<s>[INST] Why do we have $ \sin(x) = x \prod_{n=1}^{\infty} ( 1 - \frac{4}{3} \sin^2 (\frac{x}{3^n})) $ 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)$?
 [/INST] 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.
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