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 $x \in (-\frac{\pi}2+h,\frac{\pi}2+h)$.

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

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 $f$.