Question #3065c

Question

862 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-09T08:54:10

See the proof below

We need

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

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

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

Therefore,

#LHS=sin^8x-cos^8x#

#=(sin^4x+cos^4x)(sin^4x-cos^4x)#

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

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

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

#=(sin^2x-cos^2x)(sin^4x+2sin^2cos^2x+cos^4x-2sin^2xcos^2x)#

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

#=(sin^2x-cos^2x)(1-2sin^2xcos^2x)#

#=RHS#

#QED#