I found the equation $$ \frac{(y-x)\sqrt{2}}{2}=\sin\left(\frac{(y+x)\sqrt{2}}{2}\right) $$ for a sine wave rotated $45^\circ$. Based on the shape, I know the relation is a one to one function. It stands to reason, then, that you could isolate y and describe it in terms of $x$. The best I could do with my knowledge of algebra was the infinite iteration: $$ y=x+\sqrt{2}\sin(x\sqrt{2}+\sin(x\sqrt{2}+\sin(x\sqrt{2}+...))). $$
Is there a way to derive a simpler expression?
$$\frac{y-x}{\sqrt{2}}=\sin \left(\frac{y+x}{\sqrt{2}}\right)\tag 1$$
Let $$x=\rho \,\sqrt 2\,\cos^2(\theta)\qquad \text{and} \qquad y=\rho \,\sqrt 2\,\sin^2(\theta)\tag 2$$ Equation $(1)$ becomes $$\rho \cos (2 \theta)+\sin (\rho)=0 \tag 3$$ the contour plot of which being interesting (have a look here).
Equation $(3)$ is explicit in $\theta$ but implicit in $\rho$. So, for a given value of $\rho$, we have the various possible values of $\theta$ then $(x,y)$.
But, equation $(3)$ also leads to the parametric equation $$x=\frac{\rho -\sin (\rho )}{\sqrt{2}}\qquad\qquad\qquad y=\frac{\rho +\sin (\rho )}{\sqrt{2}}$$
So, for example, the arc length $$L=\int_0^t \sqrt{1+\cos ^2(\rho )}\,d\rho=\sqrt{2}\, E\left(t \left|\frac{1}{2}\right.\right)$$
Similarly, the surface area under the curve $$A=\int_0^t \frac{1}{2} (\rho +\sin (\rho )) (1-\cos (\rho ))\,d\rho=\frac{1}{8} \left(2 t^2-4 t \sin (t)-8 \cos (t)+\cos (2 t)+7\right)$$