Math, asked by nimeshkunal0077, 1 year ago

How to solve it.please tell me.I will mark him as brainlest

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Answered by sandy1816
2

Step-by-step explanation:

sin3x.sin³x+cos3x.cos³x

=(3sinx-4sin³x)sin³x+(4cos³x-3cosx)cos³x

=3(sin⁴x-cos⁴x)-4(sin^6x-cos^6x)

=3(sin²x-cos²x)-4{(sin²)³-(cos²)³}

=3(sin²x-cos²x)-4(sin²x-cos²x)(sin⁴x+sin²xcos²x+cos⁴x)

[using,a³-b³=(a-b)(a²+ab+b²)]

=3(sin²x-cos²x)-4(sin²x-cos²x)(1-sin²xcos²x)

=-3(cos²x-sin²x)+4(cos²x-sin²x)(1-sin²xcos²x)

=(cos²x-sin²x)[-3+4(1-sin²xcos²x]

=(cos²x-sin²x)(1-4sin²xcos²x)

=(cos²x-sin²x)[1-4(1-cos2x/2)(1+cos2x/2)

[using cos2x=1-2sin²x=2cos²x-1]

=(cos²x-sin²x)[1-4{1-cos²2x/4}]

=(cos²x-sin²x)(1-1+cos2x)

=cos2x.cos²2x

=cos³2x

thank you

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