Q. 79 - 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 $\theta$: $(2,-3)$

Step-by-Step Solution

Answer

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

$$\begin{gathered} \sin \theta = \frac{-3}{\sqrt{13}} \\ \cos \theta = \frac{2}{\sqrt{13}} \\ \tan \theta = \frac{-3}{2} \\ \csc \theta = \frac{\sqrt{13}}{-3} \\ \sec \theta = \frac{\sqrt{13}}{2} \\ \cot \theta = \frac{2}{-3} \end{gathered}$$

Step 1. Given information

We are given a point $(2,-3)$.

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 have the point $(2,-3)$. Here, $x=2, y=-3$

To find $r$,

$$\begin{gathered} r=\sqrt{x^2+y^2} \\ r=\sqrt{2^2+{(-3)}^2} \\ r=\sqrt{4+9} \\ r=\sqrt{13} \end{gathered}$$

Step 4. Finding the values of the six trigonometric functions

We have the point $(2,-3)$. From this, we have that $x=2, y=-3$. Also, we have found that $r=\sqrt{13}$

Then the six trigonometric functions are 

$$\begin{gathered} \sin \theta = \frac{y}{r} = \frac{-3}{\sqrt{13}} \\ \cos \theta =\frac{x}{r} = \frac{2}{\sqrt{13}} \\ \tan \theta =\frac{y}{x} = \frac{-3}{2} \\ \csc \theta = \frac{r}{y} =\frac{\sqrt{13}}{-3} \\ \sec \theta = \frac{r}{x}=\frac{\sqrt{13}}{2} \\ \cot \theta = \frac{x}{y}=\frac{2}{-3} \end{gathered}$$

Step 5. Final Answer

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

$$\begin{gathered} \sin \theta = \frac{-3}{\sqrt{13}} \\ \cos \theta = \frac{2}{\sqrt{13}} \\ \tan \theta = \frac{-3}{2} \\ \csc \theta = \frac{\sqrt{13}}{-3} \\ \sec \theta = \frac{\sqrt{13}}{2} \\ \cot \theta = \frac{2}{-3} \end{gathered}$$