geometric meaning of a trigonometric identity

Solution 1:

I haven't worked it out up to arbitrary $n$, but I've managed the cases $n=3$ and $n=4$ to follow your lead and it becomes tempting to write "+4 terms" and so on.

I believe geometric interpretation of that ultimately goes down to interpreting geometrically the cosine law, since it is the main tool in your arguments.

EDIT : I just thought of something : suppose the sides of the inscribed polygon are almost are equal, or even better, that they form a regular polygon. Consider the following image :

You can consider the squares as a prism that rose over the inscribed polygon and then got unfolded. The sum of the squares can then be thought of an approximation of the lateral area of a cylinder with height being the average of the $a_i$'s, i.e. if you let $$ \mu = \frac{\sum_{i=1}^n a_i}n, $$ we can intuitively argue that $$ \sum_{i=1}^n a_i^2 \approx 2 \pi r \mu. $$ Hope that helps,

Solution 2:

Instead of an answer to the question you asked, let me give several observations.

Firstly, by homogeneity, we can consider the case where the diameter of the circle is 2, or the radius being 1. In this case, then, basic trigonometry gives you that $a = 2 \sin \alpha$, $b = 2\sin \beta$, etc.

Thus the right hand side consists of terms of the form $$\sin \times \sin \times \cos \times \cos\times \cos + \sin \times\sin \times\sin\times \sin \times \cos + \dots $$ which should be reminiscent of the angle addition rule for cosine.

Indeed, we have that

$$ \cos \sum \theta_k = \Re \exp \sum i\theta_k = \Re \prod \left( \cos \theta_k + i \sin\theta_k\right) $$

which is similar to the RHS that you've written down, except that on your RHS you do not have the term $\prod \cos\theta_k$. This term can be recovered by what you have on the LHS.

Note that if you have an inscribed polygon, $\sum \theta_k = \pi$. So $\cos \sum\theta_k = -1$. Notice that the LHS of your equation is proportional to $\sum \sin^2\theta_k$, and observe the following:

$$\begin{align} \sum 2\sin^2\theta_k & = \sum (1 - \cos(2\theta_k))\\ & = N - \sum_{k\neq 1} \cos(2\theta_k) - \cos 2\theta_1 \end{align}$$

Now using that $\sum \theta_k = \pi$, you get that $\cos2\theta_j = \cos(2\pi - 2\theta_j) = \cos(\sum_{k\neq j} 2\theta_k)$. And we observe that

$$ \cos(\sum_{k\neq j} 2\theta_k) = 2\cos(\sum_{k\neq j}\theta_j)\cos(\pi - \theta_k) - 1$$

where for $\cos(\sum_{k\neq j}\theta_j)$ we can inductively use the polynomial expression we have already derived. Doing this for all terms you end up with

$$ \sum \sin^2 \theta_k = 2N(1+\prod_k\cos\theta_k) + P \quad\quad (\sharp)$$

where $P$ is some polynomial in $\cos\theta_k$ and $\sin\theta_k$ that contains terms with positive, even number of factors in $\sin$.

So perhaps a better interpretation of your identity is actually as the identity for the angle addition formula for cosine, coupled with identity (#) above, which holds for a list of angles $\theta_k$ that sums to $\pi$.