Chapter 4.4, Problem 12E

a.

To determine

To discuss: the exact values of the six trigonometric functions of each angle $\theta$ .

Answer to Problem 12E

  $\begin{array}{l}\sin \theta =\frac{\sqrt{10}}{10}\\ \cos \theta =\frac{3\sqrt{10}}{10}\\ \tan \theta =\frac{1}{3}\end{array}$

  $\begin{array}{l}\csc \theta =\sqrt{10}\\ \sec \theta =\frac{\sqrt{10}}{3}\\ \cot \theta =3\end{array}$

Explanation of Solution

Observe the given figure.

From above figure, we have $x=3$ , $y=1$

Then,

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ =\sqrt{{\left(3\right)}^2+{\left(1\right)}^2}\\ =\sqrt{9+1}\\ =\sqrt{10}\end{array}$

Now,

  $\begin{array}{c}\sin \theta =\frac{y}{r}\\ =\frac{1}{\sqrt{10}}\\ =\frac{\sqrt{10}}{10}\end{array}$

  $\begin{array}{c}\cos \theta =\frac{x}{r}\\ =\frac{3}{\sqrt{10}}\\ =\frac{3\sqrt{10}}{10}\end{array}$

  $\begin{array}{c}\tan \theta =\frac{y}{x}\\ =\frac{1}{3}\end{array}$

  $\begin{array}{c}\csc \theta =\frac{r}{y}\\ =\frac{\sqrt{10}}{1}\\ =\sqrt{10}\end{array}$

  $\begin{array}{c}\sec \theta =\frac{r}{x}\\ =\frac{\sqrt{10}}{3}\end{array}$

  $\begin{array}{c}\cot \theta =\frac{x}{y}\\ =\frac{3}{1}\\ =3\end{array}$

b.

To determine

To discuss: the exact values of the six trigonometric functions of each angle $\theta$ .

Answer to Problem 12E

  $\begin{array}{l}\sin \theta =\frac{\sqrt{21}}{5}\\ \cos \theta =\frac{-2}{5}\\ \tan \theta =-\frac{\sqrt{21}}{2}\end{array}$

  $\begin{array}{l}\csc \theta =\frac{5\sqrt{21}}{21}\\ \sec \theta =-\frac{5}{2}\\ \cot \theta =-\frac{2\sqrt{21}}{21}\end{array}$

Explanation of Solution

Observe the given figure.

From above figure, we have $x=-2$ , $y=\sqrt{21}$

Then,

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ =\sqrt{{\left(-2\right)}^2+{\left(\sqrt{21}\right)}^2}\\ =\sqrt{4+21}\end{array}$

  $\begin{array}{c}=\sqrt{25}\\ =5\end{array}$

Now,

  $\begin{array}{c}\sin \theta =\frac{y}{r}\\ =\frac{\sqrt{21}}{5}\end{array}$

  $\begin{array}{c}\cos \theta =\frac{x}{r}\\ =\frac{-2}{5}\end{array}$

  $\begin{array}{c}\tan \theta =\frac{y}{x}\\ =\frac{\sqrt{21}}{-2}\\ =-\frac{\sqrt{21}}{2}\end{array}$

  $\begin{array}{c}\csc \theta =\frac{r}{y}\\ =\frac{5}{\sqrt{21}}\\ =\frac{5\sqrt{21}}{21}\end{array}$

  $\begin{array}{c}\sec \theta =\frac{r}{x}\\ =\frac{5}{-2}\\ =-\frac{5}{2}\end{array}$

  $\begin{array}{c}\cot \theta =\frac{x}{y}\\ =\frac{-2}{\sqrt{21}}\\ =-\frac{2\sqrt{21}}{21}\end{array}$