Chapter 4.4, Problem 12E

(a)

To determine

To find: The exact value of six trigonometric function of each angle $\theta$ .

Answer to Problem 12E

The exact value of six trigonometric function of each angle $\theta$ are $\cos \theta =-\frac{12}{13},\sin \theta =-\frac{5}{13},\cot \theta =\frac{12}{5},\tan \theta =\frac{5}{12},$$\sec \theta =-\frac{13}{12},\csc \theta =-\frac{13}{5}$ .

Explanation of Solution

Given information:

The given coordinates are shown in figure (1).

  

Figure (1)

Calculation:

Use $\theta$ be an angle in standard position with $\left(x,y\right)$ from figure (1).

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ \ne 0\end{array}$

The trigonometric functions of any angle.

  $\begin{array}{l}\cos \theta =\frac{x}{r}\\ \sin \theta =\frac{y}{r}\\ \cot \theta =\frac{x}{y}\\ \tan \theta =\frac{y}{x}\end{array}$

  $\begin{array}{l}\sec \theta =\frac{r}{x}\\ \csc \theta =\frac{r}{y}\end{array}$

Calculate the value of $r$ and then calculate the value of the trigonometric functions.

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ =\sqrt{{\left(-12\right)}^2+{\left(-5\right)}^2}\\ =13\end{array}$

Substitute $-12$ for $x$ , $-5$ for $y$ , and $13$ for $r$ in trigonometric functions.

  $\begin{array}{l}\cos \theta =-\frac{12}{13}\\ \sin \theta =-\frac{5}{13}\\ \cot \theta =\frac{12}{5}\\ \tan \theta =\frac{5}{12}\end{array}$

  $\begin{array}{l}\sec \theta =-\frac{13}{12}\\ \csc \theta =-\frac{13}{5}\end{array}$

Therefore, the exact value of six trigonometric function of each angle $\theta$ are $\cos \theta =-\frac{12}{13},\sin \theta =-\frac{5}{13},\cot \theta =\frac{12}{5},\tan \theta =\frac{5}{12},$$\sec \theta =-\frac{13}{12},\csc \theta =-\frac{13}{5}$ .

(b)

To determine

To find: The exact value of six trigonometric function of each angle $\theta$ .

Answer to Problem 12E

The exact value of six trigonometric function of each angle $\theta$ are $\cos \theta =\frac{1}{\sqrt{2}},\sin \theta =-\frac{1}{\sqrt{2}},\cot \theta =-1,\tan \theta =-1,$$\sec \theta =\sqrt{2},\csc \theta =-\sqrt{2}$ .

Explanation of Solution

Given information:

The given coordinates are shown in figure (2).

  

Figure (2)

Calculation:

Use $\theta$ be an angle in standard position with $\left(x,y\right)$ from figure (2).

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ \ne 0\end{array}$

The trigonometric functions of any angle.

  $\begin{array}{l}\cos \theta =\frac{x}{r}\\ \sin \theta =\frac{y}{r}\\ \cot \theta =\frac{x}{y}\\ \tan \theta =\frac{y}{x}\end{array}$

  $\begin{array}{l}\sec \theta =\frac{r}{x}\\ \csc \theta =\frac{r}{y}\end{array}$

Calculate the value of $r$ and then calculate the value of the trigonometric functions.

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ =\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{2}\end{array}$

Substitute $1$ for $x$ , $-1$ for $y$ , and $\sqrt{2}$ for $r$ in trigonometric functions.

  $\begin{array}{c}\cos \theta =\frac{1}{\sqrt{2}}\\ \sin \theta =-\frac{1}{\sqrt{2}}\\ \cot \theta =-1\\ \tan \theta =-1\end{array}$

  $\begin{array}{l}\sec \theta =\sqrt{2}\\ \csc \theta =-\sqrt{2}\end{array}$

Therefore, the exact value of six trigonometric function of each angle $\theta$ are $\cos \theta =\frac{1}{\sqrt{2}},\sin \theta =-\frac{1}{\sqrt{2}},\cot \theta =-1,\tan \theta =-1,$$\sec \theta =\sqrt{2},\csc \theta =-\sqrt{2}$ .