show that：
$$\prod_{k=1}^{n-1}\sin{\dfrac{k\pi}{tn}}=\dfrac{\sqrt[t]{n}}{2^{n-1}}?(not, true),t\in N^{+}$$
maybe for $t$is real numbers also true?
I can show when $t=1$ case. because I use $$z^{n-1}-1=(z-x_{1})(z-x^2_{1})\cdots (z-x^{n-1}_{1}),x_{1}=\cos{\dfrac{2\pi}{n}}+i\sin{\dfrac{2\pi}{n}}$$ also see Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$
But for $t\neq 1$,I can't use this identity
Another reason for which the formula cannot be true is that $\log(\sin x)$ is an integrable function over $(0,1)$, hence for $t\in\mathbb{N}$ it is expected that: $$\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{t n}\approx n\int_{0}^{\pi/t}\log\sin x\,dx \approx -n\frac{1+\log\frac{t}{\pi}}{t}$$ by Riemann sums, while: $$\log\frac{\sqrt[t]{n}}{2^{n-1}}\approx -n\log 2+\frac{\log n}{t}$$ has the wrong magnitude.
To compute the product, a possibility is given by the Fourier series identity: $$-\log\sin x = -\log 2-\sum_{k\geq 1}\frac{\cos(2kx)}{k}. $$
