Calculate the exact value of $$\tan^2(5^\circ)+\tan^2(10^\circ)+\tan^2(15^\circ)+\cdots+\tan^2(85^\circ)$$
How to evaluate this sum of all these values? Is there a specific way? Thanks in advance.
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1Is this in degrees? – mrnovice Mar 30 '17 at 08:54
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Yeap, It's in degrees. – Mathxx Mar 30 '17 at 08:55
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1As $180/(10-5)=36$ use $\tan(36x)$ like http://math.stackexchange.com/questions/951522/trig-sum-tan-21-circ-tan-22-circ-tan2-89-circ – lab bhattacharjee Mar 30 '17 at 08:59
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Have a look at Cauchy's proof of Basel problem on Wikipedia. It exploits exactly such kind of trigonometric sums. – Jack D'Aurizio Mar 30 '17 at 10:25
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$$\tan^2(5^\circ)+\tan^2(10^\circ)+...+\tan^2(85^\circ) = \tan^2\bigg(\frac{5\pi}{180}\bigg) +\tan^2\bigg(\frac{10\pi}{180}\bigg)+...+\tan^2\bigg(\frac{85\pi}{180}\bigg)$$
$$= \sum_{r=1}^{17}\tan^2\bigg(\frac{r\pi}{2\cdot18}\bigg)$$
Now note this result: Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$
We have $n = 18$ so then we get:
$$S = \frac{(18-1)(2\cdot 18 -1)}{3} =\frac{595}{3}$$
