Q. 77 - 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 $\theta$: $(-3,4)$

Step-by-Step Solution

Answer

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

$$\begin{gathered} \sin \theta = \frac{4}{5} \\ \cos \theta =\frac{-3}{5} \\ \tan \theta =\frac{4}{-3} \\ \csc \theta = \frac{5}{4} \\ \sec \theta = \frac{5}{-3} \\ \cot \theta = \frac{-3}{4} \end{gathered}$$

Step 1. Given Information

The given point is $(-3,4)$.

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 the point $(-3,4)$. Here $x=-3, y=4$.

We know that,

 $$\begin{gathered} r=\sqrt{x^2+y^2} \\ r=\sqrt{{(-3)}^2+4^2} \\ r=\sqrt{9+16} \\ r=\sqrt{25} \\ r=5 \end{gathered}$$

Step 4. Finding the exact values of the six trigonometric functions

From the point $(-3,4)$, we have $x=-3, y=4$. Also, we have found that $r=5$. Therefore the trigonometric functions are,

$$\begin{gathered} \sin \theta = \frac{y}{r} = \frac{4}{5} \\ \cos \theta =\frac{x}{r} = \frac{-3}{5} \\ \tan \theta =\frac{y}{x} = \frac{4}{-3} \\ \csc \theta = \frac{r}{y} = \frac{5}{4} \\ \sec \theta = \frac{r}{x} = \frac{5}{-3} \\ \cot \theta = \frac{x}{y} =\frac{-3}{4} \end{gathered}$$

Step 5. Final Answer

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

$$\begin{gathered} \sin \theta = \frac{4}{5} \\ \cos \theta =\frac{-3}{5} \\ \tan \theta =\frac{4}{-3} \\ \csc \theta = \frac{5}{4} \\ \sec \theta = \frac{5}{-3} \\ \cot \theta = \frac{-3}{4} \end{gathered}$$