No longer require the answer to this, but if you want to, feel free to try it so I can double check my math (but also my newest question is more urgent).
Prove this identity:
1−8sin2θcos2θ=cos(4θ)
Prove:
1−8sin2θcos2θ=cos(4θ)LHS=1−2(2sinθcosθ)2LHS=1−2sin2(2θ)LHS=1−sin2(2θ)−sin2(2θ)LHS=cos2(2θ)−sin2(2θ)LHS=cos(4θ)LHS=RHSQED