Inverse Trigonometric Functions

What is an Inverse Function

Given a function $f : A \rightarrow B$, with image $D \subseteq B$, the inverse function is $f^{-1} : D \rightarrow A$ such that $f(f^{-1}(x)) = f^{-1}(f(x)) = x$. In plain English, it “undoes” the given function. Note that in this context, $f^{-1} (x) \neq (f(x))^{-1} = \frac{1}{f(x)}$. It’s a bit of an abuse of notation, but usually it’s easy to determine the intended meaning through context.

An inverse function only exists if $f$ is injective/one-to-one. This is a fancy word that means the function maps every input to a unique output. Written mathematically, we have.

\[f(x) = f(y) \quad\implies\quad x = y\]

Let’s take the function $f(x) = x^2$ as an example. This function is not injective, since $f(1) = 1 = f(-1)$, but $1 \neq -1$. Thus, an inverse function does not exist. Why? What should be the value of $f^{-1}(1)$? It needs to be both $1$ and $-1$, which means it is not a function. In complex analysis, we call this a multi-function, but these are not allowed in real analysis.


Domain Restriction of a Function

Notice that all of the trigonometric functions are not injective, since they are all periodic. Then, how do we define an inverse function? There’s a trick. If a function is not injective, then we can just take the part of it that is. To take $f(x) = x^2$ as an example again. This function is injective on the interval $[0, \infty)$.

To define this in rigorous mathematic notation, let $f : A \rightarrow B$ and $C \subseteq A$. Then the function $f$ restricted to $C$ is denoted $f \rvert_C : C \rightarrow B$. In particular, $f \rvert_C (x) = f(x) \quad \forall x \in C$ and $f \rvert_C (x)$ is left undefined for $x \in A \backslash C$.

Therefore, for $f(x) = x^2$, $f \rvert_{[0, \infty)} (x)$ is injective, so its inverse function $f^{-1} \rvert_{[0, \infty)} (x)$ exists. This inverse function is precisely $\sqrt{x}$. Notice that $\sqrt{(x^2)} = (\sqrt{x})^2 = x$ for all $x \in [0, \infty)$.


Trigonometric Inverse Function

Using domain restriction on the trigonometric functions, we can make them injective and define the inverse functions as follows.

It’s a bit of an abuse of notation to write $\sin^{-1} (x)$ since there is actually no inverse function for $\sin (x)$. Thus, many sources opt to write $\arcsin (x)$ instead. You will see both.

Inverse Sine

\[\arcsin (x) \equiv (\sin \rvert_{[-\pi/2, \pi/2]})^{-1} (x)\]
   


Cosine

\[\arccos (x) \equiv (\cos \rvert_{[0, \pi]})^{-1} (x)\]
   


Tangent

\[\arctan (x) \equiv (\tan \rvert_{[-\pi/2, \pi/2]})^{-1} (x)\]
   


Cosecant

\[\arccsc (x) \equiv (\csc \rvert_{[-\pi/2, \pi/2]})^{-1} (x)\]
   


Secant

\[\arcsec (x) \equiv (\sec \rvert_{[0, \pi]})^{-1} (x)\]
    


Cotangent

\[\arccot (x) = (\cot \rvert_{[-\pi/2, \pi/2]})^{-1} (x)\]
   

Trigonometry Series