We start by recalling the key concept: The inverse of a function. 

Definition INV Inverse Function

We consider the three core trigonometric functions $sin$, $cos$ and $tan$  and their related inverse functions $\arcsin, \arccos,$ and $ \arctan$.

Functions that have these core trigonometric functions as components either arithmetically or as compositions will have inverses using the related inverse functions. If a core trigonometric function is used, then the related inverse trigonometric function will appear in the inverse, and vice versa.

Note: The algebra of trigonometric and inverse trigonometric functions can sometimes help simplify the expressions of inverses.

As as example we consider functions of the form $f(x) = A \tan(x-h)+k$ for $x \in (-\frac{\pi}2+h,\frac{\pi}2+h)$.

This trigonometric function is constructed with compositions using the tangent  function, $\tan$ , and core linear functions. Finding the algebraic expression for the inverse of $f$ can be interpreted visually by composing the inverses of   $f_{-h}(x)=x-h$, $\tan$ , $f_A(x) =A*x$,and  $f_{+k}(x)=x+k$  in the reverse order ("socks and shoes").

Thus the inverse of $f(x) = A \tan(x-h)+k$   is
$g(x)= (f_{+h}\circ \arctan \circ f_{1/A}\circ f_{-k})(x) =  \arctan { \frac {x-k} A} +h $.

We use mapping diagrams to verify and visualize how the function $g(x) = \arctan { \frac {x-k} A} +h $ is the inverse of $f$ when $A \ne 0$ by looking at the composition functions $(f\circ g)(x) = x$ for $x \in R$ and $(g \circ f)(x) = x$  for  all $x \in (-\frac{\pi}2+h,\frac{\pi}2+h)$.

Example TRIG.INV.1 : Suppose $f(x) = 2\tan(x-1) -3$ for $x \in (-\frac{\pi}2+1,\frac{\pi}2+1)$. Verify that $g(x) = \arctan( \frac {x+3} 2) + 1$ is the inverse function for $f$ .

Inverses for $f(x) =A*trig(x)$ and  $f(x)=trig(x) + k$  are quite simple: $g(x) = arctrig( \frac x A )$ and  $g(x)= arctrig(x-k)$.

Example TRIG.INV.2 (i) Suppose $f(x) =A*trig(x)$ [$A \ne 0$]  Verify that $g(x) = arctrig( \frac x A )$ is the inverse function for $f$.
ii) Suppose  $f(x)=trig(x) + k$. Verify that $g(x)= arctrig(x-k)$  is the inverse function for $f$.

You can use this next dynamic example to examine visually $g$ as the inverse of $f$ with the mapping diagrams and the graphs of $g$ and $f$.

Example TRIG.DINV.0.1 (Not Yet Done) Dynamic Visualization of The Inverse Function for General Trigonometric Functions: Graph and Mapping Diagrams