Explicit form of $\sin(x-y/2)$ I want to know if there exist and how to evaluate the explicit form of a curve given by $$y=\sin(x-y/2)$$
My intuition tells me there should exist one, because deivative of y w.r.t. x is always finite:
$$y'(x) = \frac{2\cos(x - y/2)}{2 + \cos(x - y/2)}$$
 but I have no idea how tho evaluate this explicit form.
 A: First, solve for $x$ when $|x|\le\frac{\pi+1}2$, where the function is invert-able:
$$f(y)=\arcsin(y)+\frac12y=x$$
By Taylor expanding,
\begin{align}f(y)&=\frac12y+\sum_{k=1}^\infty\binom{2k}k\frac{2ky^{2k-1}}{4^k(2k-1)^2}\\&=\frac32y+\frac16y^3+\frac3{40}y^5+\dots\end{align}
Substitute $y=a_0x+a_1x^3+a_2x^5+\dots$ and apply series reversion to get

$$y=\frac23x-\frac2{27}x^3+\frac{239}{21870}x^5+\mathcal O(x^7),\quad|x|\le\frac{\pi+1}2$$

For $\frac{\pi+1}2\le x\le\frac{3\pi-1}2$, use
$$f(y)=-\arcsin(y)+\frac12y=x-\pi$$
From there, we apply a similar approach:
\begin{align}f(y)&=\frac12y-\sum_{k=1}^\infty\binom{2k}k\frac{2ky^{2k-1}}{4^k(2k-1)^2}\\&=-\frac12y-\frac16y^3-\frac3{40}y^5-\dots\end{align}
Substitute $y=a_0(x-\pi)+a_1(x-\pi)^3+a_2(x-\pi)^5+\dots$ and apply series reversion to get

$$y=-2(x-\pi)+\frac83(x-\pi)^3-\frac{88}{15}(x-\pi)^5+\mathcal O((x-\pi)^7),\quad \frac{\pi+1}2\le x\le\frac{3\pi-1}2$$

And finally,
$$y(x)=\begin{cases}\frac23x-\frac2{27}x^3+\frac{239}{21870}x^5+\mathcal O(x^7),&|x|\le\frac{\pi+1}2\\-2(x-\pi)+\frac83(x-\pi)^3-\frac{88}{15}(x-\pi)^5+\mathcal O((x-\pi)^7),&\frac{\pi+1}2\le x\le\frac{3\pi-1}2\\y(x\pm2\pi),&\text{else}\end{cases}$$
