Precalculus: Lecture Notes

Section 4.3 More Trigonometric Functions

What has to be overcome is not difficulty of the intellect but of the will. — Ludwig Wittgenstein, Austrian-born philosopher, 1889 – 1951

Solution

Recall that, by definition, for \(x\in \mathbb{R}\text{,}\) the number \(\cos x\) is the first coordinate of the point on the unit circle that corresponds to the number \(x\text{.}\) See Figure 4.41.

It follows that \(\cos x=0\) means that the point on the unit circle that corresponds to the number \(x\) must be the point \((0,1)\) or the point \((0,-1)\text{.}\) By our construction, i.e. by the way how we associate real numbers and points on the unit circle, we conclude

\begin{equation*} \cos x =0 \Leftrightarrow x\in\left\{ \ldots, -\frac{3\pi}{2}, -\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}, \ldots\right\}. \end{equation*}

Observe that

\begin{equation*} \left\{ \ldots, -\frac{3\pi}{2}, -\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}, \ldots\right\}=\left\{x\in \mathbb{R}:x=\frac{\pi}{2}+k\pi \text{ for some } k\in \mathbb{Z}\right\}. \end{equation*}

We write

\begin{equation*} \cos x = 0 \Leftrightarrow x=\frac{\pi}{2}+k\pi =(2k+1)\cdot\frac{\pi}{2}, \ k\in \mathbb{Z}. \end{equation*}

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Example 4.3.3. Tangent ad slope.

Consider the line $p$ is given by the equation $y=mx\text{,}$ $m\in (0,\infty)\text{.}$

Let $\alpha$ be the smaller of the two positively oriented angles with the vertex at the origin, the positive part of the $x$ axis being the initial ray, and the corresponding part of the line $p$ being the terminal ray. See Figure 4.43.

Find $\cos \alpha\text{,}$ $\sin \alpha\text{,}$ and $\tan \alpha\text{.}$

Solution

To find the coordinates of the point in the first quadrant in which the line \(y=mx\) and the unit circle intersect, we solve the system of equations

\begin{equation*} y = mx \text{ and } x^2+y^2=1, \end{equation*}

keeping in mind that \(x\gt 0\) and \(y\gt 0\text{.}\)

It follows that

\begin{equation*} x^2+(mx)^2=1\Leftrightarrow (1+m^2)x^2=1\Leftrightarrow x^2=\frac{1}{1+m^2}\Rightarrow x=\frac{1}{\sqrt{1+m^2}} \end{equation*}

and, consequently,

\begin{equation*} y=mx=\frac{m}{\sqrt{1+m^2}}. \end{equation*}

By definition

\begin{equation*} \sin \alpha= \frac{m}{\sqrt{1+m^2}}, \ \cos\alpha =\frac{1}{\sqrt{1+m^2}},\text{ and }\tan\alpha =\frac{\frac{m}{\sqrt{1+m^2}}}{\frac{1}{\sqrt{1+m^2}}}=m. \end{equation*}

1. Cotangent: The cotangent function is defined as

   $$\begin{equation*} \cot x=\frac{1}{\tan x}=\frac{\cos x}{\sin x}. \end{equation*}$$

   See Figure 4.45.

   

   All you need to know about the function $f(x)=\cot x$ … for now:

   $$\begin{equation*} \begin{array}{c|l} \amp f(x)=\cot x\\ \hline \text{Domain}\amp \{x\in \mathbb{R}:x\not=k\pi, k\in\mathbb{Z}\}=\cup_{k\in \mathbb{Z}}(k\pi, (k+1)\pi)\\ \hline \text{Range}\amp \mathbb{R}\\ \hline \text{Zeros}\amp x=(2k+1)\cdot \frac{\pi}{2}, k\in \mathbb{Z}\\ \hline \cot x\gt 0\amp \cup_{k\in \mathbb{Z}}\left(k\pi,k\pi+\frac{\pi}{2}\right)\\ \hline \cot x\lt 0\amp \cup_{k\in \mathbb{Z}}\left(k\pi+\frac{\pi}{2}, (k+1)\pi\right)\\ \hline \text{Vertical}\amp x=k\pi, \ k\in \mathbb{Z}\\ \text{Asymptotes}\amp \\ \hline \text{Parity}\amp \text{Odd}\\ \hline \text{Period}\amp T=\pi\\ \hline \text{Increasing}\amp \text{Never}\\ \hline \text{Decreasing}\amp \text{On each of the intervals}\\ \amp (k\pi,(k+1)\pi), k\in \mathbb{Z}\\ \end{array} \end{equation*}$$

2. Secant and Cosecant: The secant and cosecant functions are defined as

   $$\begin{equation*} g(x)=\sec x=\frac{1}{\cos x} \text{ and } h(x)=\csc x=\frac{1}{\sin x} \end{equation*}$$

   See Figure 4.46 and Figure 4.47.

   

   

   All you need to know about functions $g(x)=\sec x$ and $h(x)=\csc x$… for now:

   $$\begin{equation*} \begin{array}{c|l|l} \amp g(x)=\sec x\amp h(x)=\csc x\\ \hline \text{Domain}\amp \{x\in\mathbb{R}:x\not=(2k+1)\frac{\pi}{2}\}\amp \{x\in\mathbb{R}:x\not=k\pi\}\\ \hline \text{Range}\amp (-\infty)-1]\cup [1,\infty)\amp (-\infty)-1]\cup [1,\infty)\\ \hline \text{Zeros}\amp \text{No zeros}\amp \text{No zeros}\\ \hline \text{Vertical}\amp x=(2k+1)\cdot \frac{\pi}{2}, k\in \mathbb{Z}\amp x=k\pi, k\in \mathbb{Z}\\ \text{Asymptotes}\amp \amp \\ \hline y=1\amp x=2k\pi\amp x=2k\pi +\frac{\pi}{2}\\ \hline y=-1\amp x=(2k+1)\pi\amp x=(2k+1)\pi+\frac{\pi}{2}\\ \hline \text{Parity}\amp \text{Even}\amp \text{Odd}\\ \hline \text{Period}\amp T=2\pi\amp T=2\pi\\ \end{array} \end{equation*}$$