The second exercise in "Statistical Independence in Probability, Analysis and Number Theory," by Mark Kac is to prove that $$
{\sin x\over x}=\prod_{k=1}^{\infty}\frac13\left(1+2\cos{2x\over3^k}\right)\tag{1}
$$
and generalize it. This is a generalization of Vieta's formula $$
{\sin x\over x}=\prod_{k=1}^{\infty}\cos{x\over 2^k}\tag{2}
$$
which is proved in the text.
It's not hard to prove $(1)$. You just write $\sin x = \sin({x\over3}+{2x\over3})$ and plug away, but I'm having trouble seeing what the generalization is supposed to be. The only thing I've been able to come up with for the next case is $$
{\sin x\over x}=\prod_{k=1}^{\infty}
\frac12\left(\cos{x\over4^k}+\cos{3x\over4^k}\right)\tag{3}
$$
Again, this isn't hard to prove, but I'm having trouble seeing a pattern in $(1),\ (2),\ \text{and } (3).$ To derive $(3)$ I used the triple angle formula for cosine, and it seems to me that as you go forward you'll need multiple angle formulas for increasing large multiples, so I foresee a lot of complication.
Can you see a formula for the $n=4$ case that is more clearly a generalization of $(1)$ than $(3)$ is? Or do you know what generalization Kac had in mind? I'm assuming that he intends for you to come up with a formula for general $n$.
Best Answer
Write $\,a_n(x) := \sin(nx)/\sin(x)\,$ in terms of $\,\cos(kx).\,$ The first few examples are: $$ a_2(x) = 2\cos(x), $$ $$ a_3(x) = 1 + 2\cos(2x), $$ $$ a_4(x) = 2\cos(x) + 2\cos(3x), $$ $$ a_5(x) = 1 + 2\cos(2x) + 2\cos(4x), $$ $$ a_6(x) = 2\cos(x) + 2\cos(3x) + 2\cos(5x). $$ The pattern is now obvious. Thus, the general infinite product with $\,n>1\,$ is: $$ \frac{\sin x}x=\prod_{k=1}^\infty \frac1n a_n\Big(\frac{x}{n^k}\Big). $$