8.2: The Tangent Function
- Page ID
- 22686
Let
\[A=\left\{\frac{\pi}{2}+n \pi: n \in \mathbb{Z}\right\}\]
and \(D=\mathbb{R} \backslash A\). Let
\[t:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathbb{R}\]
be the inverse of the arctangent function. Note that \(t\) is increasing and differentiable on \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .\) We may extend \(t\) to a function on \(D\) as follows: For any \(x \in D,\) let
\[g(x)=\sup \left\{n: n \in \mathbb{Z},-\frac{\pi}{2}+n \pi<x\right\}\]
and define \(T(x)=t(x-g(x) \pi)\).
With the notation of the above discussion, for any \(x \in D,\) we call the value \(T(x)\) the tangent of \(x,\) which we denote \(\tan (x) .\)
The tangent function has domain \(D\) (as defined above), range \(\mathbb{R},\) and is differentiable at every point \(x \in D .\) Moreover, the tangent function is increasing on each interval of the form
\[\left(-\frac{\pi}{2}+n \pi, \frac{\pi}{2}+n \pi\right),\]
\(n \in \mathbb{Z},\) with
\[\tan \left(\left(\frac{\pi}{2}+n \pi\right)+\right)=-\infty\]
and
\[\tan \left(\left(\frac{\pi}{2}+n \pi\right)-\right)=+\infty .\]
- Proof
-
These results follow immediately from our definitions. \(\quad\) Q.E.D.
Let \(E \subset \mathbb{R}\). We say a function \(f: E \rightarrow \mathbb{R}\) is periodic if there exists a real number \(p>0\) such that, for each \(x \in E, x+p \in E\) and \(f(x+p)=f(x) .\) We say \(p\) is the period of a periodic function \(f\) if \(p\) is the smallest positive number for which \(f(x+p)=f(x)\) for all \(x \in E .\)
The tangent function has period \(\pi .\)
- Proof
-
The result follows immediately from our definitions. \(\quad\) Q.E.D.
(Addition formula for tangent)
For any \(x, y \in D\) with \(x+y \in D\),
\[\tan (x+y)=\frac{\tan (x)+\tan (y)}{1-\tan (x) \tan (y)}.\]
- Proof
-
Suppose \(y_{1}, y_{2} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) with \(y_{1}+y_{2} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .\) Let \(x_{1}=\tan \left(y_{1}\right)\) and \(x_{2}=\tan \left(y_{2}\right) .\) Note that if \(x_{1}>0,\) then \(x_{1} x_{2} \geq 1\) would imply that
\[x_{2} \geq \frac{1}{x_{1}},\]
which in turn implies that
\[\begin{aligned} y_{1}+y_{2} &=\arctan \left(x_{1}\right)+\arctan \left(x_{2}\right) \\ & \geq \arctan \left(x_{1}\right)+\arctan \left(\frac{1}{x_{1}}\right) \\ &=\frac{\pi}{2}, \end{aligned}\]
contrary to our assumptions. Similarly, if \(x_{1}<0,\) then \(x_{1} x_{2} \geq 1\) would imply that
\[x_{2} \leq \frac{1}{x_{1}},\]
which in turn implies that
\[\begin{aligned} y_{1}+y_{2} &=\arctan \left(x_{1}\right)+\arctan \left(x_{2}\right) \\ & \leq \arctan \left(x_{1}\right)+\arctan \left(\frac{1}{x_{1}}\right) \\ &=-\frac{\pi}{2}, \end{aligned}\]
contrary to our assumptions. Thus we must have \(x_{1} x_{2}<1 .\) Moreover, suppose \(u\) is a number between \(-x_{1}\) and \(x_{2} .\) If \(x_{1}>0,\) then
\[x_{2}<\frac{1}{x_{1}},\]
and so
\[u<\frac{1}{x_{1}}.\]
If \(x_{1}<0,\) then
\[x_{2}>\frac{1}{x_{1}},\]
and so
\[u>\frac{1}{x_{1}}.\]
Now let
\[x=\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}.\]
We want to show that
\[\arctan (x)=\arctan \left(x_{1}\right)+\arctan \left(x_{2}\right),\]
which will imply that
\[\frac{\tan \left(y_{1}\right)+\tan \left(y_{2}\right)}{1-\tan \left(y_{1}\right) \tan \left(y_{2}\right)}=\tan \left(y_{1}+y_{2}\right).\]
We need to compute
\[\arctan (x)=\arctan \left(\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}\right)=\int_{0}^{\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}} \frac{1}{1+t^{2}} d t.\]
Let
\[t=\varphi(u)=\frac{x_{1}+u}{1-x_{1} u},\]
where \(u\) varies between \(-x_{1},\) where \(t=0,\) and \(x_{2},\) where \(t=x .\) Now
\[\varphi^{\prime}(u)=\frac{\left(1-x_{1} u\right)-\left(x_{1}+u\right)\left(-x_{1}\right)}{\left(1-x_{1} u\right)^{2}}=\frac{1+x_{1}^{2}}{\left(1-x_{1} u\right)^{2}},\]
which is always positive, thus showing that \(\varphi\) is an increasing function, and
\[\begin{aligned} \frac{1}{1+t^{2}} &=\frac{1}{1+\left(\frac{x_{1}+u}{1-x_{1} u}\right)^{2}} \\ &=\frac{\left(1-x_{1} u\right)^{2}}{\left(1-x_{1} u\right)^{2}+\left(x_{1}+u\right)^{2}} \\ &=\frac{\left(1-x_{1} u\right)^{2}}{\left(1+x_{1}^{2}\right)\left(1+u^{2}\right)}. \end{aligned}\]
Hence
\[\begin{aligned} \arctan (x) &=\int_{-x_{1}}^{x_{2}} \frac{1}{1+u^{2}} d u \\ &=\int_{-x_{1}}^{0} \frac{1}{1+u^{2}} d u+\int_{0}^{x_{2}} \frac{1}{1+u^{2}} d u \\ &=-\int_{0}^{-x_{1}} \frac{1}{1+u^{2}} d u+\arctan \left(x_{2}\right) \\ &=-\arctan \left(-x_{1}\right)+\arctan \left(x_{2}\right) \\ &=\arctan \left(x_{1}\right)+\arctan \left(x_{2}\right). \end{aligned}\]
Now suppose \(y_{1}, y_{2} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) with \(y_{1}+y_{2}>\frac{\pi}{2} .\) Then \(y_{1}+y_{2} \in\left(\frac{\pi}{2}, \pi\right)\),\(x_{1}>0, x_{2}>0,\) and
\[x_{2}>\frac{1}{x_{1}}.\]
With \(u\) and \(x\) as above, note then that as \(u\) increases from \(-x_{1}\) to \(\frac{1}{x_{1}}, t\) increases from 0 to \(+\infty,\) and as \(u\) increases from \(\frac{1}{x_{1}}\) to \(x_{2}, t\) increases from \(-\infty\) to \(x .\)
Hence we have
\[\begin{aligned} \arctan (x)+\pi &=\int_{0}^{x} \frac{1}{1+t^{2}} d t+\int_{-\infty}^{0} \frac{1}{1+t^{2}} d t+\int_{0}^{+\infty} \frac{1}{1+t^{2}} d t \\ &=\int_{-\infty}^{x} \frac{1}{1+t^{2}} d t+\int_{0}^{+\infty} \frac{1}{1+t^{2}} d t \\ &=\int_{\frac{1}{x_{1}}}^{x_{2}} \frac{1}{1+u^{2}} d u+\int_{-x_{1}}^{\frac{1}{x_{1}}} \frac{1}{1+u^{2}} d u \\ &=\int_{-x_{1}}^{x_{2}} \frac{1}{1+u^{2}} d u \\ &=\arctan \left(x_{2}\right)-\arctan \left(-x_{1}\right) \\ &=\arctan \left(x_{2}\right)+\arctan \left(x_{1}\right). \end{aligned}\]
Hence
\[\begin{aligned} \tan \left(y_{1}+y_{2}\right) &=\tan \left(y_{1}+y_{2}-\pi\right) \\ &=\tan (\arctan (x)) \\ &=\frac{x_{1}+x_{2}}{1-x_{1} x_{2}} \\ &=\frac{\tan \left(y_{1}\right)+\tan \left(y_{2}\right)}{1-\tan \left(y_{1}\right) \tan \left(y_{2}\right)}. \end{aligned}\]
The case when \(x_{1}<0\) may be handled similarly; it then follows that the addition formula holds for all \(y_{1}, y_{2} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .\) The case for arbitrary \(y_{1}, y_{2} \in D\) with \(y_{1}+y_{2} \in D\) then follows from the periodicity of the tangent function. \(\quad\) Q.E.D.