cos 6x=32cos6 x-48cos" x + 18 cos' x-1. – InterviewSolution

1.

cos 6x=32cos6 x-48cos" x + 18 cos' x-1.

Answer»

cos6x=cos[2(3x)]=2cos^2(3x)-1 cos2x=2cos^2x-1

=2[cos(3x)]^2-1 cos3x=4cos^3x-3cosx

=2[4cos^3x-3cosx]^2-1 (a-b)^2=a^2+b^2-2ab

=2[16cos^6x+9cos^2x-24cos^4x]-1

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

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