Q.Prove the following: $\sin6x+\sin4x-\sin2x=4\cos x\sin2x\cos3x$

Factorization (also called transformation) formulae convert a SUM or DIFFERENCE of two sines or cosines into a PRODUCT, and, conversely, convert a PRODUCT of two sines/cosines into a sum or difference. Substituting $A=\dfrac{C+D}{2}$, $B=\dfrac{C-D}{2}$ into the four compound-angle identities gives $\sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$, $\sin C-\sin D=2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$, $\cos C+\cos D=2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$ and $\cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$. Reading the same four identities the other way around gives the product-to-sum forms $2\sin A\cos B=\sin(A+B)+\sin(A-B)$, $2\cos A\sin B=\sin(A+B)-\sin(A-B)$, $2\cos A\cos B=\cos(A+B)+\cos(A-B)$ and $2\sin A\sin B=\cos(A-B)-\cos( …