Generalizing a Trigonometric Infinite Product of Vieta

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 $n$.

• I see (1) as an average of three evenly spaced cosines, of -2x/3^k, 0, and +2x/3^k. And I see (3) as an average of four evenly spaced cosines, of -3x/4^k, -x/4^k, +x/4^k, and +3x/4^k. Does that help? Jul 28 '18 at 5:07
• @mjqxxxx I think you're onto something. I remember many years ago, when I first saw Vieta's formula for $2/\pi,$ I proved it by converting to a Riemann sum. Thanks. Jul 28 '18 at 5:13
• I think it is a wonderful monograph. For me, it connected many threads together in a very succinct way. Jul 28 '18 at 5:36

You want to find an expression for $$\ a_n(x) := \sin(nx)/\sin(x) \$$ written in terms of cosines of multiple angles. The first examples are: $$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).$$ The pattern is now obvious. The general infinite product is: $$\frac{\sin x}x = \prod_{k=1}^ \infty \frac1n a_n\Big(\frac{x}{n^k}\Big).$$