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Question -

Prove that:┬аsin 2x┬а+ 2sin 4x┬а+ sin 6x┬а= 4cos2┬аx┬аsin4x



Answer -


By furthersimplification

= 2 sin 4x┬аcos (тАУ2x) + 2 sin 4x

It can be written as

= 2 sin 4x┬аcos2x┬а+ 2 sin 4x

Taking common terms

= 2 sin 4x┬а(cos2x┬а+ 1)

Using the formula

= 2 sin 4x┬а(2 cos2┬аx┬атАУ1 + 1)

We get

= 2 sin 4x┬а(2 cos2┬аx)

= 4cos2┬аx┬аsin4x

= R.H.S.

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