Simplifying (\sin^6 x+\cos^6 x)-(\sin^4 x+\cos^4 x)+\sin^2 x \cos^2 x

2alr8w

2alr8w

Answered question

2022-01-28

Simplifying (sin6x+cos6x)(sin4x+cos4x)+sin2xcos2x

Answer & Explanation

Aiden Cooper

Aiden Cooper

Beginner2022-01-29Added 14 answers

(sin6x+cos6x)(sin4x+cos4x)+sin2xcos2x
=sin6x+cos6xsin4xcos4x+sin2xcos2x
=cos6x(tan6x+1tan4xsec2{x}sec2x+tan2xsec2x)
=cos6x(tan6x+1tan4x(1+tan2x)(1+tan2x)+tan2x(1+tan2x))
=cos6x(tan6x+1tan4xtan6x1tan2x+tan2x+tan4x)
=cos6x(0)
=0
pripravyf

pripravyf

Beginner2022-01-30Added 12 answers

hint: sin2x+cos2x=1 if you square that, you get
sin4x+2sin2xcos2x+cos4x=1
which you can turn into
sin4x+cos4x=12sin2xcos2x
and that lets you get rid of the 4th powers in the left hand side.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?