XCAS wins

[Albert Chan]

Found a nice proof from the book "An Imaginary Tale, The Story of √ (-1)", by Paul Nahin, p71-72
Note: book proof actually does product of cos, then flip to sin, using \(\cos(\pi/2-x) = \sin(x)\)

Proof: \(\displaystyle\prod _{k=1}^{n-1} \sin \left({k \pi \over 2n}\right) = { \sqrt{n} \over 2^{n-1}}\)

For \(z^{2n} = 1\), we have 2 real roots ±1, and 2(n-1) complex roots, spread out evenly over complex unit circle.
Each complex root, paired with its conjugate, producing quadratic factor: \((z - r)(z - 1/r) = z^2 - (r+1/r)z + 1\)

\(z^{2n} - 1 = (z^2-1) \displaystyle\prod _{k=1}^{n-1}(z^2 - 2\cos\left({k \pi \over n}\right)z + 1)\)

Divide both side by \(z^n\)

\(z^n - 1/z^n = (z-1/z) \displaystyle\prod _{k=1}^{n-1}((z+1/z) - 2\cos\left({k \pi \over n}\right))\)

Let \(z = e^{i θ}\), we have:

\(2i\sin (nθ) = (2i\sin θ) \displaystyle\prod _{k=1}^{n-1}(2\cos θ - 2\cos\left({k \pi \over n}\right))\)

Now, this is the trick to remove θ. Divide both side by \(2i\sin θ\), and take the limit when θ → 0

\(n = 4^{n-1} \displaystyle\prod _{k=1}^{n-1}{1-2\cos\left({k \pi \over n}\right) \over 2}
= 4^{n-1} \displaystyle\prod _{k=1}^{n-1}\sin^2\left({k \pi \over 2n}\right)
\)

Divide both side by 4^n-1, and take the square root, the proof is done.