A generalisation of the Pythagorean trigonometric identity Is there a theorem which states that
$\forall n \in \mathbb Z^+\ \ \exists c \in \mathbb R\ \ \forall x \in \mathbb R$
$$\sum_{k=1}^{2n} \sin^{2n}\left(x + \frac{k \pi}{2n}\right) = c$$
(essentially, a generalisation of $\sin^2x + \cos^2x = 1$ to all positive even powers)?
I have confirmed this holds for $n \in \{1, 2, 3, 4, 5\}$, but that's obviously not good enough.
Also, if such a result has been proven, is the relation between $n$ and $c$ known?
 A: A standard trick when dealing with sums of trigonometric functions is to use complex numbers. Specifically, we will use the identitiy $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$. Using this identity, the binomial theorem and interchanging the order of summation we get:
$$\begin{align*}
&\sum \limits_{k = 1}^{2n} \sin^{2n}(x + k\pi/(2n)) = \sum \limits_{k = 1}^{2n} (2i)^{-2n} \left\{\exp(i(x + k\pi/(2n))) - \exp(-i(x + k\pi/(2n)\right\}^2 \\
={}& (-4)^{-n} \sum \limits_{k = 1}^{2n} \sum \limits_{j = 1}^{2n} \binom{2n}{j} (-1)^{2n - j} \exp(ij(x + k\pi/(2n))) \exp(-i(2n - j)(x + k\pi/(2n))) \\
={}& (-4)^{-n} \sum \limits_{j = 1}^{2n} \binom{2n}{j} (-1)^{2n - j} \exp(2i(j - n)x) \sum \limits_{k = 1}^{2n}\exp(i(j - n)k\pi/n) \\
\end{align*}$$
Now how is this term simpler than what we began with? Well, the inner sum is simply a geometric sum and we can use the formula
$$\sum \limits_{k=1}^{2n} q^k = q \sum \limits_{k = 0}^{2n - 1} q^k=
\begin{cases}
q\frac{q^{2n} - 1}{q - 1} & q \ne 1 \\
2n & q = 1
\end{cases}.$$
In our formula, we have $q = \exp(i(j/n -1)\pi)$. Since $1 \le j \le 2n$, this will be $1$ iff $j = n$. For $j \ne n$ we have however $q^{2n} - 1 = \exp(2i\pi (j - n)) - 1 = 0$, which means
$$\sum \limits_{k = 1}^{2n}\exp(i(j - n)k\pi/n) = 
\begin{cases}
0 & n \ne j \\
2n & n = j
\end{cases}$$
But this will simplify our sum enormously, since all the summands with $j \ne n$ vanish:
$$\sum \limits_{k = 1}^{2n} \sin^{2n}(x + k\pi/(2n)) = (-4)^{-n} \binom{2n}{n} (-1)^{2n - n} \exp(2i(n - n)x)2n = 2n \binom{2n}{n} 4^{-n}.$$
Bonus: Stirling's approximation yields $2n \binom{2n}{n} 4^{-n} \approx 2\sqrt{\frac{n}{\pi}}$.
