I've recently taken interest in infinite products, and I'm having trouble with a proof I found in this PDF file: "Infinite Products and Elementary Functions":
An intermediate step in finding an infinite product to represent $\sin(x)$ is given as follows:
$$\sin x=x\lim_{n\to \infty}\sum_{k=0}^{(n-1)/2}(-1)^k\binom{n}{2k+1}\frac{x^{2k}}{n^{2k+1}} \tag{2.8} $$ Of all the stages
in Euler’s procedure, which, as a whole, represents a real work of
art, the next stage is perhaps the most critical and decisive.
Factoring the polynomial in (2.8) into the trigonometric form: $$\sin x=x\lim_{n \to \infty}\prod_{k=1}^{(n-1)/2}\left(1-\dfrac{(1+\cos(2k\pi/n))x^2}{(1-\cos(2k\pi/n))n^2}\right)$$
I've tried a proof by induction, but could only manage to check a couple of iterations, that hinted to a pattern that looks like this: $$\prod_{k=1}^{\frac{n-1}{2}} (1-a_{k,n}\frac{x^2}{n^2})$$
but this is a dead end since I couldn't find a trigonometric expression of $a_{k,n}$. I can't figure out where the trigonometric functions came from in the right hand side of the second equation.
My last attempt was to turn the product into a sum using the natural logarithm, but again this complicated things further.
The author seems to hint at an elementary factorization, but I'm starting to have some doubts. Is there really an elementary approach? Is there some identity I'm missing? Any answer, or ("Socratic") hint is appreciated.
Best Answer
If we don't insist on writing it with the $\dfrac{1+\cos\theta}{1-\cos\theta}$ form, a factorisation is easy to be had. We have
$$\sum_{k = 0}^{\left\lfloor \frac{n-1}{2}\right\rfloor} (-1)^k \binom{n}{2k+1} \frac{x^{2k}}{n^{2k+1}} = \frac{\bigl(1 + i\frac{x}{n}\bigr)^n- \bigl(1 - i\frac{x}{n}\bigr)^n}{2ix}.$$
Writing
$$1 \pm i \frac{x}{n} = \sqrt{1 + \frac{x^2}{n^2}}\cdot \exp \Biggl(\pm i\arctan \frac{x}{n}\biggr)$$
we see that
$$\frac{\bigl(1 + i\frac{x}{n}\bigr)^n- \bigl(1 - i\frac{x}{n}\bigr)^n}{2ix} = \biggl(1+\frac{x^2}{n^2}\biggr)^{n/2}\cdot \frac{\sin\bigl(n\arctan \frac{x}{n}\bigr)}{x},$$
so the zeros of the polynomial are
$$\pm n\tan \frac{k\pi}{n},\quad 1 \leqslant k \leqslant \biggl\lfloor\frac{n-1}{2}\biggr\rfloor.$$
Grouping the zeros in pairs of negatives, we obtain the factorisation
$$\sum_{k = 0}^{\left\lfloor \frac{n-1}{2}\right\rfloor} (-1)^k \binom{n}{2k+1} \frac{x^{2k}}{n^{2k+1}} = \prod_{k = 1}^{\left\lfloor \frac{n-1}{2}\right\rfloor}\biggl( 1 - \frac{x^2}{n^2\tan^2 \frac{k\pi}{n}}\biggr),$$
since the constant term of the polynomial is $1$.
I unfortunately don't see a natural way that leads to the form
$$\prod_{k = 1}^{\left\lfloor \frac{n-1}{2}\right\rfloor} \biggl( 1 - \frac{(1+\cos (2k\pi/n))x^2}{(1 - \cos (2k\pi/n))n^2}\biggr).$$