$\newcommand\De\Delta$Elementarily: The relation $f=\De_h^{-1}g$ means that $f(x+h)-f(x)=g(x)$ for all $x$.

For a function $g$ and each real $r$, we know $f_r:=\De_1^{-1}g_r$, where $g_r(x):=g(rx)$, so that $$f_r(x+1)-f_r(x)=g(rx) \tag{10}\label{10}$$

For real $h\ne0$, real $r$, and all real $x$, let now $$f_{r,h}(x):=f_{rh}(x/h). \tag{20}\label{20}$$ Then $$f_{r,h}(x+h)-f_{r,h}(x)=f_{rh}(x/h+1)-f_{rh}(x/h)=g((rh)(x/h))=g(rx)$$ for all real $x$. That is, $f_{r,h}=\De_h^{-1}g_r$. $\quad\Box$

To illustrate these simple considerations, consider the example from your post, with $g_r(x)=\sin rx$ and $f_r(x)=\dfrac{-\cos(r(x-1/2))}{2\sin(r/2)}$. Then, by \eqref{20}, for $\De_h^{-1}g_r=f_{r,h}$ we have $$(\De_h^{-1}g_r)(x)=f_{r,h}(x)=\dfrac{-\cos((rh)(x/h-1/2))}{2\sin(rh/2)} =\dfrac{-\cos(r(x-h/2))}{2\sin(rh/2)}. \quad\Box$$

$\newcommand\De\Delta$Elementarily: The relation $f=\De_h^{-1}g$ means that $f(x+h)-f(x)=g(x)$ for all $x$.

For a function $g$ and each real $r$, we know $f_r:=\De_1^{-1}g_r$, where $g_r(x):=g(rx)$, so that $$f_r(x+1)-f_r(x)=g(rx) \tag{10}\label{10}$$

For real $h\ne0$, real $r$, and all real $x$, let now $$f_{r,h}(x):=f_{rh}(x/h). \tag{20}\label{20}$$ Then $$f_{r,h}(x+h)-f_{r,h}(x)=f_{rh}(x/h+1)-f_{rh}(x/h)=g((rh)(x/h))=g(rx)$$ for all real $x$. That is, $f_{r,h}=\De_h^{-1}g_r$. $\quad\Box$

Note that we do not need or impose any conditions on the functions $g$ and $f_r$ (such as measurability, continuity, growth, etc.) except for \eqref{10}.

To illustrate these simple considerations, consider the example from your post, with $g_r(x)=\sin rx$ and $f_r(x)=\dfrac{-\cos(r(x-1/2))}{2\sin(r/2)}$. Then, by \eqref{20}, for $\De_h^{-1}g_r=f_{r,h}$ we have $$(\De_h^{-1}g_r)(x)=f_{r,h}(x)=\dfrac{-\cos((rh)(x/h-1/2))}{2\sin(rh/2)} =\dfrac{-\cos(r(x-h/2))}{2\sin(rh/2)}. \quad\Box$$

$\newcommand\De\Delta$Elementarily: The relation $f=\De_h^{-1}g$ mean that $f(x+h)-f(x)=g(x)$ for all $x$.

For a function $g$ and each real $r$, we know $f_r:=\De_1^{-1}g_r$, where $g_r(x):=g(rx)$, so that $$f_r(x+1)-f_r(x)=g(rx)$$

For real $h\ne0$, real $r$, and all real $x$, let now $$f_{r,h}(x):=f_{rh}(x/h).$$ Then $$f_{r,h}(x+h)-f_{r,h}(x)=f_{rh}(x/h+1)-f_{rh}(x/h)=g((rh)(x/h))=g(rx)$$ for all real $x$. That is, $f_{r,h}=\De_h^{-1}g_r$. $\quad\Box$

$\newcommand\De\Delta$Elementarily: The relation $f=\De_h^{-1}g$ means that $f(x+h)-f(x)=g(x)$ for all $x$.

For a function $g$ and each real $r$, we know $f_r:=\De_1^{-1}g_r$, where $g_r(x):=g(rx)$, so that $$f_r(x+1)-f_r(x)=g(rx) \tag{10}\label{10}$$

For real $h\ne0$, real $r$, and all real $x$, let now $$f_{r,h}(x):=f_{rh}(x/h). \tag{20}\label{20}$$ Then $$f_{r,h}(x+h)-f_{r,h}(x)=f_{rh}(x/h+1)-f_{rh}(x/h)=g((rh)(x/h))=g(rx)$$ for all real $x$. That is, $f_{r,h}=\De_h^{-1}g_r$. $\quad\Box$

To illustrate these simple considerations, consider the example from your post, with $g_r(x)=\sin rx$ and $f_r(x)=\dfrac{-\cos(r(x-1/2))}{2\sin(r/2)}$. Then, by \eqref{20}, for $\De_h^{-1}g_r=f_{r,h}$ we have $$(\De_h^{-1}g_r)(x)=f_{r,h}(x)=\dfrac{-\cos((rh)(x/h-1/2))}{2\sin(rh/2)} =\dfrac{-\cos(r(x-h/2))}{2\sin(rh/2)}. \quad\Box$$