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I often fail to find trigonometric sums such as the one in the question shown in the following.

When I tried the question, I first led $z=e^{i\pi/180}$.

After simple calculations, I obtained

$\sum_{n=0}^{14}\tan(12n+1)=\sum_{n=0}^{14}\dfrac{z^{24n+2}-1}{z^{24n}+z^2}$

How can I proceed calculation further?
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1 Answer 1

up vote 3 down vote accepted

This is a special case of the more general identity: $$\sum_{k=0}^{n-1} \tan\left(\theta + \frac{k\pi}{n}\right) = -n\cot \left(\frac{n\pi}{2} + n\theta \right)$$ In your case of course $\theta = 1^\circ$ and $n=15$. A proof of this identity may be found as an answer to a previous question on this site, here.

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How beautiful the formula is! Though I knew that a certain sum of sin's whose angle increases linearly like that is proportional or inversely proportional to cot, I didn't know this formula. If I read your page, I may find the closed form of sum of not only tan's but also other trigonometric functions. Thanks so much. –  Aran Komatsuzaki Nov 1 '13 at 1:50

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