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It follows from the law of cosines that if $a,b,c$ are the lengths of the sides of a triangle with respective opposite angles $\alpha,\beta,\gamma$, then $$ a^2+b^2+c^2 = 2ab\cos\gamma + 2ac\cos\beta + 2bc\cos\alpha. $$

For a cyclic (i.e. inscribed in a circle) polygon, consider the angle "opposite" a side to be the angle between adjacent diagonals whose endpoints are those of that side (it doesn't matter which vertex of the polygon serves as the vertex of that angle because of the Inscribed Angle Theorem). Then for a cyclic quadrilateral with sides $a,b,c,d$ and opposite angles $\alpha,\beta,\gamma,\delta$, one can show that $$ \begin{align} a^2+b^2+c^2+d^2 = {} & 2ab\cos\gamma\cos\delta + 2ac\cos\beta\cos\delta + 2ad\cos\beta\cos\gamma \\ & {} +2bc\cos\alpha\cos\delta+2bd\cos\alpha\cos\gamma+2cd\cos\alpha\cos\beta \\ & {}-4\frac{abcd}{(\text{diameter})^2} \end{align} $$ And for a cyclic pentagon, with sides $a,b,c,d,e$ and respective opposite angles $\alpha,\beta,\gamma,\delta,\varepsilon$, $$ \begin{align} a^2 + \cdots + e^2 = {} & 2ab\cos\gamma\cos\delta\cos\varepsilon+\text{9 more terms} \\& {} - 4\frac{abcd}{(\text{diameter})^2}\cos\varepsilon+ \text{4 more terms} \end{align} $$ And for a cyclic $n$-gon with sides $a_i$ and opposite angles $\alpha_i$, $$ \begin{align} \sum_{i=1}^n a_i^2 = {} & \text{a sum of }\binom{n}{2}\text{ terms each with coefficient 2} \\ & {} - \text{a sum of }\binom{n}{4}\text{ terms each with coefficient 4} \\ & {} + \text{a sum of }\binom{n}{6}\text{ terms each with coefficient 6} \\ & {} - \cdots \end{align} $$ The number of terms depends on $n$ and the power of the diameter on the bottom is in each case what is needed to make the term homogeneous of degree $2$ in the side lengths ("dimensional correctness" if you like physicists' language), and the alternation of signs continues.

I showed this by induction. It should work for infinitely many sides, too, by taking limits. Each term would then have a product of infinitely many cosines.

My question is: Is there some reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle?

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    This was [crossposted at MO](http://mathoverflow.net/questions/71691/geometric-meaning-of-a-trigonometric-identity). Michael, please wait some time before posting your question in multiple fora, and when you do, provide links to the other posts (there is already an answer on the MO one by Noam Elkies).2011-07-31
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    I mildly object to the notion that the first identity follows from the law of cosines: in fact, the usual trigonometric proof proceeds by multiplying $c = a\cos{\beta} + b\cos{\alpha}$ by $c$. Doing this for each side and writing out $a^2 + b^2 - c^2$ gives you the law of cosines, writing out $a^2 + b^2 + c^2$ gives you the identity you ask about.2011-07-31
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    No geometric interpretation, but an observation: Assuming the polygon convex (a requirement for each $\alpha_i$ to be uniquely determined), we have $a_i = 2 r \sin{\alpha_i}$, where $r$ is the radius of the circle. Moreover, the arcs determined by the polygon's sides (obviously) comprise the full circle; hence the inscribed angles sum to $\pi$. So, dividing your formula by $4 r^2$ leaves a trig identity for the sum of squares of sines of $n$ (non-negative) angles that total $\pi$.2011-07-31
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    @Theo: Here's a picture. :) http://math.stackexchange.com/questions/803/what-is-the-most-elegant-proof-of-the-pythagorean-theorem/1336#13362011-07-31

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