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: \(-1 \leq x \leq 1\)
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*}
