Question

Prove that : 2^5 (sinθ)^4 (cosθ)^2 =cos6θ -2cos4θ -cos2θ +2

Answer 1

Answer:

omg such a beautiful sum u have got

Answer 2

Answer:-

LHS=2(sin6θ+cos6θ)−3(sin4θ+cos4θ)+1

=2{(sin2θ+cos2θ)3−3sin2θcos2θ(sin2θ+cos2θ)}−3(sin2θ+cos2θ)2−2(sin2θcos2θ)}+1

We know, [sin²x+cos²x=1]

=2{1−3sin2θcos2θ}−3{1−2sin2θcos2θ}+1

=2−6sin2θcos2θ−3+6sin2θcos2θ+1

=0

=RHS