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I will very much appreciate if someone can give me a formula for the sum of lengths of all the sides and all the diagonals of a regular n-gon inscribed in a unit circle. Thank you.

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See this post from a day ago; it discusses a formula for the length of a side of any given 2n-gon inscribed in a circle of radisus R. So you'd use R = 1 for a unit circle. (Note: this only helps with regular polygons with an even number of sides, and is recursive, so it might help to put Andre's info (comment above) and Ross's answer to work!)… – amWhy Oct 29 '12 at 20:53

Hint: The sides of the regular $n$-gon are chords of the circle, with central angle $\frac {2\pi}n$. The diagonals are also chords, with central angles some multiple of that. Think about how many diagonals come from one vertex.

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Without loss of generality we may assume that the circle has radius $1$ (if it has radius $r$ we can multiply by $r$). We can also assume that one of the vertices of the polygon is at $(1,0)$. Let $\theta=\frac{2\pi}{n}$. Then the next vertex (counterclockwise) makes an angle of $\theta$ with the $x$-axis, and the next one an angle of $2\theta$, then $3\theta$, all the way to $(n-1)\theta$.

By basic trigonometry, the distance from $(1,0)$ to the first vertex is $\sin(\theta/2)$. The distance from $(1,0)$ to the second vertex is $\sin(2\theta/2)$. Continue. The distance from $(1,0)$ to the last vertex is $\sin((n-1)\theta/2)$. Add up. We get $$\sin \alpha+\sin 2\alpha+\sin 3\alpha+\cdots +\sin (n-1)\alpha,$$ where $\alpha=\dfrac{\pi}{n}$.

There is a pleasantly simple formula for this kind of sum: see here. This formula is most easily proved by using complex numbers, for then we are just dealing with the sum of two geometric progressions. In that formula, let $\varphi=0$, and use $n-1$ instead of $n$. Note that the already quite simple formula simplifies further.

Now repeat this calculation at all $n$ vertices. We get the same number each time. But if we multiply the sum at each vertex by $n$, we will have counted the length of each edge and each diagonal twice. So we take the sum at the first vertex, multiply by $n$ and then divide by $2$.

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Very nice Andre. Thank you very much! The sum can be simplified to $n\cot(\pi/2n)$. – Martin Oct 29 '12 at 21:53
As I had mentioned, for this special case where we are summing "all the way," the sum of an arithmetic progression of sines simplifies. – André Nicolas Oct 29 '12 at 21:57

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