Real values $b, c, x, y, z$ are such that $x+y+z = \pi/3+b+c = \pi$
The problem is to find trigonometric function $F(x, y, z, b, c)$ such that
For $(x, y, z)\not= (0, 0, \pi), (0, \pi, 0), (\pi, 0, 0)$
$$\begin{align} \frac{F(x, y, z, b,c)}{F(x, z, y, c, b)} &= \frac{\sin(x-z)}{\sin(x-y)}, && \text{if }b=c=\pi/3 \\[4pt] \frac{F(x, y, z, b,c)}{F(x, z, y, c, b)} &= \frac{\cos(y-c)\cos(z-\pi/3)}{\cos(z-b)\cos(y-\pi/3)}, &&\text{if }x=0\\[4pt] \frac{F(x, y, z, b,c)}{F(x, z, y, c, b)} &= \frac{\cos(x+b+2\pi/3)}{\cos c}, &&\text{if }y=0\\[4pt] \frac{F(x, y, z, b,c)}{F(x, z, y, c, b)} &= \frac{\cos b}{\cos(x+c+2\pi/3)}, &&\text{if }z=0. \end{align}$$
In general I am interested in trigonometric function $F(x, y, z, a, b, c)$ such that
For $x+y+z=a+b+c=\pi$ and $(x, y, z)\not= (0, 0, \pi), (0, \pi, 0), (\pi, 0, 0)$
$$\begin{align} \frac{F(x, y, z, a, b,c)}{F(x, z, y, a, c, b)} &= \frac{\sin(x-z)}{\sin(x-y)}, && \text{if }a=b=c=\pi/3 \\[4pt] \frac{F(x, y, z, a, b,c)}{F(x, z, y, a, c, b)} &= \frac{\cos(y-c)\cos(z-a)}{\cos(z-b)\cos(y-a)}, &&\text{if }x=0\\[4pt] \frac{F(x, y, z, a, b,c)}{F(x, z, y, a, c, b)} &= \frac{\cos(x+b+2a)}{4\cos^2 a\cos c}, &&\text{if }y=0\\[4pt] \frac{F(x, y, z, a, b,c)}{F(x, z, y, a, c, b)} &= \frac{4\cos^2 a\cos b}{\cos(x+c+2a)}, &&\text{if }z=0. \end{align}$$
By trigonometric function $F(x, y, z, b, c)$ I mean function of the form $\prod_i\sin(\lambda_{i}x+\ldots+\mu_ic+\gamma_i)$ with constants $\lambda_i,\ldots,\gamma_i$.
I think that there is no such formula in variables $a, b, c, x, y, z$ but if we consider a triangle $ABC$ and point $P$ with $\angle BPC = \pi-x$, $\angle APC = \pi-y$, $\angle APB = \pi -z$ and add additional angles $A=\angle CAP$, $A'=\angle PAB$, $B=\angle ABP$, $B'=\angle PBC$, $C=\angle BCP$, $C'=\angle PCA$ then the formula
$$F/F=\\\frac{\sin(x-y)^2\cos(z-a) \sin z \sin B' \sin^4 c \cos b }{\sin(x-z)^2\cos(y-a) \sin y \sin C \sin^4 b \cos c}\times\\\times\frac{ \left(\frac{\tan A }{ \tan C'} - \frac{\tan A' }{\tan C }\right) \left(\frac{\tan A }{ \tan C' } - \frac{\tan a }{\tan c }\right) \left(\frac{\tan A' }{ \tan C } - \frac{\tan a }{\tan c }\right)}{ \left(\frac{\tan A }{ \tan B' } - \frac{\tan A' }{ \tan B }\right) \left(\frac{\tan A }{\tan B' }-\frac{\tan a }{ \tan b }\right) \left( \frac{\tan A' }{\tan B }-\frac{\tan a }{ \tan b }\right)}$$
is correct.
I found it by fixing triangle $ABC$ and looking at the zeroes and poles of boundary conditions as a functions on $P$ and guessing the trigonometric function with the same zeroes and poles. By complex analysis theorem about meromorphic function we can conclude that this function will be equal to our boundary conditions up to constant. Finally I guessed the constant value.