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

1 Answer
Mar 9, 2018

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