Trigonometry (11th Edition) Clone

$$8\sin7x\sin9x=4\cos2x-4\cos16x$$
$$A=8\sin7x\sin9x$$ Again, we switch the position of $\sin 7x$ and $\sin 9x$ to avoid later complexity. $$A=8\sin9x\sin7x$$ The product-to-sum identity that will be applied here is $$\sin X\sin Y=\frac{1}{2}[\cos(X-Y)-\cos(X+Y)]$$ Therefore, A would be $$A=8\times\frac{1}{2}[\cos(9x-7x)-\cos(9x+7x)]$$ $$A=4(\cos2x-\cos16x)$$ $$A=4\cos2x-4\cos16x$$