Inverse Trigonometric Functions

Section B.3 Inverse Trigonometric Functions

In order to construct inverse trigonometric functions we first have to restrict their domains so as to make them one-to-one (or injective). We do this as shown below

\begin{equation*} \sin\theta \end{equation*}

\begin{equation*} \cos \theta \end{equation*}

\begin{equation*} \tan \theta \end{equation*}

Domain: $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$

Domain: $0 \leq \theta \leq \pi$

Domain: $-\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}$

Range: $-1 \leq \sin \theta \leq 1$

Range: $-1 \leq \cos \theta \leq 1$

Range: all real numbers

\begin{equation*} \arcsin x \end{equation*}

\begin{equation*} \arccos x \end{equation*}

\begin{equation*} \arctan x \end{equation*}

Domain: $-1 \leq x \leq 1$

Domain: all real numbers

Range: $-\frac{\pi}{2} \leq \arcsin x \leq \frac{\pi}{2}$

Range: $0 \leq \arccos x \leq \pi$

Range: $-\frac{\pi}{2} \lt \arctan x \lt \frac{\pi}{2}$

Since these functions are inverses of each other we have

\begin{align*} \arcsin(\sin \theta) &= \theta & -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\\ \arccos(\cos \theta) &= \theta & 0 \leq \theta \leq \pi\\ \arctan(\tan \theta) &= \theta & -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \end{align*}

and also

\begin{align*} \sin(\arcsin x) &= x & -1 \leq x \leq 1\\ \cos(\arccos x) &= x & -1 \leq x \leq 1\\ \tan(\arctan x) &= x & \text{any real $x$} \end{align*}

We can read other combinations of trig functions and their inverses, like, for example, $\cos(\arcsin x)\text{,}$ off of triangles like

We have chosen the hypotenuse and opposite sides of the triangle to be of length 1 and $x\text{,}$ respectively, so that $\sin(\theta)=x\text{.}$ That is, $\theta = \arcsin x\text{.}$ We can then read off of the triangle that

\begin{align*} \cos(\arcsin x) &= \cos(\theta) = \sqrt{1-x^2} \end{align*}

We can reach the same conclusion using trig identities, as follows.

Write $\arcsin x=\theta\text{.}$ We know that $\sin(\theta)=x$ and we wish to compute $\cos(\theta)\text{.}$ So we just need to express $\cos(\theta)$ in terms of $\sin(\theta)\text{.}$
To do this we make use of one of the Pythagorean identities
\begin{align*} \sin^2\theta + \cos^2\theta &=1\\ \cos\theta &= \pm \sqrt{1-\sin^2\theta} \end{align*}
Thus
\begin{gather*} \cos(\arcsin x) = \cos\theta = \pm\sqrt{1-\sin^2\theta} \end{gather*}
To determine which branch we should use we need to consider the domain and range of $\arcsin x\text{:}$
\begin{align*} \text{Domain: } -1 \leq x \leq 1 && \text{Range: } -\frac{\pi}{2} \leq \arcsin x \leq \frac{\pi}{2} \end{align*}
Thus we are applying cosine to an angle that always lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}\text{.}$ Cosine is non-negative on this range. Hence we should take the positive branch and
\begin{align*} \cos(\arcsin x) &= \sqrt{1-\sin^2\theta}= \sqrt{1-\sin^2(\arcsin x)}\\ &= \sqrt{1-x^2} \end{align*}

In a very similar way we can simplify $\tan(\arccos x)\text{.}$

Write $\arccos x=\theta\text{,}$ and then
\begin{align*} \tan( \arccos x) &= \tan \theta = \frac{\sin\theta}{\cos \theta} \end{align*}
Now the denominator is easy since $\cos \theta = \cos \arccos x = x\text{.}$
The numerator is almost the same as the previous computation.
\begin{align*} \sin\theta &= \pm \sqrt{1-\cos^2\theta}\\ &= \pm \sqrt{1-x^2} \end{align*}
To determine which branch we again consider domains and and ranges:
\begin{align*} \text{Domain: } -1 \leq x \leq 1 && \text{Range: } 0 \leq \arccos x \leq \pi \end{align*}
Thus we are applying sine to an angle that always lies between $0$ and $\pi\text{.}$ Sine is non-negative on this range and so we take the positive branch.
Putting everything back together gives
\begin{align*} \tan(\arccos x) &= \frac{\sqrt{1-x^2}}{x} \end{align*}

Completing the 9 possibilities gives:

\begin{align*} \sin( \arcsin x ) &= x & \sin( \arccos x ) &= \sqrt{1-x^2} & \sin( \arctan x ) &= \frac{x}{\sqrt{1+x^2}}\\ \cos( \arcsin x ) &= \sqrt{1-x^2} & \cos( \arccos x ) &= x & \cos( \arctan x ) &= \frac{1}{\sqrt{1+x^2}}\\ \tan( \arcsin x ) &= \frac{x}{\sqrt{1-x^2}} & \tan( \arccos x ) &= \frac{\sqrt{1-x^2}}{x} & \tan( \arctan x ) &= x \end{align*}

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