Verify Cos^2x-cos^4x=sin^2x-sin^4x?

3 Answers
May 10, 2018

Please see the proof below

Explanation:

We need

cos^2x+sin^2x=1

Therefore,

LHS=cos^2x-cos^4x

=cos^2x(1-cos^2x)

=cos^2xsin^2x

=(1-sin^2x)sin^2x

=sin^2x-sin^4x

=RHS

QED

May 10, 2018

Kindly refer to Explanation.

Explanation:

cos^2x-cos^4x,

=(1-cos^2x)cos^2x,

=(sin^2x)(1-sin^2x),

=sin^2x-sin^4x, as desired!

Aliter :

We have, cos^4x-sin^4x,

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

=cos^2x-sin^2x, i.e.,

cos^4x-sin^4x=cos^2x-sin^2x," equivalently, "

sin^2x-sin^4x=cos^2x-cos^4x.

May 10, 2018

Read below

Explanation:

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

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