# 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?

• I can't do the calculations right now, but if the identity holds true, plugging in $\sin(x) = (e^{ix}-e^{-ix})/2$ should do the trick. Commented Nov 14, 2016 at 14:03
• If i'm not mistaken, $c=2n-1$ holds. I will post the calculations later when I have time. You just need to consider my first comment, change the order of summation and apply the geometric sum formula. Commented Nov 14, 2016 at 14:14
• That doesn't seem right, as (according to WolframAlpha), for $n = 1, 2, 3, 4, 5$, $c$ is $1, \frac{3}{2}, \frac{15}{8}, \frac{35}{16}, \frac{315}{128}$, respectively. Commented Nov 14, 2016 at 14:17
• I found my mistake, $c=2n\binom{2n}{n}4^{-n}$. Commented Nov 14, 2016 at 14:45

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 $2n \binom{2n}{n} 4^{-n} \approx 2\sqrt{\frac{n}{\pi}}$.

• It was indeed a bit over my head - I'm a computer scientist and this is relevant to me because it facilitates designing overlapping constant-sum window functions. But I'm just curious - this isn't a very important result, is it? I wasn't able to find anything relevant online. Commented Nov 14, 2016 at 17:38
• At least I haven't seen this result before. Its importance is maybe that of a lemma in a book about fourier series. There you can find many similar rquations. Commented Nov 14, 2016 at 17:58