Question

cos^8@-tan^8@=(cos^2@-sin^2@)(1-2sin^2@cos^@)

Answer 1

Let's start form the Left Hand Side("LHS"):

=>" LHS "=sin^2xcos^2x =(2/2sinxcosx)^2

Remember that, 2sinxcosx=sin2x

=>" LHS "=(1/2sin2x)^2 = 1/4sin^2(2x)

Use the half angle identity, cos2theta=1-2sin^2theta

Replace theta by 2x =>cos4x=1-2sin^2(2x)

Rearrange that " ; "sin^2(2x)=1/2(1-cos4x)

=> " LHS"=1/4(1/2(1-cos4x))

=1/8(1-cos4)x=(1-cos(4x))/8=" RHS"

And that's it!