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\begin{document}
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\chapter{Graph of Trigonometric and Inverse Trigonometric Functions and Solution of Trigonometric Equations}
Trigonometric functions are usually defined either with the help of unit circle or right angled triangles. We also study their properties with a special emphasis on their graphs.
\section{Introduction}
The domain of trigonometric functions are the set of angles, rather than real numbers. We can however, make the domains of \\
trigonometric functions, subsets of real numbers, by defining them on the unit circle, that is a circle whose radius is 1.\\
Let $\theta$ be a central angle of the unit circle and $P(x,y)$ be the point as show in figure, then $r=OP=\sqrt{x^2+y^2}$ and the six \textbf{trigonometric functions} or \textbf{circular functions} are defined as:
\begin{center}
$\sin\theta=\dfrac{y}{1}=y$\\
$\cos\theta=\dfrac{x}{1}=x$\\
$\tan\theta=\dfrac{y}{x},$\quad$x\neq0$\\
$cosec\theta=\dfrac{1}{y},$ \quad$y\neq0$\\
$\sec\theta=\dfrac{1}{x},$ \quad$x\neq0$\\
$\cot\theta=\dfrac{x}{y},$ \quad$y\neq0.$
\end{center}
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\end{center}
Since any real number can represent the \\
length of exactly one arc on the unit circle. If $t$ is a positive number, we can find the arc of length $t$ by measuring a distance $t$ counterclockwise direction along an arc of the unit circle beginning at $C(1,0).$ So, we get arc $CP$ of length $t.$\\
If $t$ is negative number, we can find the arc of length $t,$ by measuring a distance $t$ clockwise direction an arc of the unit circle beginning at the point $C(1,0).$ In each case we get a unique point $P(x,y)$ that corresponds to a real number $t.$\\
We also know if $s$ is an arc which subtends an angle $\theta$ at the center of circle with radius $r,$ we have $s=r\theta$ where $\theta$ is in radians.\\
Let $s=t$ and $r=1$ then above equation reduce to $t=\theta$ or $\theta=t.$\\
Thus we obtain:
\begin{center}
$\sin\theta=\sin t,$\quad\quad
$\cos\theta=\cos t$\\
$\tan\theta=\tan t$\quad\quad
$cosec\theta=cosec t$\\
$\sec\theta=\sec t$\quad\quad
$\cot\theta=\cot t.$
\end{center}
Thus we can think of each trigonometric expression as being either a trigonometric function of an angle measured in radians or as a trigonometric function of a real number $t.$\\
Thus the trigonometric functions can be thought of as functions that have domains and ranges that are subsets of real numbers.
\section{Domain and Range of Trigonometric Functions}
\subsection*{Domain and Range of Sine and \\Cosine Functions}
We know from above discussion that \\
$\sin\theta=y$\quad\quad\quad$\cos\theta=x$\\
Domain of sine and cosine the set of real numbers $\mathbb{R}.$\
Since any $P(x,y)$ is on the unit circle \\
$\therefore$\quad\quad$-1\leq y\leq1$\quad\quad and\quad$-1\leq x\leq1.$\\
or\thinspace$-1\leq \sin\theta\leq1$\quad and\quad$-1\leq \cos\theta\leq1.$\\
Thus the range of sine and cosine functions are $[-1,1].$
\subsection*{Domain and Range of Tangent and Cotangent Functions}
We know that:\\
$\tan\theta=\dfrac{y}{x},$ \quad\quad$x\neq0.$\\
When $x=0,$ then terminal side can not coincide with $OY$ or $\acute{OY};$ in other words \\
$\theta\neq \pm\dfrac{\pi}{2},\pm\dfrac{3\pi}{2},\pm\dfrac{5\pi}{2},...$ or\\ $\theta\neq\dfrac{(2n+1)\pi}{2};n\in\mathbb{Z}.$\\
$\therefore$\quad Domain=$\mathbb{R}-\biggl\{t|t=(2n+1)\dfrac{\pi}{2};n\in\mathbb{Z}\biggr\}.$\\
and Range=$\mathbb{R}(\text{Set of real numbers}.)$\\
Since \quad$\cot\theta=\dfrac{x}{y}, \quad$ $y\neq0$\\
when $y\neq0$ then terminal side OP does not coincide with $OX$ or $\acute{OX},$ in other words\\
$\theta\neq0 \pm\pi,\pm2\pi,\pm3\pi,...$ or \\
$\theta\neq n\pi;\quad n\in\mathbb{Z}.$\\
$\therefore$\quad Domain=$\mathbb{R}-\biggl\{t|t=n\pi;n\in\mathbb{Z}\biggr\}.$\\
and Range=$\mathbb{R}(\text{Set of real numbers}).$
\subsection*{Domain and Range of Secant and Cosecant Functions}
Since \quad$cosec\theta=\dfrac{1}{y}, \quad y\neq0$\\
If $y\neq0,$ then as we seen in case of $\cot\theta, \quad$\\
$\theta\neq n\pi;\quad n\in\mathbb{Z}.$\\
$\therefore$\quad Domain=$\mathbb{R}-\{t|t=n\pi\quad;n\in\mathbb{Z}\}.$\\
Since $|y|=\sqrt{y^2}\leq\sqrt{x^2+y^2}=1$\\
$\Rightarrow$$\dfrac{1}{|y|}\leq1.$ Thus either $\dfrac{1}{y}\geq1$ or $\dfrac{1}{y}\leq-1$ that is $cosec\theta\geq1$ or $cosec\theta\leq-1$\\
That is $cosec\theta$ attains all values except those that lie between $-1$ and $1.$\\
Hence Range=$\mathbb{R}-\{t|t\in(-1,1)\}.$\\
Now we know that $\sec\theta=\dfrac{1}{x}\quad$ $x\neq0,$ thus as seen in case of $\tan\theta$\\
$\theta\neq(2n+1)\dfrac{\pi}{2}\quad n\in\mathbb{Z}.$\\
$\therefore$\quad Domain=$\mathbb{R}-\bigg\{t|t=(2n+1)\dfrac{\pi}{2};n\in\mathbb{Z}\biggr\}.$\\
Also $|x|=\sqrt{x^2}\leq\sqrt{x^2+y^2}=1$\\
$\Rightarrow$$|x|\leq1$\\
$\Rightarrow$$\dfrac{1}{|x|}\geq1$\\
Thus either $\dfrac{1}{x}\geq1$ or $\dfrac{1}{x}\leq-1$ that is \\
$\sec\theta\geq1$ or $\sec\theta\leq-1.$\\
That is $\sec\theta$ attains all values except those which lie between $-1$ and $1.$\\
$\therefore$\quad Range=$\mathbb{R}-\{t|t\in(-1,1)\}.$
\begin{ex}
Find the domain of the each of the following functions.\\
\textbf{(i)}\quad$\sec3x$\quad\textbf{(ii)} \quad$\tan\dfrac{1}{5}x$\quad\textbf{(iii)}\quad$cosec\dfrac{1}{2}x.$\\
\textbf{Solution:} \textbf{(i)} We know that the domain of $\sec t$ is:\quad $-\infty<t<\infty,$ \quad $t\neq(2n+1)\dfrac{\pi}{2},\thinspace n\in\mathbb{Z}.$\quad
If $t=3x,$ then the domain \\
$-\infty<3x<\infty,$ \quad $3x\neq(2n+1)\dfrac{\pi}{2},\thinspace n\in\mathbb{Z}$ \\
$\Rightarrow$$-\infty<x<\infty$ \thinspace$x\neq(2n+1)\dfrac{\pi}{6};\quad n\in\mathbb{Z}.$\\
$\therefore$ Domain of $\sec3x=\mathbb{R}-\bigg\{t|t=(2n+1)\dfrac{\pi}{6}\bigg\}.$\\
\\
\textbf{(ii)} Domain of $\tan t$ is: \\
$-\infty<t<\infty,$ \quad $t\neq(2n+1)\dfrac{\pi}{2},\quad n\in\mathbb{Z}.$\\ If $t=\dfrac{1}{5}x$ then dom of $\tan\dfrac{1}{5}x$ is\\
$-\infty<\dfrac{1}{5}x<\infty,$\quad$\dfrac{1}{5}x\neq(2n+1)\dfrac{\pi}{2}$\\
$\Rightarrow$ $-\infty<x<\infty$\quad $x\neq(2n+1)\dfrac{5\pi}{2}.$\\
$\therefore$ Domain of {\small $\tan\dfrac{1}{5}x=\mathbb{R}-\bigg\{x|x=5(2n+1)\dfrac{\pi}{2}\bigg\}.$}\\
\textbf{(iii)} Domain of $cosec t$ is $-\infty<t<\infty,$\quad$t\neq n\pi\quad n\in\mathbb{Z}.$\\
Let $t=\dfrac{1}{x}$ then domain of $cosec\dfrac{1}{2}x$ is:\\ $-\infty<\dfrac{1}{2}x<\infty,$\quad$\dfrac{1}{2}x\neq n\pi\quad n\in\mathbb{Z}.$\\
$\Rightarrow$$-\infty<x<\infty,$\quad$x\neq2n\pi\quad n\in\mathbb{Z}.$\\
$\therefore$\quad Domain of $cosec\dfrac{1}{2}x\\
=\mathbb{R}-\{x|x=2n\pi;n\in\mathbb{Z}\}.$\\
\end{ex}
\begin{ex}
Find the range of the each function.\\
\textbf{(i)}\quad $\cos3x$\quad\textbf{(ii)}\quad$3\tan2x$\quad\textbf{(iii)} $2cosec\dfrac{1}{3}x$\\
\textbf{Solution:} \textbf{(i)} We know that for all \\
$t\in \text{domain of}\cos t,$
$-1\leq\cos t\leq1.$\\
Let $t=3x$ then $-1\leq\cos3x\leq1.$\\
Hence range of $\cos3x$ is the closed interval $[-1,1].$\\
\textbf{(ii)} Since for all $t$ in domain of $\tan t,$\quad \\
$-\infty<\tan t<\infty.$\\
Let $t=2x$ then
$-\infty<\tan2x<\infty$\\
$\Rightarrow$$-\infty<3\tan2x<\infty.$\\
Thus the range of $3\tan2x$ is $\mathbb{R}.$\\
\textbf{(iii)} Since for all $t$ in domain of $cosec t$\\
$cosect\leq-1$ or $cosect\geq1.$\\
Let $t=\dfrac{1}{3}x$ then\\
$cosec\dfrac{1}{3}x\leq-1$ or $cosec\dfrac{1}{3}x\geq1$\\
Hence $2cosec\dfrac{1}{3}x\leq-2$ or $2cosec\dfrac{1}{3}x\geq2.$\\
Hence the range of \\
$2cosec\dfrac{1}{3}x=\mathbb{R}-\{p|p\in(-2,2)\}.$
\end{ex}
\section*{Even and Odd Trigonometric Functions}
\textbf{(a)} A function $f(x)$ is said to be even \\
if\quad $f(-x)=f(x).$\\
\textbf{(b)} A function $f(x)$ is said to be odd \\
if\quad $f(-x)=-f(x).$\\
\textbf{Note:} \textbf{(i)} Trigonometric functions are either even or odd.
The trigonometric functions $\sin\theta, cosec\theta, \tan\theta, \cot\theta$ are odd while $\cos\theta$ and $\sec\theta$ are even functions.\\
\textbf{(ii)} The sum of an odd function and an even function is neither even nor odd.
\section{Periodicity of Trigonometric Functions}
A function $f(x)$ is said to be periodic if there exists a smallest positive number $k$ such that\quad
$f(x+k)=f(x),$ \\
the smallest such positive number is called period of the function $f(x).$\\
\textbf{Note:} All six trigonometric functions are periodic functions, because they repeat their values after their periods. This behaviour of trigonometric functions is called periodicity.
\end{document}
章节名称太长。它与页码和章节中的部分混淆。如何解决这个问题！
答案1
分段命令（例如\chapter）接受可选参数。如果您给出可选参数，那么这就是目录和标题中显示的内容：因此，使用\chapter[short]{long title}，您会在目录和标题中获得“短”，但章节标题为“长标题”。