Q. 83 - 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 $\theta$ : $\left(\frac{1}{3},\frac{1}{4}\right)$

Step-by-Step Solution

Answer

The exact value of the six trigonometric functions are, 

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

Step 1. Given information

We are given a point $\left(\frac{1}{3},\frac{1}{4}\right)$

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 $\left(\frac{1}{3},\frac{1}{4}\right)$. Here, $x=\frac{1}{3}, y=\frac{1}{4}$

To find $r$, we have

$$\begin{gathered} r=\sqrt{x^2+y^2} \\ r=\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{4}\right)}^2} \\ r=\sqrt{\frac{1}{9}+\frac{1}{16}} \\ r=\sqrt{\frac{25}{144}} \\ r=\frac{5}{12} \end{gathered}$$

Step 4. Finding the values of the six trigonometric functions

We are given a point $\left(\frac{1}{3},\frac{1}{4}\right)$. Here, $x=\frac{1}{3}, y=\frac{1}{4}$. Aslo, we have that  $r=\frac{5}{12}$

$$\begin{gathered} \sin \theta =\frac{y}{r} \\ = \frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{5}{12}}}=\frac{12}{20}=\frac{3}{5} \\ \cos \theta =\frac{x}{r} \\ =\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{5}{12}}}=\frac{12}{15}=\frac{4}{5} \\ \tan \theta =\frac{y}{x} \\ =\frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{3}}}=\frac{3}{4} \\ \csc \theta = \frac{r}{y} \\ =\frac{{\displaystyle \frac{5}{12}}}{{\displaystyle \frac{1}{4}}}=\frac{20}{12}=\frac{5}{3} \\ \sec \theta = \frac{r}{x} \\ =\frac{{\displaystyle \frac{5}{12}}}{{\displaystyle \frac{1}{3}}}=\frac{15}{12}=\frac{5}{4} \\ \cot \theta =\frac{x}{y} \\ =\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{4}}}= \frac{4}{3} \end{gathered}$$

Step 5. Final Answer

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

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