Math, asked by coolk3901, 10 months ago

Cos 6theta=32cos^6theta-48cos^4theta+18cos^2theta-1

Answers

Answered by abhi1824

Step-by-step explanation:

cos6x=cos[2(3x)]=2cos2(3x)-1 cos2x=2cos2x-1. ... cos6x=cos[2(3x)]=2cos2(3x)-1 cos2x=2cos2x-1 =2[cos(3x)]2-1 cos3x=4cos3x-3cosx =2[4cos3x-3cosx]2-1 (a-b)2=a2+b2-2ab =2[16cos6x+9cos2x-24cos4x]-1 =32cos6x-48cos4x+18cos2x-1. ... ==> cos(6x) = cos 2(3x) = 2cos2(3x) – 1

hope it's helpful to you

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