Q. 80 - Precalculus Enhanced with Graphing Utilities

Question

A point on the terminal side of an angle $\theta$ in standard position is given. Find the exact value of each of the six trigonometric functions of $\theta$: $(-1,-2)$

Step-by-Step Solution

Answer

The exact value of the trigonometric functions of  $\theta$ is,

$$\begin{gathered} \sin \theta =\frac{-2}{\sqrt{5}} \\ \cos \theta =\frac{-1}{\sqrt{5}} \\ \tan \theta =2 \\ \csc \theta =\frac{\sqrt{5}}{-2} \\ \sec \theta =-\sqrt{5} \\ \cot \theta =\frac{1}{2} \end{gathered}$$

Step 1. Given information

We are given a point $(-1,-2)$.

We need to find the exact value of each of the six trigonometric functions.

Step 2. Concept

Let $(x,y)$ be a point on the terminal side of the angle $\theta$ in the standard position, which is also on the circle $x^2+y^2=r^2$. Then,

$$\begin{gathered} \sin \theta = \frac{y}{r} \\ \cos \theta =\frac{x}{r} \\ \tan \theta =\frac{y}{x} \\ \csc \theta = \frac{r}{y} \\ \sec \theta = \frac{r}{x} \\ \cot \theta = \frac{x}{y} \end{gathered}$$ 

Step 3. Finding the value of r

We are given a point $(-1,-2)$. Here, $x=-1, y=-2$.

To find $r$,

$$\begin{gathered} r=\sqrt{x^2+y^2} \\ r=\sqrt{{(-1)}^2+{(-2)}^2} \\ r=\sqrt{1+4} \\ r=\sqrt{5} \end{gathered}$$

Step 4. Finding the exact values of the trigonometric functions

The given point is $(-1,-2)$. Here, $x=-1, y=-2$. Also, we have that $r=\sqrt{5}$.

The values of the trigonometric functions are,

$$\begin{gathered} \sin \theta = \frac{y}{r}=\frac{-2}{\sqrt{5}} \\ \cos \theta =\frac{x}{r}=\frac{-1}{\sqrt{5}} \\ \tan \theta =\frac{y}{x}=\frac{-2}{-1}=2 \\ \csc \theta = \frac{r}{y}=\frac{\sqrt{5}}{-2} \\ \sec \theta = \frac{r}{x}=\frac{\sqrt{5}}{-1}=-\sqrt{5} \\ \cot \theta = \frac{x}{y}=\frac{-1}{-2}=\frac{1}{2} \end{gathered}$$

Step 5. Final Answer

The exact values of the six trigonometric functions of $\theta$ are,

$$\begin{gathered} \sin \theta =\frac{-2}{\sqrt{5}} \\ \cos \theta =\frac{-1}{\sqrt{5}} \\ \tan \theta =2 \\ \csc \theta =\frac{\sqrt{5}}{-2} \\ \sec \theta =-\sqrt{5} \\ \cot \theta =\frac{1}{2} \end{gathered}$$