Chapter 13.5, Problem 47STP

To determine

To find: Exact values of the six Trigonometric functions of $\theta$ .

Answer to Problem 47STP

  $\begin{array}{l}\sin \left(\theta \right)=\frac{-9}{\sqrt{117}}\\ \cos \left(\theta \right)=\frac{6}{\sqrt{117}}\\ \tan \left(\theta \right)=\frac{-3}{2}\\ \cos ec\left(\theta \right)=-\frac{\sqrt{117}}{9}\\ \sec \left(\theta \right)=\frac{\sqrt{117}}{6}\\ \cot \left(\theta \right)=\frac{-2}{3}\end{array}$

Explanation of Solution

Given information:

  $\left(x,y\right)=\left(6,-9\right)$

Concept:

If the point in the angle’s terminal side is $P=\left(x,y\right)$ then the Trigonometric functions can be calculated as:

  $\begin{array}{l}\sin \left(\theta \right)=\frac{y}{r}\\ \cos \left(\theta \right)=\frac{x}{r}\\ \tan \left(\theta \right)=\frac{y}{x}\\ \cos ec\left(\theta \right)=\frac{r}{y}\\ \sec \left(\theta \right)=\frac{r}{x}\\ \cot \left(\theta \right)=\frac{x}{y}\end{array}$

Where $r=\sqrt{x^2+y^2}$

Given

  $\left(x,y\right)=\left(6,-9\right)$

Therefore,

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \Rightarrow r=\sqrt{6^2+{\left(-9\right)}^2}\\ \Rightarrow r=\sqrt{36+81}\\ \Rightarrow r=\sqrt{117}\end{array}$

Therefore,

On substituting values

We get

  $\begin{array}{l}\sin \left(\theta \right)=\frac{y}{r}\\ \Rightarrow \sin \left(\theta \right)=\frac{-9}{\sqrt{117}}\end{array}$

  $\begin{array}{l}\cos \left(\theta \right)=\frac{x}{r}\\ \Rightarrow \cos \left(\theta \right)=\frac{6}{\sqrt{117}}\end{array}$

  $\begin{array}{l}\tan \left(\theta \right)=\frac{y}{x}\\ \Rightarrow \tan \left(\theta \right)=\frac{-9}{6}\\ \Rightarrow \tan \left(\theta \right)=\frac{-3}{2}\end{array}$

  $\begin{array}{l}\cos ec\left(\theta \right)=\frac{r}{y}\\ \Rightarrow \cos ec\left(\theta \right)=\frac{\sqrt{117}}{-9}\\ \Rightarrow \cos ec\left(\theta \right)=-\frac{\sqrt{117}}{9}\end{array}$

  $\begin{array}{l}\sec \left(\theta \right)=\frac{r}{x}\\ \Rightarrow \sec \left(\theta \right)=\frac{\sqrt{117}}{6}\end{array}$

  $\begin{array}{l}\cot \left(\theta \right)=\frac{x}{y}\\ \Rightarrow \cot \left(\theta \right)=\frac{6}{-9}\\ \Rightarrow \cot \left(\theta \right)=\frac{-2}{3}\end{array}$