Express cos4x as powers of cosx. ?

2 Answers
May 18, 2018

cos4x=cos2(2x)=color(red)[2cos^2(2x)-1

Explanation:

cos2(2x)=cos^2(2x)-sin^2(2x)

=cos^2(2x)-1+cos^2(2x)=color(red)[2cos^2(2x)-1]

=2[cos2x*cos2x]-1=2[(cos^2x-sin^2x)*(cos^2x-sin^2x)]-1

=2[cos^4x-sin^2x*cos^2x-sin^2x*cos^2x+sin^4x]-1

=[2cos^4x-4sin^2x*cos^2x+2sin^4x]-1

May 18, 2018

rarrcos4x=8cos^4x-8cos^2x+1

Explanation:

rarrcos4x

=cos2*(2x)

=cos^2(2x)-sin^2(2x)

=[cos^2x-sin^2x]^2-[2sinx*cosx]^2

=cos^4x-2cos^2x*sin^2x+(sin^2x)^2-4sin^2x*cos^2x

=cos^4x-2cos^2x(1-cos^2x)+(1-cos^2x)^2-4(1-cos^2x)*cos^2x

=cos^4x-2cos^2x+2cos^4x+1-2cos^2x+cos^4x-4cos^2x+4cos^4x

=8cos^4x-8cos^2x+1