Establish the identity: sin (2theta)[costheta+cos(3theta)]=-cos(2theta)[sintheta+sin(3theta)] what are the steps to getting that answer? Solution sin(2)[cos+cos(3)] = sin(2)[cos(3)+cos]...................................................................since cosA+cosB=2cos((A+B)/2)cos((A-B)/2) = sin(2)[2cos((3+)/2)cos((3-)/2)] = sin(2)[2cos(2)cos()] = cos(2)[2sin(2)cos()]...................................................................since sinAcosB=(1/2)[sin(A+B)+sin(A-B)] = cos(2)[2(1/2)[sin(2+)+sin(2-)]] = cos(2)[sin(3)+sin] = cos(2)[sin+sin(3)] therefore sin(2)[cos+cos(3)] =cos(2)[sin+sin(3)] sin(2)[cos+cos(3)] - cos(2)[sin+sin(3)].