$\begingroup$How do you prove that $8 \cos{(x)}\cos{(2x)}\cos{(3x)} - 1 = \dfrac{\sin{(7x)}}{\sin{(x)}}$?
$\endgroup$12 Answers
$\begingroup$We have
$$8 \sin x\cos{(x)}\cos{(2x)}\cos{(3x)} =4\sin(2x)\cos(2x)\cos(3x)=2\sin(4x)\cos(3x)$$ Moreover
$$\sin x+\sin(7x)=2\sin\left(\frac{x+7x}{2}\right)\cos\left(\frac{7x-x}{2}\right)=\cdots$$ and the result follows easily.
$\endgroup$1$\begingroup$Multiply both sides by $\sin x$, then
$$2\cos x \sin x = \sin 2x$$
$$2\cos 2x \sin 2x = \sin 4x$$
$$2\sin 4x \cos 3x = \sin 7x + \sin x$$
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