Find the sum $\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 \left(\frac{k \pi}{2 \cdot 2005}\right)}.$
Solution 1
Since $\sin^2\theta+\cos^2\theta=1$ and $\cos(\frac{\pi}{2}-\theta)=\sin\theta$, $$\sum_{k=1}^{2004}\cos^2\frac{k\pi}{2\cdot 2005}=\sum_{k=1}^{1002}\left(\cos^2\frac{k\pi}{2\cdot 2005}+\cos^2\frac{(2005-k)\pi}{2\cdot 2005}\right)=\color{red}{1002}.$$
Solution 2
$$\begin{align} \sum_{n=1}^{2004}\frac{1}{1+\tan^2\left(\frac{n\pi}{2\cdot 2005}\right)}&=\sum_{n=1}^{2004}\cos^2\left(\frac{n\pi}{2\cdot 2005}\right)\\\\ &=\frac12\sum_{n=1}^{2004}(1+\cos(n\pi/2005))\\\\ &=1002+\frac12 \text{Re}\left(\sum_{n=1}^{2004}e^{in\pi/2005}\right)\\\\ &\bbox[5px,border:2px solid #C0A000]{=1002} \end{align}$$
since
$$\text{Re}\left(\sum_{n=1}^{2004}e^{in\pi/2005}\right)=\text{Re}\left(i\cot\left(\frac{\pi}{2\cdot 2005}\right)\right)=0$$
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Updated on July 22, 2022Comments
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Find the sum $$\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 \left(\frac{k \pi}{2 \cdot 2005}\right)}.$$