Prove that sin2 6x – sin2 4x = sin 2x sin 10x

Asked by Pragya Singh | 1 year ago |  107

##### Solution :-

We know that,

sinA+sinB=2sin$$( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})$$

And

sinA-sinB=2cos$$( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})$$

L.H.S.=sin26x-sin24xa

=[sin6x+sin4x)(sin6x-sin4x)

$$[ (2sin \dfrac{6x+4x}{2})cos( \dfrac{6x-4x}{2})]$$

$$[ (2cos \dfrac{6x+4x}{2}).sin( \dfrac{6x-4x}{2})]$$

=(2sin5xcosx)(2cos5xsinx)

Now we know that, sin2x=2sinxcosx ,

Therefore we have,

L.H.S=(2sin5xcos5)(2sinxcosx)

=sin10xsin2x

= R.H.S.

Answered by Abhisek | 1 year ago

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