Q. 78 - 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 $\theta$ : $(5,-12)$

Step-by-Step Solution

Answer

The exact value of each of the six trigonometric functions is,

$$\begin{gathered} \sin \theta = \frac{-12}{13} \\ \cos \theta =\frac{5}{13} \\ \tan \theta = \frac{-12}{5} \\ \csc \theta = \frac{13}{-12} \\ \sec \theta = \frac{13}{5} \\ \cot \theta = \frac{5}{-12} \end{gathered}$$

Step 1. Given information

We are given a point $(5,-12)$.

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

The given point is $(5,-12)$. Here, $x=5, y=-12$

To find $r$, we have $x^2+y^2=r^2$.

Therefore,

$$\begin{gathered} r=\sqrt{x^2+y^2} \\ r=\sqrt{5^2+{(-12)}^2} \\ r=\sqrt{25+144} \\ r=\sqrt{169} \\ \therefore r=13 \end{gathered}$$

Step 4. Finding the values of the six trigonometric functions

The given point is $(5,-12)$. Here, $x=5, y=-12$. Also we know that $r=13$.

Then the values of the six trigonometric functions are,

$$\begin{gathered} \sin \theta = \frac{y}{r}=\frac{-12}{13} \\ \cos \theta =\frac{x}{r}=\frac{5}{13} \\ \tan \theta =\frac{y}{x}=\frac{-12}{5} \\ \csc \theta = \frac{r}{y}=\frac{13}{-12} \\ \sec \theta = \frac{r}{x}=\frac{13}{5} \\ \cot \theta = \frac{x}{y}=\frac{5}{-12} \end{gathered}$$

Step 5. Final Answer

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

$$\begin{gathered} \sin \theta = \frac{-12}{13} \\ \cos \theta =\frac{5}{13} \\ \tan \theta = \frac{-12}{5} \\ \csc \theta = \frac{13}{-12} \\ \sec \theta = \frac{13}{5} \\ \cot \theta = \frac{5}{-12} \end{gathered}$$