Can $\sin\left(\frac{x+y}{\sqrt{2}}\right)=\frac{y-x}{\sqrt{2}}$be written as a function of $x$ I started by noticing that the derivative of $\sin(x)$ is always between $1$ and $-1$. Therefore if i have a line that intersects the $x$ axis at $45°$ it will aways pass through the line once or be tangent to it. In other words $\sin(x)=\pm x-a$ has only one solution for any $a$ and sign for $x$. So using the rotation matrix to rotate $\sin(x)$ $-45°$ around the origin I get $$\sin\left(\frac{x+y}{\sqrt{2}}\right)=\frac{y-x}{\sqrt{2}}$$ I got stuck on trying to write this as a function of $x$ but i do believe it can be. And as an add on question can the same be done for $\sin(x+y)=y-x$

Well, one way to approach this is to use the power series expansion around the origin of $$sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}- \cdots$$ A first approximation would then be $sin(x+y)=x+y-\frac{(x+y)^3}{3!}=y-x$, which yields $y=-x+\sqrt[3]{12x}$. This provides a nice plot (Desmos):

![enter image description here](

You could take more powers of the in the Taylor-expansion into account of course.