Question

Question #ce805

Answer 1

`LHS=Cos2thetacos(theta/2)-cos3thetacos(theta/2)`

`=cos(theta/2)(cos2theta-cos3theta)`

`=cos(theta/2)2sin(theta/2)sin((5theta)/2)`

`=2sin(theta/2)cos(theta/2)sin((5theta)/2)`

`=sinthetasin((5theta)/2)=RHS`.