Prove that: $\cos 4A-\cos 4B=8(\cos A-\cos B)(\cos A + \cos B)(\cos A - \sin B)(\cos A + \sin B)$

Question

Prove that: $\cos 4A-\cos 4B=8(\cos A-\cos B)(\cos A + \cos B)(\cos A - \sin B)(\cos A + \sin B)$

My Attempt:

$$L.H.S=\cos 4A - \cos 4B$$ $$=2\sin \dfrac {4A+4B}{2}.\sin \dfrac {4B-4A}{2}$$ $$=2\sin (2A+2B).\sin (2B-2A)$$ How do I proceed further?

Answer 1

HINT:

From the RHS,

use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

and by Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B=1-\cos^2A-(1-\cos^2B)$$

Can you take it from here?

Answer 2

Start on the right: \begin{align} &8(\cos{A}-\cos{B})(\cos{A}+\cos{B})(\cos{A}-\sin{B})(\cos{A}+\sin{B}) \\ &= 2(2\cos^2{A}-2\cos^2{B})(2\cos^2{A}-2\sin^2{B}) \\ &= 2(\cos{2A}-\cos{2B})(\cos{2A}+\cos{2B}) \\ &= 2\cos^2{2A}-2\cos^2{2B} \\ &= \cos{4A}-\cos{4B}, \end{align} using $\cos{2\theta}=2\cos^2{\theta}-1=1-2\sin^2{\theta}$ liberally.