Evaluate $\displaystyle \sum_{n=1}^{89}\tan^2\bigg(\frac{n\pi}{180}\bigg)$.

Question

Find $\displaystyle \sum_{n=1}^{89}\tan^2\bigg(\frac{n\pi}{180}\bigg)$.

I didn't have any idea to solve this problem :/ Can someone give a hint for me?

What is the best strategy to solve problems of the type $\displaystyle \sum_{k=1}^{n}\tan^2(k\theta)?$

Answer 1

$\mathcal{Hint}$

$$0=\tan(180x)=\binom{180}{1}\tan x-\binom{180}{3}\tan^3x+...+\binom{180}{177}\tan^{177}x-\binom{180}{179}\tan^{179}x$$

Where $x=\frac{n\pi}{180}$ for $n=0,1..89,91,...,179$ i.e $179$ values .

Divde by $\tan x$ and consider $\tan^2x=y$.

Answer 2

Since the user @user355088 demonstrated a beautiful solution, let me provide an ugly solution. Write $\omega = e^{2\pi i/180} $ for simplicity. Then

\begin{align*} \sum_{n=1}^{89} \tan^2 (n\alpha/2) = \frac{1}{2} \sum_{n=-89}^{89} \tan^2 (n\alpha/2) = - \frac{1}{2} \sum_{n=-89}^{89} \frac{(1-\omega^n)^2}{(1+\omega^n)^2} = - \frac{1}{2} \lim_{\epsilon \to 0^+} S(\epsilon), \end{align*}

where $S(\epsilon)$ is defined by

$$ S(\epsilon) := \sum_{n=-89}^{89} \frac{(1-e^{-2\epsilon}\omega^n)^2}{(1+e^{-2\epsilon}\omega^n)^2}. $$

Now we use the identity $\frac{(1-z)^2}{(1+z)^2} = 1+4\sum_{k=1}^{\infty} (-1)^k k z^{k} $ to expand

\begin{align*} S(\epsilon) &= 179 + 4 \sum_{k=1}^{\infty} (-1)^k k e^{-2k\epsilon} \Biggl( \sum_{n=-89}^{89} \omega^{kn} \Biggr) \\ &= 179 + 4 \sum_{k=1}^{\infty} (-1)^k k e^{-2k\epsilon} \Biggl( (-1)^{k-1} + \sum_{n=-89}^{90} \omega^{kn} \Biggr) \end{align*}

By the sifting property of the sum of the roots of unity, we know:

$$ \sum_{n=-89}^{90} \omega^{kn} = \begin{cases} 180, & \text{180 divides $k$}; \\ 0, & \text{otherwise}; \end{cases} $$

So, by using $\frac{1}{\sinh^2 x} = \frac{4e^{-2x}}{(1-e^{-2x})^2} = 4 \sum_{k=1}^{\infty} k e^{-2kx}$ for $x > 0$,

\begin{align*} S(\epsilon) &= 179 + 4 \sum_{k=1}^{\infty} (-1)^k k e^{-2k\epsilon} \left( (-1)^{k-1} + 180 \cdot \mathbf{1}_{\{\text{180 divides $k$}\}} \right) \\ &= 179 + 4 \left( \sum_{k=1}^{\infty} 180^2 k e^{-360k\epsilon} - \sum_{k=1}^{\infty} k e^{-2k\epsilon} \right) \\ &= 179 + \left( \frac{180^2}{\sinh^2(180\epsilon)} - \frac{1}{\sinh^2(\epsilon)} \right). \end{align*}

Since $\frac{1}{\sinh^2 x} = \frac{1}{x^2} - \frac{1}{3} + \mathcal{O}(x^2)$, we obtain

\begin{align*} S(\epsilon) &= 179 - \frac{180^2 - 1}{3} + \mathcal{O}(\epsilon^2), \end{align*}

and taking $\epsilon \to 0^+$ proves that

$$ \sum_{n=1}^{89} \tan^2 (n\alpha/2) = -\frac{1}{2}\left( 179 - \frac{180^2 - 1}{3} \right) = \frac{15931}{3}, $$

which is the same as @user355088's answer $ \binom{180}{177}\big/\binom{180}{179} = \frac{15931}{3} $.