Let $\displaystyle m$ be a non-negative integer. Prove that
$\displaystyle \prod_{k=1}^{m} \tan\left(\frac{\pi k}{2m+1}\right) = \sqrt{2m+1}$
This is a proof I wrote which uses the same idea as an elementary proof of the Basel problem which I've seen, but it is somewhat different.
Lemma 1: For any odd positive a, $\displaystyle \sin(ax)$ can be expressed as a polynomial in $\displaystyle \sin(x)$ of degree a with exclusively odd powers. Example: $\displaystyle \sin(3x) = -4\sin^3(x) + 3\sin(x)$
This can be proven by induction using the identity $\displaystyle \sin(nx) = 2\cos(x)\sin((n-1)\cdot x) - \sin((n-2)\cdot x)$. I won't prove it here, it is a known fact and easy to look up.
Let $\displaystyle n = 2m+1$, where m is the variable in the original problem, and $\displaystyle f(x)$ be defined as follows:
$\displaystyle f(x) = \frac{1}{x} \cdot (1+x^2)^{\frac{n}{2}}\cdot\sin(n\cdot\arctan(x))$
We make the claim that f(x) is a polynomial of n-1 degree, when x≠0, and prove it as follows.
According to lemma 1, $\displaystyle \sin(n\cdot\arctan(x))$ can be expressed as an n-degree polynomial in $\displaystyle \sin(\arctan(x))$, which includes only odd powers. After simplifying,$\displaystyle \sin(\arctan(x)) = \frac{x}{\sqrt{1+x^2}}$ at all real numbers. Therefore $\displaystyle \sin(n\cdot\arctan(x))$ can be written as a sum of terms of the form $\displaystyle a_r \left (\frac{x}{\sqrt{1+x^2}} \right )^r$, where r is a positive odd integer. So f(x) can be written as a sum of terms of the form $\displaystyle a_r \cdot x^{r-1}\cdot(1+x^2)^{\frac{n-r}{2}}$, each of which is a polynomial of degree n-1, because (n-r) is even and r ≠ 0. Thus f(x) is of degree n-1.
The roots of f(x) occur precisely where $\displaystyle n \cdot \arctan(x)$ is an integer multiple of $\displaystyle \pi$, with the exception of zero, because it cancels with the 1/x term, and f(0) is undefined. Because tan(x) is one-to-one on the interval $\displaystyle (-\frac{\pi}{2},\frac{\pi}{2})$ and ranges from -infinity to infinity, each root can be expressed uniquely as the tangent of a number in this interval. So the zeroes occur when $\displaystyle x = \tan\left (\pm\frac{k\pi}{2m+1}}\right ), 1 \leq k \leq m$ (Recall that (2m+1) = n). So there are 2m = n - 1 roots, as expected by the degree of the polynomial.
The product of all roots of an even-degree polynomial p(x) is obtained by dividing p(0) by the leading coefficient (because a polynomial can be expressed as $\displaystyle c_0(x-r_1)(x-r_2)....(x-r_p).$)
f(0) is undefined, but because at all other points it coincides with a polynomial which is continuous everywhere, the limit at 0 must exist. So we proceed with:
$\displaystyle \lim_{x \to 0} \left(\frac{1}{x} \cdot (1+x^2)^{\frac{n}{2}}\cdot\sin(n\cdot\arctan(x)) \right ) = n$
By simplification and L'Hôpital's. The leading coefficient of f(x) can be calculated by
$\displaystyle \lim_{x \to \infty} \frac {f(x)}{x^{n-1}}$
$\displaystyle = \lim_{x \to \infty} \frac{\frac{1}{x} \cdot (1+x^2)^{\frac{n}{2}}\cdot\sin(n\cdot\arctan(x))}{ x^{n-1}}$
$\displaystyle = \lim_{x \to \infty} \frac{(1+x^2)^{\frac{n}{2}}}{x^n} \cdot\sin(n\frac{\pi}{2})$
$\displaystyle = 1 \cdot \pm1 = \pm1$
Therefore the product of the roots is $\displaystyle \frac{f(0)}{\pm 1} = \pm n = \pm(2m+1)$. But the absolute value of this is simply the square of the desired product in the original problem, because each positive root has a negative counterpart. Thus the desired product is $\displaystyle \sqrt{2m +1}$, and this completes the proof.
Well, I'm sure my proof is not the standard one. You find similar identities for sine and coside. For sine, you transform the product of sines into a product of gamma functions (using the reflection formula) and then Gauss' multiplication formula. You'll obtain
$\displaystyle \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$
for positive integers $\displaystyle n$. Then for cosine I don't remember but the identity is
$\displaystyle \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}} $
Next, dividing them out you have
$\displaystyle \prod_{k=1}^{n-1} \tan\left(\frac{k\pi}{n}\right) = \frac{n}{\sin(\pi n/2)}$
and for odd $\displaystyle n = 2m+1$ you get the above fomula. I'm not entirely sure how All in all, I'll need to look through my notebooks to find it. If anyone wants to prove it using my way, feel free to go ahead and type it up.
PS. In fact, I was the one that put up these above formulae on the wiki page List of trigonometric identities - Wikipedia, the free encyclopedia