Trigonometry (11th Edition) Clone

$$\sin x+\sin 3x+\sin5x+\sin7x=4\cos x\cos2x\sin4x$$ The equation is an identity, after proving that 2 sides are equal.
$$\sin x+\sin 3x+\sin5x+\sin7x=4\cos x\cos2x\sin4x$$ We start from the left side. $$X=\sin x+\sin 3x+\sin5x+\sin7x$$ $$X=(\sin x+\sin3x)+(\sin5x+\sin7x)$$ Apply the sum-to-product identity for sine here, which states $$\sin A+\sin B=2\sin\Big(\frac{A+B}{2}\Big)\cos\Big(\frac{A-B}{2}\Big)$$ we have $1)$ About $\sin x+\sin3x$ $$\sin x+\sin3x=2\sin\Big(\frac{x+3x}{2}\Big)\cos\Big(\frac{x-3x}{2}\Big)$$ $$\sin x+\sin3x=2\sin\frac{4x}{2}\cos\frac{-2x}{2}$$ $$\sin x+\sin3x=2\sin2x\cos(-x)$$ - However, from negative-angle identities: $\cos(-x)=\cos x$. Thus, $$\sin x+\sin3x=2\sin2x\cos x$$ $2)$ Now, about $\sin5x+\sin7x$ $$\sin5x+\sin7x=2\sin\Big(\frac{5x+7x}{2}\Big)\cos\Big(\frac{5x-7x}{2}\Big)$$ $$\sin5x+\sin7x=2\sin\frac{12x}{2}\cos\frac{-2x}{2}$$ $$\sin5x+\sin7x=2\sin6x\cos(-x)$$ - Again, as $\cos(-x)=\cos x$, $$\sin5x+\sin7x=2\sin6x\cos x$$ $3)$ Combine back to $X$ $$X=2\sin2x\cos x+2\sin6x\cos x$$ $$X=2\cos x(\sin 2x+\sin6x)$$ We apply the sum-to-product identity once more $\sin 2x+\sin 6x$: $$X=2\cos x\Big[2\sin\Big(\frac{2x+6x}{2}\Big)\cos\Big(\frac{2x-6x}{2}\Big)\Big]$$ $$X=4\cos x\sin4x\cos(-2x)$$ $$X=4\cos x\cos(-2x)\sin4x$$ - Also, from negative-angle identity for cosine, $\cos(-2x)=\cos 2x$ $$X=4\cos x\cos 2x\sin 4x$$ Therefore, $$\sin x+\sin 3x+\sin5x+\sin7x=4\cos x\cos2x\sin4x$$ The equation is an identity as a result.