Tunjukkan bahwa cos(6x)=32cos^6(x)-48cos^4(x)+18cos^2(x) -1

cos6x=cos[2(3x)]=2cos2(3x)-1 cos^2x=2cos^2x-1

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

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