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