Answer
$$\sin x+\sin 3x+\sin5x+\sin7x=4\cos x\cos2x\sin4x$$
The equation is an identity, after proving that 2 sides are equal.
Work Step by Step
$$\sin x+\sin 3x+\sin5x+\sin7x=4\cos x\cos2x\sin4x$$
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.
