use [tex] a^2-b^2=(a+b)(a-b)[/tex]
to get [tex] (\cos^3A-\sin^3A)(\cos^3A+\sin^3A)[/tex]
then use [tex] a^3+b^3=(a+b)(a^2+b^2-ab)[/tex]
and [tex]a^3-b^3=(a-b)(a^2+b^2+ab)[/tex]
also, [tex] \sin^2\theta+\cos^2\theta=1[/tex]
to get [tex](\cos A-\sin A)(1+\sin A\cos A)(\cos A+ \sin A)(1-\sin A\cos A)[/tex]
then again use the first identity In both pairs, i.e.
[tex](\cos A-\sin A)(\cos A+ \sin A) \cdot (1+\sin A\cos A)(1-\sin A\cos A)[/tex]
to get [tex] \cos 2A (1-\sin^2A\cos^2A)[/tex]
multiply and divide by 4 to get the RHS.
because, [tex] \sin(2A)= 2\sin A \cos A[/tex]
squaring both sides, [tex] \sin^2 (2A)=4\sin^2A\cos^2A[/tex]
