## Trigonometry (11th Edition) Clone

$$6\sin4x\sin5x=3\cos x-3\cos9x$$
$$A=6\sin4x\sin5x$$ One thing you can do here is to switch the position of $\sin 4x$ and $\sin 5x$ so that later we might not encounter negative values inside trigonometric functions. What I mean is this $$A=6\sin5x\sin4x$$ 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 $$A=6\times\frac{1}{2}[\cos(5x-4x)-\cos(5x+4x)]$$ $$A=3(\cos x-\cos 9x)$$ $$A=3\cos x-3\cos9x$$