I have $F_X(x)= \frac{1}{2} + \frac{1}{\pi}arctan(x)$ And I know $Y = aX+b$
Does that make $F_Y(y) = \frac{1}{2} + \frac{1}{\pi}arctan(\frac{y-b}{a})$
Pretty sure $F_Y(y) = F_X(g^{-1}(x)$ where we have $g(x)=aX+b$?
One further question that I can't copy and past in the comment of the current only answer.
Yes. $F_Y(y)=P(Y \le y)=P(aX+b \le y)=P(aX \le y-b)=P(X \le \frac{y-b}{a})=F_X(\frac{y-b}{a})$