos 4x┬а= 1 тАУ 8sin2┬аx┬аcos2┬аx



Answer -

Consider

LHS = cos 4x

We can write it as

= cos 2(2x)

Using the formula cos2A┬а= 1 тАУ 2 sin2┬аA

= 1 тАУ 2 sin2┬а2x

Again by using theformula sin2A┬а= 2sin┬аA┬аcos┬аA

= 1 тАУ 2(2 sin┬аx┬аcos┬аx)┬а2

So we get

= 1 тАУ 8 sin2x┬аcos2x

= R.H.S.

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