Show that $\prod_{k=1}^{n}\sqrt{1+8\sin^2(k\pi/n)}=2^n-1$ for all positive integers $n$.

Question

Let $P(n)=\prod\limits_{k=1}^{n}\sqrt{1+8\sin^2\left(\frac{k\pi}{n}\right)}$

Show that $P(n)=2^n-1$ for all positive integers $n$.

Context

You may have heard of the following remarkable fact about the unit circle: If $n$ equally spaced points are chosen on a unit circle, and line segments are drawn from one of the points to each of the other points, then the product of the lengths of these $n-1$ line segments, equals $n$ (proof).

I wondered, if instead, line segments are drawn from a fixed point outside the plane of the circle, to each of the $n$ chosen points on the circle, then what is the product of the lengths of the $n$ line segments? I experimented with different locations of the fixed point, and different size circles, to see if any interesting results can be found.

If the circle has equation $x^2+y^2=\sqrt2^2, z=0$, and one of the $n$ evenly spaced points on the circle is $(\sqrt2,0,0)$, and the fixed point is $(\sqrt2,0,1)$, then the product of the lengths of the $n$ line segments can be expressed as $P(n)=\prod\limits_{k=1}^{n}\sqrt{1+8\sin^2\left(\frac{k\pi}{n}\right)}$. Numerical investigation suggests that $P(n)=2^n-1$ for all positive integers $n$. This is a rather elegant result, so I'm trying to find a way to prove it.

My attempt

Squaring $P(n)$, and using $\sin^2\theta\equiv\frac{1-\cos 2\theta}{2}$, we have

$$P(n)^2=\prod_{k=1}^n\left(5-4\cos\left(\frac{2k\pi}{n}\right)\right)=(-2)^n\prod_{k=1}^n\left(2\left(-\frac54\right)+2\cos\left(\frac{2k\pi}{n}\right)\right)$$

Then using an identity (listed under "Finite products of trigononetric functions"), we have

$$P(n)^2=(-2)^n\left(2\left(T_n\left(-\frac54\right)-(-1)^n\right)\right)$$

where $T_n$ is the Chebyshev polynomial. This gives

$$P(n)=\sqrt{2^{n+1}\left(\cosh\left(n\cosh^{-1}\left(\frac54\right)\right)-1\right)}$$

But I don't know why this is equal to $2^n-1$.

Fun fact

If the circle is $x^2+y^2=1, z=0$, and one of the $n$ evenly spaced points on the circle is $(1,0,0)$, and the fixed point is $(1,0,1)$, then it can be shown that, as $n\to\infty$, the geometric mean of the lengths of the $n$ line segments approaches the golden ratio, $\frac{1+\sqrt5}{2}$.

Answer 1

The following is inspired by “product of factors” on AoPS.

For a fixed positive integer $n$ let $\omega_k = \exp(2 \pi i k/n) $ be the $n$-th roots of unity. Note that $\omega_k \cdot \omega_{n-k} = 1$ and $\omega_k + \omega_{n-k} = 2 \cos(2 \pi k/n)$. Then $$ (x^n-1)^2 = \prod_{n=1}^n (x - \omega_k)(x - \omega_{n-k}) \\ = \prod_{k=1}^n \left( x^2 - 2 x\cos\left(\frac{2k\pi}{n} \right) + 1\right) \, . \tag{$*$} $$ Setting $x=2$ gives $$ (2^n-1)^2 = \prod_{k=1}^n \left(5 - 4\cos\left(\frac{2k\pi}{n} \right)\right) = \prod_{k=1}^n \left(1+8\sin^2\left(\frac{k\pi}{n}\right) \right)\, . $$

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The same approach can be used to compute the product of distances of an arbitrary point in $\Bbb R^3$ to $n$ equally distributed points on a circle.

The product $P(n)$ of the distances of a point $(r \cos(\phi), r \sin(\phi), z)$ to the $n$ points $(r \cos(2 \pi k/n), r \sin(2 k \pi/n), 0)$, $k=1, \ldots, n$, satisfies $$ \begin{align} P(n)^2 &= (Rr)^n \left( x^n + x^{-n} - 2\cos(n \phi) \right) \\ &= (Rr)^n \left (2 T_n \left( \frac{R^2+r^2+z^2}{2Rr}\right) - 2\cos(n \phi) \right) \, , \end{align} $$ where $x > 0$ is a solution of $$ x + \frac 1x = \frac{R^2+r^2+z^2}{Rr} $$ and $T_n$ is the $n$-th Chebyshev polynomial of the first kind.

Proof: The square of the distance to the $k$-th point on the circle is $$ \left( R \cos\left(\frac{2 k \pi}{n}\right) - r\cos(\phi)\right)^2 + \left( R \sin\left(\frac{2 k \pi}{n}\right) - r\sin(\phi)\right)^2 + z^2 \\ = R^2 + r^2 + z^2 - 2Rr \cos\left(\frac{2 k \pi}{n}-\phi\right) \\ = \frac{Rr}{x} \left( x^2 - 2x \cos\left(\frac{2 k \pi}{n}-\phi\right) + 1\right) \, . $$

Using an identity similar to $(*)$ above (compare Prove $\prod\limits_{k=0}^{n-1} \left(x^2-2x\cos \left(\alpha+\frac{2k\pi}{n}\right)+1\right)=x^{2n}-2x^n\cos(n\alpha)+1$) it follows that the square of the product of the distances is $$ P(n)^2 = \left( \frac{Rr}{x}\right)^n (x^{2n} - 2x^n \cos(n \phi) + 1) = (Rr)^n \left( x^n - 2\cos(n \phi) + x^{-n} \right) \,, $$ which is the first formula. For the second formula we substitute $x = e^u$ (compare Solve $g(x + \frac{1}{x}) = x^y + \frac{1}{x^y}$), then $$ 2 \cosh(u) = x + \frac 1x = \frac{R^2+r^2+z^2}{Rr} \\ \implies u = \cosh^{-1} \left( \frac{R^2+r^2+z^2}{2Rr}\right) \\ \implies x^n + x^{-n} = 2 \cosh (nu) = 2 T_n \left( \frac{R^2+r^2+z^2}{2Rr}\right) \, . $$

Example 1: For the “fun fact” case at the end of the question we have $R=r=z=1$ and $\phi = 0$. Solving $x+1/x=3$ gives $x = (3+\sqrt 5)/2 = ((1+\sqrt 5)/2)^2$, i.e. $x$ is the square of the golden ratio. Then $$ P(n)^2 = x^n + x^{-n} = \left( \frac{1+\sqrt 5}{2}\right)^{2n} + \left( \frac{1+\sqrt 5}{2}\right)^{-2n} -2 \, , $$ confirming the limit $$ P(n)^{1/n} \to \frac{1+\sqrt 5}{2} $$ for $n \to \infty$. Here $P(n)^2 = L_{2n} - 2$ with $L_n$ being the Lucas numbers, in particular all $P(n)^2$ are integers.

Example 2: If the “outside point” is in same plane as the circle, i.e. $z=0$, then $x$ can be chosen as $x=R/r$, and the square of the product of the distances is $$ P(n)^2 = R^{2n} - 2 R^nr^n \cos(n\phi) + r^{2n} \, . $$ In particular is $|R^n - r^n| \le P(n) \le R^n + r^n$ in this configuration.

Answer 2

You want to prove that $$ (-2)^n\left(2\left(T_n\left(-\frac54\right)-(-1)^n\right)\right) = (2^n-1)^2 \, . $$ Expanding the terms and using the symmetry $T_n(-x) = (-1)^n T_n(x)$ that is equivalent to $$ T_n\left(\frac54\right) = \frac 12 \left( 2^n + \frac{1}{2^n}\right) \, , $$ which can be proven with induction. It is correct for $n=0$ and $n=1$ by direct verification, and the induction step uses the recursion formula for Chebyshev polynomials $$ T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x) $$ with $x =5/4$. Alternatively use the explicit expression $$ T_{n}(x)= \frac 12 \left( \left(x - \sqrt{x^2-1}\right)^n + \left(x + \sqrt{x^2-1}\right)^n\right) $$ with $x=5/4$.

Answer 3

Hint: $$\cosh\left( n\cosh^{-1}\left(\frac{5}{4}\right)\right) = \frac{2^{n}+2^{-n}}{2}$$ This comes from $\cosh(x)= \frac{e^{x}+e^{-x}}{2}$ when you inverse the function and solve for $x$ then $\cosh^{-1}(x)=\ln(x+\sqrt{x^2-1})$

$$P(n)=\sqrt{2^{n+1}\left(\cosh\left(n\cosh^{-1}\left(\frac54\right)\right)-1\right)}$$ $$P(n)=\sqrt{2^{n+1}\left(\frac{2^{n}+2^{-n}}{2}\right)-1}$$

Answer 4

All of your derivation was correct for positive integers $n$.

The final blow comes from the easy fact that \begin{align*}\cosh^{-1}\frac{5}{4}&=\ln\left(\frac{5}{4}+\sqrt{\left(\frac{5}{4}\right)^{2}-1}\right)\\&=\ln\left(\frac{5}{4}+\frac{3}{4}\right)\\&=\ln 2\end{align*} and thus \begin{align*}(P(n))^{2}&=2^{n+1}\left(\cosh\left(n\cosh^{-1}\frac{5}{4}\right)-1\right)\\&=2^{n+1}\left(\cosh(n\ln2)-1\right)\\&=2^{n+1}\left(2^{-n-1}+2^{n-1}-1\right)\\&=1+2^{2n}-2^{n+1}\\&=(2^{n}-1)^{2}\end{align*}

Answer 5

(Self-answering)

Taking the hint from @EDX's comment, we can use $\cosh^{-1}x=\ln\left(x+\sqrt{x^2-1}\right)$, as well as $\cosh x=\frac12\left(e^x+e^{-x}\right)$.

Thus, $\cosh\left(n\cosh^{-1}x\right)=\frac12\left(\left(x+\sqrt{x^2-1}\right)^n+\left(x+\sqrt{x^2-1}\right)^{-n}\right)$.

Then, continuing from the OP's attempt, we have

$$P(n)=\sqrt{2^{n+1}\left(\frac12\left(\left(\frac54+\sqrt{\left(\frac54\right)^2-1}\right)^n+\left(\frac54+\sqrt{\left(\frac54\right)^2-1}\right)^{-n}\right)-1\right)}=2^n-1$$