Q. 81 - 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 $\theta$ : $(-2,-2)$

Step-by-Step Solution

Answer

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

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

Step 1. Given information

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

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 $(-2,-2)$. Here $x=-2, y=-2$

To find $r$,

$$\begin{gathered} r=\sqrt{x^2+y^2} \\ r=\sqrt{{(-2)}^2+{(-2)}^2} \\ r=\sqrt{4+4} \\ r=\sqrt{8} \\ r=2\sqrt{2} \end{gathered}$$

Step 4. Finding the exact values of the trigonometric function

The given point is $(-2,-2)$. Here, $x=-2, y=-2$. We also have that, $r=2\sqrt{2}$

Then the six trigonometric functions are,

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

Step 5. Final Answer

The exact value of the six trigonometric functions of $\theta$ are

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