TeachingCalculus – Inverse functions – Inverse trigonometric functions
Definition: \(\arcsin(x)=y\quad\Longleftrightarrow\quad\sin(y)=x\quad\text{with}\quad-\frac{1}{2}\pi\leq y\leq\frac{1}{2}\pi\).
Definition: \(\arccos(x)=y\quad\Longleftrightarrow\quad\cos(y)=x\quad\text{with}\quad0\leq y\leq\pi\).
Definition: \(\arctan(x)=y\quad\Longleftrightarrow\quad\tan(y)=x\quad\text{with}\quad-\frac{1}{2}\pi < y < \frac{1}{2}\pi\).
Theorem: \(\sin(\arcsin(x))=x\) for all \(x\in[-1,1]\) and \(\arcsin(\sin(y))=y\) for all \(y\in[-\frac{1}{2}\pi,\frac{1}{2}\pi]\).
Theorem: \(\cos(\arccos(x))=x\) for all \(x\in[-1,1]\) and \(\arccos(\cos(y))=y\) for all \(y\in[0,\pi]\).
Theorem: \(\tan(\arctan(x))=x\) for all \(x\in\mathbb{R}\) and \(\arctan(\tan(y))=y\) for all \(y\in(-\frac{1}{2}\pi,\frac{1}{2}\pi)\).
Note that the first formula in each theorem above holds for all \(x\) in the domain. The second formula only holds for \(y\) in a certain interval.
Examples:
\[\arcsin(\sin(\tfrac{13}{8}\pi))=\arcsin(\sin(-\tfrac{3}{8}\pi))=-\tfrac{3}{8}\pi,\quad\arccos(\cos(\tfrac{15}{7}\pi))=\arccos(\cos(\tfrac{1}{7}\pi))=\tfrac{1}{7}\pi,\] \[\arctan(\tan(\tfrac{12}{5}\pi))=\arctan(\tan(\tfrac{2}{5}\pi))=\tfrac{2}{5}\pi\quad\text{and}\quad\arctan(\tan(\tfrac{13}{9}\pi))=\arctan(\tan(\tfrac{4}{9}\pi))=\tfrac{4}{9}\pi.\]Here we used the fact that \(\sin(x+2\pi)=\sin(x)\), \(\cos(x+2\pi)=\cos(x)\) and \(\tan(x+\pi)=\tan(x)\) to reach the correct interval. However, sometimes we also need that \(\sin(\pi-x)=\sin(x)\) and \(\cos(2\pi-x)=\cos(x)\) to reach the correct interval. For instance:
\[\arcsin(\sin(\tfrac{5}{8}\pi))=\arcsin(\sin(\tfrac{3}{8}\pi))=\tfrac{3}{8}\pi\quad\text{and}\quad\arccos(\cos(\tfrac{8}{7}\pi))=\arccos(\cos(\tfrac{6}{7}\pi))=\tfrac{6}{7}\pi.\]Using \(\sin(\frac{1}{2}\pi-x)=\cos(x)\) and \(\cos(\frac{1}{2}\pi-x)=\sin(x)\) we also have results such as
\[\arcsin(\cos(\tfrac{3}{8}\pi))=\arcsin(\sin(\tfrac{1}{8}\pi))=\tfrac{1}{8}\pi\quad\text{and}\quad\arccos(\sin(\tfrac{2}{7}\pi))=\arccos(\cos(\tfrac{3}{14}\pi))=\tfrac{3}{14}\pi.\]Examples of the other way around are:
\[\cos(\arcsin(\tfrac{3}{5}))=\tfrac{4}{5}\quad\text{and}\quad\sin(\arccos(\tfrac{5}{13}))=\tfrac{12}{13}.\]To see this, we usually take a rectangular triangle; in the first example with rectangular sides of length \(3\) and \(4\) and hypotenuse of
length \(5\) and in the second example with rectangular sides of length \(5\) and \(12\) and hypotenuse of length \(13\). Note that \(3^2+4^2=5^2\)
and \(5^2+(12)^2=(13)^2\).
Similarly we have:
Furthermore, we have:
\[\sin(\arctan(\tfrac{3}{4}))=\tfrac{3}{5},\quad\cos(\arctan(\tfrac{3}{4}))=\tfrac{4}{5},\quad\sin(\arctan(\tfrac{5}{12})=\tfrac{5}{13}\quad\text{and}\quad\cos(\arctan(\tfrac{5}{12}))=\tfrac{12}{13}.\]More general:
Stewart §1.6, Example 13: Simplify the expression \(\cos(\arctan(x))\).
In this picture we have
that \(\arctan(x)=y\) or \(\tan(y)=x\) with \(0 < y < \frac{1}{2}\pi\).
So, in that case we conclude that \(\cos(\arctan(x))=\cos(y)=\displaystyle\frac{1}{\sqrt{1+x^2}}\).
How to prove the formula in general? For \(-\frac{1}{2}\pi < y < 0\) we might use that \(y=\arctan(x)\) implies that \(-y=\arctan(-x)\) with \(0 < -y < \frac{1}{2}\pi\). Then we can use the same picture with \(-x\) instead of \(x\) and \(-y\) instead of \(y\). Since \((-x)^2\) is the same as \(x^2\) this leads to the same result. Finally, the formula trivially holds if \(y=0\) since then \(x=0\) as well.
Alternatively, we can prove the result directly using the definition. Start by writing
\[\arctan(x)=y\quad\Longleftrightarrow\quad\tan(y)=x\quad\text{with}\quad -\tfrac{1}{2}\pi < y < \tfrac{1}{2}\pi.\]Now we need to find a formula for \(\cos(y)\). Is it possible to write such a formula in terms of \(\tan(y)=x\)? Note that
\[\tan^2(y)=\frac{\sin^2(y)}{\cos^2(y)}=\frac{1-\cos^2(y)}{\cos^2(y)}\quad\Longrightarrow\quad\frac{1}{\cos^2(y)}=1+\tan^2(y) \quad\text{or equivalently}\quad\cos^2(y)=\frac{1}{1+\tan^2(y)}=\frac{1}{1+x^2}.\]This implies that \(\cos(y)=\pm\displaystyle\frac{1}{\sqrt{1+x^2}}\). Note that for \(-\frac{1}{2}\pi < y < \frac{1}{2}\pi\) we have that \(\cos(y)>0\). So we conclude that \(\cos(y)=\displaystyle\frac{1}{\sqrt{1+x^2}}\).
Finding angles:
The dot product of two vectors has the property that
\[\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos(\theta),\quad 0\leq\theta\leq\pi.\]Here \(\theta\) denotes the angle between the vectors \(\mathbf{a}\) and \(\mathbf{b}\). There we noticed that there is only one value for \(\theta\in[0,\pi]\) such that
\[\cos(\theta)=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|},\]provided that \(\mathbf{a}\) and \(\mathbf{b}\) are nonzero vectors. Now we may write
\[\theta=\arccos\left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right).\]Note that the interval \([0,\pi]\) is exactly the range of the function arccos.
Last modified on March 1, 2021
