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I've found the time and thought I should post this as I had a little breakthrough. This isn't an answer to the question but is an answer to a question posted in the comments. If the result holds, does it hold for complex values? I am being brief here and certainly not rigorous as I thought it would be a nice quip to add; nonetheless the result should follow if one wishes to fill in the gaps. If we assume the answer to the OP's question is yes, then

$$\frac{1}{\pi}\int_0^\pi \dfrac{\sin^{z}(t)}{\sin^{k}(\frac{k}{z}t)\sin^{z-k}(\frac{z-k}{z}t)}\,dt = \dbinom{z}{k}$$

This is rather involved (and would be too involved if I chose to make it rigorous) so pay close attention. Consider firstly a consequence of Ramanujan's master theorem

If $f_1(z)$ and $f_2(z)$ are holomorphic for $\Re(z) > 0$ and if $|f_{12}(x+iy)| < C e^{\tau|y|+\rho|x|}$ for $\tau < \pi$ and $\rho>0$ then

$$f_1 \Big{|}_{\mathbb{N}} = f_2\Big{|}_{\mathbb{N}} \Rightarrow f_1 = f_2$$

So essentially what we are going to do is show this in two steps. Firstly that

$$f_k(z) = \frac{1}{\pi}\int_0^\pi \dfrac{\sin^{z}(t)}{\sin^{k}(\frac{k}{z}t)\sin^{z-k}(\frac{z-k}{z}t)}\,dt$$

is bounded so that Ramanujan's master theorem will prevail and necessarily $f_k(z) = \dbinom{z}{k}$ since $\dbinom{z}{k}$ is equally so bounded.

Taking the function $g(z) = \sup_{t \in [0,\pi]} \Big{|}\dfrac{\sin^{z}(t)}{\sin^{k}(\frac{k}{z}t)\sin^{z-k}(\frac{z-s}{z}t)}\Big{|}$ for $\Re(z) > k$ we can show that this function is properly bounded. For each $t$ we know $\sin(t)^{z}$ is bounded as required as $y \to \infty$ for $\epsilon < t < \pi - \epsilon$; because this is exponentiation with a positive real value base--it is periodic. As $x \to \infty$ it just tends to $0$ so all good there. Now $\sin^{k-z}(\frac{z-s}{z}t)$ is exponentiation of a value which tends to $\sin(t)$. This is a little tricky but

$$\sin^{k}(t - \frac{k}{z}t)$$ is bounded and now all that's left is the troublesome

$$\sin^{-z}(t - \frac{k}{z}t)$$

which clearly grows like $\frac{1}{\sin^{x}(t)}$ as $\Re(z) = x \to \infty$. As $\Im(z) = y\to\infty$ it is not periodic, but it is eventually bounded by $\sin^{-z}(t\pm i\delta)$ though not exactly. This bound is of type $\tau < \pi$. This works for all $t\in [\epsilon,\pi-\epsilon]$ and so as $\epsilon \to 0$ it will follow taking close care to observe the end points tend to $1$ as $t \to 0,\pi$. Therefore $g(z) < Ce^{\tau|y| + \rho|x|}$, $f_k$ is of a Ramanujan bound for $\Re(z) > k$ and necessarily

$$f_k(z) = \frac{1}{\pi}\int_0^\pi \dfrac{\sin^{z}(t)}{\sin^{k}(\frac{k}{z}t)\sin^{z-k}(\frac{z-k}{z}t)}\,dt = \dbinom{z}{k}$$

This is all rather hand waivey because I don't want to take up too much space, the amount of epsilons and deltas is exhausting; plus this is more of an extended comment.

Taking $f_s(z)$ is much trickier. Performing the same procedure in the opposite direction is impossible, this is because $\dbinom{z}{s}$ is not bounded in $s$ in the sense described above. It grows like $\sin(\pi s)$ which isn't subject to Ramanujan's master theorem. I thought I could trick it into working but I've had no luck.