More Trigonometric Functions

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

Reminder: Solve

\begin{equation*} \cos x=0. \end{equation*}

Figure 4.40.

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.

Figure 4.41. From \(x\) to \(\cos x\) and \(\sin x\text{.}\)
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*}

Definition.The tangent function is defined as

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

Example 4.3.1. Cosines, sines, and tangents.

Complete the table:

\begin{equation*} \begin{array}{c|c|c|c} x\amp \sin x\amp \cos x\amp \tan x\\ \hline 0\amp 0\amp 1\amp 0\\ \hline \frac{\pi}{6}\amp \frac{1}{2}\amp \frac{\sqrt{3}}{2}\amp \\ \hline \frac{\pi}{4}\amp \amp \amp \\ \hline \frac{\pi}{3}\amp \amp\amp \\ \hline \frac{\pi}{2}\amp \amp \amp \\ \hline \frac{2\pi}{3}\amp \amp \amp\\ \hline \frac{3\pi}{4}\amp \amp \amp \\ \hline \frac{5\pi}{6}\amp \amp \amp\\ \hline \pi\amp \amp \amp \\ \hline \frac{3\pi}{2}\amp \amp \amp \\ \hline 2\pi\amp \amp \amp \\ \end{array} \end{equation*}

Solution

Recall that, by definition, \(\tan x=\frac{\sin x}{\cos x}\text{.}\)

\begin{equation*} \begin{array}{c|c|c|c} x\amp \sin x\amp \cos x\amp \tan x\\ \hline 0\amp 0\amp 1\amp 0\\ \hline \frac{\pi}{6}\amp \frac{1}{2}\amp \frac{\sqrt{3}}{2}\amp \frac{\sqrt{3}}{3}\\ \hline \frac{\pi}{4}\amp \frac{\sqrt{2}}{2}\amp \frac{\sqrt{2}}{2}\amp 1\\ \hline \frac{\pi}{3}\amp \frac{\sqrt{3}}{2}\amp \frac{1}{2}\amp \sqrt{3}\\ \hline \frac{\pi}{2}\amp 1\amp 0\amp \text{Not defined}\\ \hline \frac{2\pi}{3}\amp \frac{\sqrt{3}}{2}\amp -\frac{1}{2}\amp -\sqrt{3}\\ \hline \frac{3\pi}{4}\amp \frac{\sqrt{2}}{2}\amp -\frac{\sqrt{2}}{2}\amp -1\\ \hline \frac{5\pi}{6}\amp \frac{1}{2}\amp -\frac{\sqrt{3}}{2}\amp -\frac{\sqrt{3}}{3}\\ \hline \pi\amp 0\amp -1\amp 0\\ \hline \frac{3\pi}{2}\amp -1\amp 0\amp \text{Not defined}\\ \hline 2\pi\amp 0\amp 1\amp 0\\ \end{array} \end{equation*}

Function \(f(x)=\tan x\text{:}\) See Figure 4.42

Figure 4.42. \(x\mapsto y=\tan x\text{.}\)

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

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

Example 4.3.2. Given \(\cos \theta\) evaluate \(\tan\theta\).

Suppose that the real number \(\theta\) is such that

\begin{equation*} \frac{\pi}{2}\lt \theta\lt \pi \text{ and }\cos \theta =-\frac{2}{3}. \end{equation*}

Find the values of \(\sin \theta\) and \(\tan \theta\text{.}\)

Solution

From the fact that \(\sin ^2\theta+\cos^2\theta=1\) it follows that

\begin{equation*} \sin ^2\theta+\left(-\frac{2}{3}\right)^2=1\Leftrightarrow \sin ^2\theta+\frac{4}{9}=1\Leftrightarrow \sin ^2\theta=\frac{5}9. \end{equation*}

Since \(\frac{\pi}{2}\lt \theta\lt \pi \) we conclude that

\begin{equation*} \sin \theta =\frac{\sqrt{5}}{3} \text{ and } \tan \theta=\frac{\sin\theta}{\cos \theta}=\frac{\frac{\sqrt{5}}{3}}{-\frac{2}{3}}=-\frac{\sqrt{5}}{2}. \end{equation*}

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*}

IMPORTANT Observation: The slope of the line \(y=mx\) equals the tangent of a positively oriented angle between the line and the positive part of the \(x\)–axis. See Figure 4.44.

Figure 4.44. Observe: \(m=\tan \alpha\text{.}\)

More trigonometric functions.

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.

Figure 4.45. \(\cot x =\frac{\cos x}{\sin x}\text{.}\)
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*}
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.

Figure 4.46. \(\sec x=\frac{1}{\cos x}\text{.}\)

Figure 4.47. \(\csc x=\frac{1}{\sin x}\text{.}\)

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*}