We need
sin^2A+cos^2A=1
cos2A=2cos^2A-1
sin2A=2sinAcosA
sin3A=sin2AcosA+cos2AsinA
=2sinAcos^2A+sinA(2cos^2A-1)
=sinA(-1+4cos^2A)
cos3A=cos2AcosA-sin2AsinA
=(2cos^2A-1)cosA-2sinAcosAsinA
=2cos^3A-cosA-2cosA(1-cos^2A)
=4cos^3A-3cosA
Therefore,
LHS=(sin2A)/(2sin3A)*(4(cos2A)^2-1)
=(cancel(2sinA)cosA)/(cancel(2sinA)(-1+4cos^2A))*(4(2cos^2A-1)^2-1)
=cosA/((-1+4cos^2A))*(16cos^4A-16cos^2A+4-1)
=cosA/(cancel((4cos^2A-1)))*cancel((4cos^2A-1))(4cos^2A-3)
=cosA(4cos^2A-3)
=4cos^3A-3cosA
=cos3A
=RHS
QED