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$$ S=\sum_{k=0}^{10}\sin\left(\frac{(2+4k)\pi}{23}\right) =\sum_{k=0}^{10}e^\left(i\frac{(2+4k)\pi}{23}\right) =e^{i\frac{2\pi}{23}}\sum_{k=0}^{10}e^{i\frac{4k\pi}{23}} =e^{iu}\sum_{k=0}^{10}\left(e^{2iu}\right)^k $$

For example, I did this by hand, and got an answer as $\frac{1}{2} \tan\left( \frac{\pi}{23} \right)$

But how do I use Mathematica to check?

Sum[Sin[Pi/23*(2 + 4*k)], {k, 0, 10}] Sum[Sin[Pi/23*(2 + 4*k)], {k, 0, 10}] - Tan[Pi/23]/2 // Simplify

I have tried some functions like Simplify,TrigFactor,TrigToExp. But I am not sure how to guide Mathematica to the final answer.

And similarly, how do I simplify 

Sum[Sin[(-1)^k*Pi/23*(2 + 4*k)], {k, 0, 10}]

which I got as $-\frac{1}{2} \tan\left( \frac{2\pi}{23} \right)$.

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  • $\begingroup$ (Sum[Sin[Pi/23*(2 + 4*k)], {k, 0, 10}] - Tan[Pi/23]/2 )// FullSimplify produces 0 in version 12.0. $\endgroup$ Commented Feb 2, 2020 at 12:30
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    $\begingroup$ @user64494 I think what he want is , how to get $\frac{1}{2} \tan \left(\frac{\pi }{23}\right)$ by mma. (Sum[Sin[Pi/23*(2 + 4*k)], {k, 0, 10}] - Tan[Pi/23]/2 )// FullSimplify is for check the answer. $\endgroup$ Commented Feb 2, 2020 at 12:36
  • $\begingroup$ I don't understand your first equality as the LHS is real while the rhs is complex. What am I missing? $\endgroup$ Commented Apr 27, 2024 at 1:42

2 Answers 2

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One way is:

Sum[Sin[Pi/23*(2 + 4*k)], {k, 0, n}] /. n -> 10 // Simplify (* 1/2 Tan[\[Pi]/23] *)
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    $\begingroup$ The second sum confuses Mathematica in its original form and has to be tweaked a little: Sum[(-1)^k*Sin[Pi/23*(2 + 4*k)], {k, 0, n}] /. n -> 10 $\endgroup$ Commented Feb 2, 2020 at 12:43
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    $\begingroup$ That's just so weird!! Why wouldn't Sum[Sin[Pi/23*(2 + 4*k)], {k, 0, 10}]//Simplify work straight away..... $\endgroup$ Commented Feb 2, 2020 at 12:46
  • $\begingroup$ @CasperYC. I don't really know, maybe because the algorithm is written in this way,can't find simpler form. Using Maple also can't . $\endgroup$ Commented Feb 2, 2020 at 12:52
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    $\begingroup$ The reason is probably that simplifying such expressions requires guessing the pattern, finding the formula for x_i and then doing the sum the usual way, guessing the pattern being the hardest part of the problem. This would greatly slow down the Simplify function. $\endgroup$ Commented Feb 2, 2020 at 12:55
  • $\begingroup$ @MariuszIwaniuk HAHA. That's fine. I gave up Maple ten years ago ... $\endgroup$ Commented Feb 2, 2020 at 13:10
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Surprisingly Wolfram Alpha does quite well for free:

Sum[Sin[Pi/23*(2 + 4*k)], {k, 0, n}] sum_(k=0)^n sin(1/23 π (4 k + 2)) = csc((2 π)/23) sin^2(2/23 π (n + 1)) Sum[Sin[Pi/23*(2 + 4*k)], {k, 0, n}] /. n -> 10 // Simplify Limit lim_(n->10) sum_(k=0)^n sin(1/23 π (2 + 4 k)) = csc((2 π)/23) sin^2(π/23) Alternate forms 1/2 tan(π/23)
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  • $\begingroup$ That was nearly 4 years ago now. $\endgroup$ Commented Jun 2, 2024 at 9:01

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