8.1: The Arctangent Function
- Page ID
- 22685
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)For any \(x \in \mathbb{R},\) we call
\[\arctan (x)=\int_{0}^{x} \frac{1}{1+t^{2}} d t\]
the arctangent of \(x .\)
The arctangent function is differentiable at every \(x \in \mathbb{R} .\)
Moreover, if \(f(x)=\arctan (x),\) then
\[f^{\prime}(x)=\frac{1}{1+x^{2}}.\]
- Proof
-
The result follows immediately from Theorem \(7.5 .4 .\) \(\quad\) Q.E.D.
The arctangent is increasing on \(\mathbb{R}\).
- Proof
-
The result follows immediately from the previous proposition and the fact that
\[\frac{1}{1+x^{2}}>0\]
for every \(x \in \mathbb{R}\). \(\quad\) Q.E.D.
\(\pi=2 \lim _{x \rightarrow+\infty} \arctan (x)=2 \int_{0}^{+\infty} \frac{1}{1+t^{2}} d t\)
Note that \(0<\pi<4\) by Example \(7.7 .1 .\)
The following proposition says that the arctangent function is an odd function.
For any \(x \in \mathbb{R},\) arctan \((x)=-\arctan (-x)\).
- Proof
-
Using the substitution \(t=-u,\) we have
\[\begin{aligned} \arctan (x)=\int_{0}^{x} \frac{1}{1+t^{2}} d t=-\int_{0}^{-x} \frac{1}{1+u^{2}} d u=-\arctan (-x). & \end{aligned}\]
Q.E.D.
It now follows that
\[\lim _{x \rightarrow-\infty} \arctan (x)=-\lim _{x \rightarrow-\infty} \arctan (-x)=-\frac{\pi}{2}.\]
Hence the range of the arctangent function is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
If \(x>0,\) then
\[\arctan (x)+\arctan \left(\frac{1}{x}\right)=\frac{\pi}{2}.\]
- Proof
-
Using the substitution \(t=\frac{1}{u},\) we have
\[\begin{aligned} \arctan \left(\frac{1}{x}\right) &=\int_{0}^{\frac{1}{x}} \frac{1}{1+t^{2}} d t \\ &=\int_{+\infty}^{x} \frac{1}{1+\frac{1}{u^{2}}}\left(-\frac{1}{u^{2}}\right) d u \\ &=-\int_{+\infty}^{x} \frac{1}{1+u^{2}} d u \\ &=\int_{x}^{+\infty} \frac{1}{1+u^{2}} d u \\ &=\frac{\pi}{2}-\int_{0}^{x} \frac{1}{1+u^{2}} d u \\ &=\frac{\pi}{2}-\arctan (x). \end{aligned}\]
Q.E.D.
If \(x<0\), then
\[\arctan (x)+\arctan \left(\frac{1}{x}\right)=-\frac{\pi}{2}.\]
- Proof
-
The result follows immediately from the preceding proposition and the fact that arctangent is an odd function.
Show that arctan \((1)=\frac{\pi}{4}\) and \(\arctan (-1)=-\frac{\pi}{4}\).