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Mathematics LibreTexts

8.2: The Tangent Function

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    22686
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    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 \(T(x)=t(x-g(x) \pi)\).

    Definition

    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 \(\tan (x) .\)

    Proposition \(\PageIndex{1}\)

    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.

    Definition

    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 \(x \in E .\)

    Proposition \(\PageIndex{2}\)

    The tangent function has period \(\pi .\)

    Proof

    The result follows immediately from our definitions. \(\quad\) Q.E.D.

    Proposition \(\PageIndex{3}\)

    (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.

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