Chapter 5.3, Problem 48E

To determine

To find: The six trigonometric functions of $\theta$ .

Answer to Problem 48E

  $\begin{array}{l}\sin \theta =\frac{3\sqrt{10}}{10},\csc \theta =\frac{\sqrt{10}}{3}\\ \cos \theta =\frac{-\sqrt{10}}{10},\sec \theta =-\sqrt{10}\\ \tan \theta =-3,\cot \theta =\frac{-1}{3}\end{array}$

Explanation of Solution

Given information: Critical Thinking: The terminal side of an angle $\theta$ in standard position coincides with the line $y=-3x$ and lies in Quadrant II.

Calculation:

For a unit circle the equation is $x^2+y^2=1$ you also have the line, $y=-3x$

This is two equations, two unknowns, solve for the $\left(x,y\right)$ pair they have in common

  $\begin{array}{l}{x}^2+{\left(-3x\right)}^2=1\\ x^2+9x^2=1\\ 10x^2=1/10\\ x=\pm \frac{1}{\sqrt{10}}=\pm \frac{\sqrt{10}}{10}\end{array}$

  $y=-3x=-3\left(\frac{\sqrt{10}}{10}\right)$

So the point $\left(\frac{\sqrt{10}}{10},-3\frac{\sqrt{10}}{10}\right)$ is the endpoint of the terminal side and it is in Quadrant II. But THIS WON’T WORK!!! The y term must be positive in quadrant two so the $\pm$ options for x, we must choose the negative value of x. So it must be

  $y=-3x=-3\left(\frac{\sqrt{10}}{10}\right)=3\left(\frac{\sqrt{10}}{10}\right)$

So our point is $\left(-\frac{\sqrt{10}}{10},3\frac{\sqrt{10}}{10}\right)$

So we have our sides

  $\text{opposite}=\frac{3\sqrt{10}}{10},\text{ adjacent}=\frac{-\sqrt{10}}{10}\text{ and hypotenuse}=1$

  

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