Prove that for m = 2, 3, ... sin π/m sin 2π/m sin 3π/m ... sin (m - 1)π/m = m/2^m-1

Question

Prove that for m = 2, 3, ...

[latex] \sin \frac{\pi}{m} \sin \frac{2 \pi}{m} \sin \frac{3 \pi}{m} \cdots \sin \frac{(m-1) \pi}{m}=\frac{m}{2^{m-1}} [/latex]

Accepted Answer

The roots of [latex]z^m=1[/latex] are [latex]z=1, e^{2 \pi i / m}, e^{4 \pi i / m}, \ldots, e^{2(m-1) \pi i / m}[/latex]. Then, we can write

[latex]z^m-1=(z-1)\left(z-e^{2 \pi i / m}\right)\left(z-e^{4 \pi i / m}\right) \cdots\left(z-e^{2(m-1) \pi i / m}\right)[/latex]

Dividing both sides by z-1 and then letting z=1 [realizing that [latex]\left(z^m-1\right) /(z-1)=1+z+z^2+\cdots+z^{m-1}[/latex] ], we find

[latex]m=\left(1-e^{2 \pi i / m}\right)\left(1-e^{4 \pi i / m}\right) \cdots\left(1-e^{2(m-1) \pi i / m}\right) (1)[/latex]

Taking the complex conjugate of both sides of (1) yields

[latex]m=\left(1-e^{-2 \pi i / m}\right)\left(1-e^{-4 \pi i / m}\right) \cdots\left(1-e^{-2(m-1) \pi i / m}\right) (2)[/latex]

Multiplying (1) by (2) using [latex]\left(1-e^{2 k \pi i / m}\right)\left(1-e^{-2 k \pi i / m}\right)=2-2 \cos (2 k \pi / m)[/latex], we have

[latex]m^2=2^{m-1}\left(1-\cos \frac{2 \pi}{m}\right)\left(1-\cos \frac{4 \pi}{m}\right) \cdots\left(1-\cos \frac{2(m-1) \pi}{m}\right) (3)[/latex]

Since [latex]1-\cos (2 k \pi / m)=2 \sin ^2(k \pi / m)[/latex], (3) becomes

[latex]m^2=2^{2 m-2} \sin ^2 \frac{\pi}{m} \sin ^2 \frac{2 \pi}{m} \cdots \sin ^2 \frac{(m-1) \pi}{m} (4)[/latex]

Then, taking the positive square root of both sides yields the required result.

Question 1.52: Prove that for m = 2, 3, ... sin π/m sin 2π/m sin 3π/m ... s...