Why do we have $ \sin(x) = x \prod_{n=1}^{\infty} ( 1 - \frac{4}{3} \sin^2 (\frac{x}{3^n})) $

Question

Show that

$$ \sin(x) = x \prod_{n=1}^{\infty} ( 1 - \frac{4}{3} \sin^2 (\frac{x}{3^n})) $$

I know $ 1 - \sin^2x = (1 - \sin x) (1 + \sin x) $ but I'm not sure how to apply it.

Or maybe we need $ 2 \sin^2 (x) = 1 - \cos(2x)$?

Answer 1

Remember that $$ \sin 3x = 3\sin x-4\sin^3 x \implies \frac{\sin 3x}{3\sin x} = 1- \frac 43 \sin^2 x$$

Now you can see the telescoping product.