Question

How do you prove that `cos^2theta - cos^2 3theta = sin4thetasin2theta`?

Answer 1

`LHS=cos^2theta-cos^2 3theta`
`=1/2(1+cos2theta)-1/2(1+cos6theta)`

`=1/2(cos2theta-cos6theta)`

`=1/2(2sin((2theta+6theta)/2)sin((6theta-2theta)/2))`

`=sin4thetasin2theta=RHS`