Prove that #Cos^6(x)+sin^6(x)=1/8(5+3cos4x)#?

Question

32087 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-09T17:04:57

We'll use

#rarra^3+b^3=(a+b)(a^2-ab+b^2)#

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

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

#rarr2cos^2x=1+cos2x# and

#rarr2sin^2x=1-cos2x#

#LHS=cos^6(x)+sin^6(x)#

#=(cos^2x)^3+(sin^2x)^3#

#=[cos^2x+sin^2x][(cos^2x)^2-cos^2x*sin^2x+sin^2x)^2]#

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

#=[cos^2(2x)+cos^2x*sin^2x]#

#=1/4[4cos^2(2x)+4cos^2x*sin^2x]#

#=1/4[2(1+cos4x)+sin^2(2x)]#

#=2/(4*2)[2+2cos4x+sin^2(2x)]#

#=1/8[4+4cos4x+2sin^2(2x)]#

#=1/8[4+4cos4x+1-cos4x]#

#=1/8[5+3cos4x]=RHS#