# How to prove identity $\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$?

Looking at Jolley, Summation of Series, formula 445: $$\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$$

How can one prove this?

Considering $$\left( \sum^{n-1}_{j=0}Z_j\right)^2 = \sum^{n-1}_{j=0} Z_j^2 +\sum^{n-1}_{j=0}\sum^{n-1}_{i\neq j} Z_jZ_i$$ from here

I thought one could take the log differential of the well known sine product:

\begin{align} \prod_{k=0}^{n-1}\sin\left(\theta + \frac{k\pi}{n}\right) &= 2^{1-n} \sin(n\theta) \end{align}

to get

$$n \cot(n\theta) = \sum_{k=0}^{n-1}\cot\left(\theta+\frac{k\pi}{n}\right)$$

or

$$-n \cot(\frac{n\pi}{2}+n\theta) = \sum_{k=0}^{n-1}\tan\left(\theta+\frac{k\pi}{n}\right)$$

then square to get

$$n^2 \cot(\frac{n\pi}{2}+n\theta)^2 = (\sum_{k=0}^{n-1}\tan\left(\theta+\frac{k\pi}{n})\right)^2$$

but I cannot see how to show the

$$\sum _{j=0}^{n-1} \sum _{k=0}^{n-1} \tan \left(\theta +\frac{\pi j}{n}\right) \tan \left(\theta +\frac{\pi k}{n}\right) [j\neq k]$$

necessary for the $$n(n-1)$$ part of the identity.

Perhaps somebody could show a more successful method.

• Commented Dec 10, 2022 at 14:12

## 2 Answers

So, here is my idea, it occurred to me when I saw this well known sine product, that it will appear if we integrate the function

$$f\left(\theta\right) = \sum_{k=0}^{n-1} \tan^{2}\left(\theta+\frac{k\pi}{n}\right).$$

Since we are going to differentiate in the end, I won't be bothering to write integration constant:

$$\int f\left(\theta\right)d\theta = \int \sum_{k=0}^{n-1} \left(\sec^{2}\left(\theta+\frac{k\pi}{n}\right) - 1\right) d\theta = \sum_{k=0}^{n-1} \tan\left(\theta+\frac{k\pi}{n}\right) - n\theta$$

$$\int f\left(\theta\right)d\theta = - n\theta - \sum_{k=0}^{n-1} \cot\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right) = - n\theta - \sum_{k=0}^{n-1} \frac{\frac{d}{d\theta}\sin\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right)}{\sin\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right)}$$

$$\int f\left(\theta\right)d\theta = - n\theta - \frac{\frac{d}{d\theta}\Pi_{k=0}^{n-1}\sin\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right)}{\Pi_{k=0}^{n-1}\sin\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right)} = -n\theta -\frac{\frac{d}{d\theta}2^{1-n}\sin\left(n\left(\theta-\frac{\pi}{2}\right)\right)}{2^{1-n}\sin\left(n\left(\theta-\frac{\pi}{2}\right)\right)}$$

$$\int f\left(\theta\right)d\theta = - n\theta - n\frac{\cos\left(n\theta-\frac{n\pi}{2}\right)}{\sin\left(n\theta-\frac{n\pi}{2}\right)} = -n\theta -n\cot\left(n\theta-\frac{n\pi}{2}\right).$$

If you sum $$n\pi$$ in the argument of $$\cot$$ it won't change its value, so:

$$\int f\left(\theta\right)d\theta = -n\theta -n\cot\left(n\theta+\frac{n\pi}{2}\right),$$

now differentiating we get:

$$f\left(\theta\right) = -n +n^{2}\csc^{2}\left(n\theta+\frac{n\pi}{2}\right) = -n +n^{2}\left(1+\cot^{2}\left(n\theta+\frac{n\pi}{2}\right)\right)$$

which give us the equality desired.

• Thank you for the answer, I'm liking the idea. It might even extend to higher powers of tan too. I'll give a chance for others that know more than me to comment. Commented Dec 10, 2022 at 8:52

If $$\frac{2n\theta}{\pi}$$ is an odd integer, one $$\tan^2$$ value in the sum is undefined (or "infinity"), so assume that's not the case.

$$z = \tan \left(\theta + \frac{k\pi}{n}\right); 0 \leq k \leq n-1$$

satisfy $$\arg(z+i) = \frac{\pi}{2} + \theta + \frac{k\pi}{n}$$. Defining $$\alpha = \theta + \frac{\pi}{2}$$ just for brevity, this is $$\arg(z+i) = \alpha + \frac{k\pi}{n}$$. These $$n$$ values of $$z$$ also satisfy

$$\operatorname{Im}\left(\left((z+i)e^{-i\alpha}\right)^n\right) = 0$$ $$\left(((z+i)e^{-i\alpha}\right)^n - \left((z-i)e^{i\alpha}\right)^n = 0$$

Since this is a polynomial in $$z$$ of degree $$n$$ (call it $$p$$), there are no other solutions, and

\begin{align*} p(z) &= (z+i)^n e^{-in\alpha} - (z-i)^n e^{in\alpha} \tag{1}\label{1} \\ p(z) &= (e^{-in\alpha} - e^{in\alpha})\prod_{k=0}^{n-1} \left(z-\tan\left(\theta+\frac{k\pi}{n}\right)\right) \\ p(z) &= -2i \sin(n\alpha) \prod_{k=0}^{n-1} \left(z-\tan\left(\theta+\frac{k\pi}{n}\right)\right) \tag{2}\label{2} \end{align*}

Now augment the polynomial's set of roots by the negatives of its $$n$$ roots, to get polynomial $$q(z) = (-1)^n p(z)p(-z)$$. From $$\ref{1}$$:

\begin{align*} (-1)^n q(z) &= \left((z+i)^n e^{-in\alpha} - (z-i)^n e^{in\alpha}\right)\left((-z+i)^n e^{-in\alpha} - (-z-i)^n e^{in\alpha}\right) \\ q(z) &= (z^2+1)^n e^{-2in\alpha} - (z-i)^{2n} - (z+i)^{2n} + (z^2+1)^n e^{2in\alpha} \\ q(z) &= 2(z^2+1)^n \cos(2n\alpha) - (z+i)^{2n} - (z-i)^{2n} \tag{3}\label{3} \end{align*}

From $$\ref{2}$$: \begin{align*} (-1)^n q(z) &= -4 \sin^2(n\alpha) \prod_{k=0}^{n-1} \left(z-\tan\left(\theta+\frac{k\pi}{n}\right)\right) \left(- z - \tan\left(\theta + \frac{k\pi}{n}\right)\right) \\ q(z) &= -4 \sin^2(n\alpha) \prod_{k=0}^{n-1} \left(z^2 - \tan^2\left(\theta + \frac{k\pi}{n}\right) \right) \tag{4}\label{4} \end{align*}

Now to get the sum of the $$\tan^2$$ values, equate the coefficients of $$z^{2n-2}$$ in $$\ref{4}$$ and $$\ref{3}$$:

\begin{align*} a_{2n-2} &= 4 \sin^2(n\alpha) \sum_{k=0}^{n-1} \tan^2\left(\theta + \frac{k\pi}{n}\right) \\ a_{2n-2} &= 2{n \choose 1} \cos(2n\alpha) - {2n \choose 2}i^2 - {2n \choose 2}(-i)^2 \\ &= 2n(1-2\sin^2(n\alpha)) + 2n(2n-1) \\ &= -4n \sin^2(n\alpha) + 4n^2 \end{align*}

\begin{align*} \sum_{k=0}^{n-1} \tan^2\left(\theta + \frac{k\pi}{n}\right) &= n^2 \csc^2(n\alpha) - n \\ &= n^2 (1+\cot^2(n\alpha)) - n \\ &= n^2 \cot^2\left(\frac{n\pi}{2} + n\theta\right) + n(n-1) \end{align*}

• thank you for your solution. Do you think this could be adapted to work with different theta's say $\theta_0 \,\theta_1$ maybe \begin{align*} -4 \sin^2(n\alpha) \prod_{k=0}^{n-1} \left(z-\tan\left(\theta_0 + \frac{k\pi}{n}\right)\right) \left(- z - \tan\left(\theta_1 + \frac{k\pi}{n}\right)\right) \end{align*} can be work through to provide a solution for \begin{align*} \sum_{k=0}^{n-1} \tan\left(\theta_0 + \frac{k\pi}{n}\right) \tan\left(\theta_1 + \frac{k\pi}{n}\right) \end{align*} Commented Dec 15, 2022 at 15:04
• I don't immediately see a way to work with that sum. The polynomial you suggest doesn't work out since it can't use the $(z-t)(z+t)=z^2-t^2$ pieces, so there will be other unwanted terms in the coefficient of $z^{2n-2}$. Commented Dec 15, 2022 at 15:18