integral of a "sin-omial" coefficients=binomial I find the following averaged-integral amusing and intriguing, to say the least. Is there any proof?

For any pair of integers $n\geq k\geq0$, we have
  $$\frac1{\pi}\int_0^{\pi}\frac{\sin^n(x)}{\sin^k(\frac{kx}n)\sin^{n-k}\left(\frac{(n-k)x}n\right)}dx=\binom{n}k. \tag1$$ 

I also wonder if there's any reason to relate these with an MO question that I just noticed. Perhaps by inverting?
AN UPDATE.  I'm extending the above to a stronger conjecture shown below.

For non-negative reals with $r\geq s$, a generalization is given by
  $$\frac1{\pi}\int_0^{\pi}\frac{\sin^r(x)}{\sin^s(\frac{sx}r)\sin^{r-s}\left(\frac{(r-s)x}r\right)}\,dx
=\binom{r}{s}. \tag2$$

 A: Tonight I read here [the answer by esg to another your question] that $\frac1{2\pi}\int_{-\pi}^\pi e^{-ik t}(1+e^{it})^ndt=\binom{n}{k}$, which is, well, obvious at least when both $n$ and $k$ are positive integers: just expand the binomial $(1+e^{it})^n$ and integrate. Denoting $\alpha=k/n$ we may rewrite this as $\frac1{2\pi}\int_{-\pi}^\pi (f(t))^n dt=\binom{n}{\alpha n}$, where the function $f(t)=(1+e^{it})e^{-i\alpha t}$ is complex-valued. For making it real-valued, we change the path between the points $-\pi$ and $\pi$. The value of the integral does not change (since $f^n$ is analytic between two paths, for integer $n$ it is simply entire function.) On the second path $f$ takes real values. Namely, for $t\in (-\pi,\pi)$ we define $s(t)=\ln \frac{\sin (1-\alpha)t}{\sin \alpha t}$. It is straightforward (some elementary high school trigonometry) that $$f(t+is(t))=\frac{\sin t}{\sin^{\alpha} \alpha t\cdot \sin^{1-\alpha}(1-\alpha)t},$$
so we replace the path from $(-\pi,\pi)$ to $\{t+s(t)i:t\in (-\pi,\pi)\}$ (limit values of $s(t)$ at the endpoints are equal to 0) and take only the real part of the integral (this allows to replace $d(t+s(t)i)$ to $dt$ in the differential). We get
$$
\frac1{2\pi}\int_{-\pi}^\pi \frac{\sin^n t}{\sin^{\alpha n} \alpha t\cdot \sin^{(1-\alpha)n}(1-\alpha)t}dt=\binom{n}{\alpha n}
$$
as desired.
A: (not an answer)
I've found a way to convert the integral evaluation to a binomial sum identity.
Incidentally, this gives the details leading up to the follow-up question here.
I have highlighted (in bold) where we need some potential rigor to make this argument complete.
Let $\zeta=e^{\pmb{i}x/n}, \pmb{i}=\sqrt{-1}$. Equation (1) becomes an integral along an arc on the unit circle
$$\frac{n}{\pmb{i}\pi}\int_1^{e^{\pmb{i}\pi/n}}\frac{(\zeta^n-\zeta^{-n})^n}
{(\zeta^k-\zeta^{-k})^k(\zeta^{n-k}-\zeta^{-(n-k)})^{n-k}}\frac{d\zeta}{\zeta}=\binom{n}k. \tag3$$
Define the rational complex functions (with a pole at the origin) $f_m(z)=(z^m-z^{-m})^m$ and 
$$F_{n,k}(z)=
\frac{f_n(z)}{f_k(z)f_{n-k}(z)}=\frac{(1-z^{2n})^nz^{-2k(n-k)}}{(1-z^{2k})^k(1-z^{2n-2k})^{n-k}}
=\frac{(1-z^{2n})^n}{(1-z^{2k})^k(z^{2k}-z^{2n})^{n-k}}.$$ 
To verify (3), compute a contour integral around the unit circle $\mathcal{C}$ (oriented positively)
$$\frac{n}{\pmb{i}\pi}\int_1^{e^{\pmb{i}\pi/n}}F_{n,k}(z)\frac{dz}z=
\pmb{\frac{2n}{2\pmb{i}\pi}\int_1^{e^{\pmb{i}\pi/n}}F_{n,k}(z)\frac{dz}z=
\frac1{2\pmb{i}\pi}\int_{\mathcal{C}}F_{n,k}(z)\frac{dz}z}=\text{Res}(F_{n,k}(z);0).$$
This is equivalent to determining the constant term in $F_{n,k}(z)$, which in turn reduces to the identity
$$\sum (-1)^a\binom{n}a\binom{k+b-1}b\binom{n-k+c-1}c=\binom{n}k\tag4$$
where the sum runs through $a,b,c\geq0$ such that $(a+c)n+(b-c)k=k(n-k)$.
It remains to prove (4).
A: (not an answer)
Denote $\alpha=k/n$, $f(x)=(\frac{\sin x}{\sin \alpha x})^\alpha (\frac{\sin x}{\sin (1-\alpha) x})^{1-\alpha}$. Then your claim may be rewritten as $\pi^{-1}\int_0^\pi f^n(x)dx=\frac{\Gamma(n+1)}{\Gamma(\alpha n+1)\Gamma((1-\alpha)n+1)}$, and it looks to be true without additional assumption that $\alpha n$ is integer (I checked for $\alpha=0.3;n=7$ or $n=7.4$ on WolframAlpha). We may multiply this by Beta-function $\int_0^1 t^{\alpha n}(1-t)^{(1-\alpha)n}dt=\frac{\Gamma(\alpha n+1)\Gamma((1-\alpha)n+1)}{\Gamma(n+2)}$, and we have to prove that $\int_0^\pi\int_0^1 h^n(t,x)dtdx=\frac{\pi}{n+1}$, where $h(t,x)=f(x)t^\alpha(1-t)^{1-\alpha}$. That is, our function $h$ on the rectangular $[0,\pi]\times [0,1]$ (with the normalized Lebesgue measure) should be equidistributed with the function $t$ on $[0,1]$. Another similar approach could be multiplying by two $\Gamma$-functions $\int_0^{\infty} y^{\alpha n}e^{-y}dy=\Gamma(\alpha n+1)$, $\int_0^{\infty} z^{(1-\alpha) n}e^{-z}dz=\Gamma((1-\alpha) n+1)$. On the probabilistic language, we get the following equivalent 
Claim. Let EXP denote the exponential law (with density $e^{-t}dt$, $t>0$). Let $Y,Z$ be independent random variables distributed by EXP, and let $X$ be a third independent (of $Y,Z$) random variable distributed uniformly on $[0,\pi]$. Then for any fixed $\alpha\in (0,1)$ we have $$\left(Y\frac{\sin X}{\sin \alpha X}\right)^\alpha \left(Z\frac{\sin X}{\sin (1-\alpha) X}\right)^{1-\alpha}\in \text{EXP}.$$
A: 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.
A: Here is another way to prove it. Surprisingly, for $n$ integral and $k$ real, the integral in question can be written down as an indefinite integral. This gives a direct proof for non-integer $k$, though obviously less clear than the contour method. (In fact, it is convenient to avoid integer $k$ in this method, and extend to integer $k$ by continuity.)
Writing $y=x/n$ and $l=n-k$, we have for example for $n=2$:
  $$\int\frac{\sin^2(2y)}{\sin^k(ky)\sin^l(ly)}dy=
  \frac{\frac{2}{l-k}\sin((l-k)y)+\sin(2y)}
  {kl\sin^{k-1}(ky)\sin^{l-1}(ly)}$$
In general for $n$ even ($n$ odd is similar with cosines):
  $$I_{n,k}(y)=\int\frac{\sin^n(ny)}{\sin^k(ky)\sin^l(ly)}dy=
  \frac{\sum_{r=0}^{n-1}\sum_{s=0}^{n-1}\lambda_{r,s}\sin(((n-1-2r)k+(n-1-2s)l)y)}
  {kl\sin^{k-1}(ky)\sin^{l-1}(ly)}$$
where
  $$\lambda_{r,s}=\begin{cases}
  \displaystyle
  \frac{(-1)^r(n-1)^{\underline{r}}}{(r-l)^{\underline{\smash{r-s}}}s!}\lambda_{0,0},\;\;r\ge s\\
  \displaystyle
  \frac{(-1)^s(n-1)^{\underline{s}}}{(s-k)^{\underline{\smash{s-r}}}r!}\lambda_{0,0},\;\;s\ge r
  \end{cases}
  $$
  $$\lambda_{0,0}=(-1)^{n/2+1}2^{1-n},$$
  and $x^{\underline{r}}$ denotes the falling power $x(x-1)\ldots(x-r+1)$.
It is easy to check the derivative, $I_{n,k}'(y)$ is correct by considering the coefficient
  of $\cos((ak+bl)y)/(\sin^k(ky)\sin^l(ly))$ for each $a$, $b$. If $a\neq b$ then you get zero,
  otherwise for $a=b=n-2r$, $(ak+bl)y=(n-2r)ny$ and you get $\frac{1}{2}(-1)^r\binom{n}{r}\lambda_{0,0}\cos((n-2r)ny)$. Then $\sin^n(ny)$ arises from the binomial expansion:
  $$(-1)^{n/2}2^{-n}\sum_{r=0}^n (-1)^r\binom{n}{r}\cos((n-2r)ny)=\sin^n(ny).$$
Note that $I_{n,k}(0)=0$ because, being the integral of something well-behaved at $0$,
  $I_{n,k}(y)$ must be continuous at $0$, so its numerator must vanish to order $n-2$
  like its denominator. Using L'H\^{o}pital, taking $n-2$ derivatives of the numerator gives only sines, which themselves vanish at 0.
  To evaluate $I_{n,k}(\pi/n)$, note that $\sin(k\pi/n)=\sin(l\pi/n)$ and
  $\sin(((n-1-2r)k+(n-1-2s)l)y)=\sin(2(r-s)k\pi/n)$.
  Conditioning on $r-s=d>0$, you get (e.g., by considering partial fractions in $k$)
  $$\sum_{s=0}^{n-1-d}\lambda_{s+d,s}=
  \frac{(-1)^d(n-1)!\binom{n-2}{d-1}}{(k-1)^{\underline{\smash{n-1}}}}\lambda_{0,0},$$
  and similarly for $d<0$ with the opposite sign, and using $-d$ in place of $d$.
  Using the binomial expansion of $(1-e^{2\pi ik/n})^{n-2}$, you get
  $$\sum_{d=1}^{n-1}(-1)^d\binom{n-2}{d-1}\sin\left(\frac{2dk\pi}{n}\right)=
     (-1)^{n/2}2^{n-2}\sin^{n-2}\left(\frac{k\pi}{n}\right)\sin(k\pi)$$
  So finally, putting the pieces together,
  $$\frac{n}{\pi}I_{n,k}\left(\frac{\pi}{n}\right)=\frac{n\sum_{r=0}^{n-1}\sum_{s=0}^{n-1}\lambda_{r,s}\sin\left(\frac{2(r-s)k\pi}{n}\right)}
  {\pi kl\sin^{k-1}\left(\frac{k\pi}{n}\right)\sin^{l-1}\left(\frac{l\pi}{n}\right)}=\frac{n!\sin(k\pi)}{\pi k^{\underline{\smash{n+1}}}}$$
  which (for even $n$) we recognise as $\binom{n}{k}$ by the reflection formula for factorials.
