Simplify #sin^4x-cos^4x+cos^2x#?

Question

11847 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-17T06:55:18

The answer is #=sin^2x#

We need

#sin^2x+cos^2x=1#

#a^2-b^2=(a+b)(a-b)#

#a^4-b^4=(a^2+b^2)(a^2-b^2)#

The expression is

#sin^4x-cos^4x+cos^2x#

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

#=(sin^2x-cancelcos^2x)+cancelcos^2x#

#=sin^2x#

Answer 2 · 2017-04-17T06:55:48

#sin^4x-cos^4x+cos^2x=sin^2x#

#sin^4x-cos^4x+cos^2x#

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

= #sin^2x-cos^2x+cos^2x#

= #sin^2x#