Chapter 13.3, Problem 1CYP

To determine

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

Answer to Problem 1CYP

  $\begin{array}{l}\sin \left(\theta \right)=\frac{1}{\sqrt{10}}\\ \cos \left(\theta \right)=\frac{-3}{\sqrt{10}}\\ \tan \left(\theta \right)=-\frac{1}{3}\\ \cos ec\left(\theta \right)=\sqrt{10}\\ \sec \left(\theta \right)=-\frac{\sqrt{10}}{3}\\ \cot \left(\theta \right)=-3\end{array}$

Explanation of Solution

Given information:

  $\left(x,y\right)=\left(-6,2\right)$

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

  $\left(x,y\right)=\left(-6,2\right)$

Therefore,

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \Rightarrow r=\sqrt{{\left(-6\right)}^2+2^2}\\ \Rightarrow r=\sqrt{36+4}\\ \Rightarrow r=\sqrt{40}\\ \Rightarrow r=2\sqrt{10}\end{array}$

Therefore,

On substituting values

We get

  $\begin{array}{l}\sin \left(\theta \right)=\frac{y}{r}\\ \Rightarrow \sin \left(\theta \right)=\frac{2}{2\sqrt{10}}\\ \Rightarrow \sin \left(\theta \right)=\frac{1}{\sqrt{10}}\end{array}$

  $\begin{array}{l}\cos \left(\theta \right)=\frac{x}{r}\\ \Rightarrow \cos \left(\theta \right)=\frac{-6}{2\sqrt{10}}\\ \Rightarrow \cos \left(\theta \right)=\frac{-3}{\sqrt{10}}\end{array}$

  $\begin{array}{l}\tan \left(\theta \right)=\frac{y}{x}\\ \Rightarrow \tan \left(\theta \right)=\frac{2}{-6}\\ \Rightarrow \tan \left(\theta \right)=-\frac{1}{3}\end{array}$

  $\begin{array}{l}\cos ec\left(\theta \right)=\frac{r}{y}\\ \Rightarrow \cos ec\left(\theta \right)=\frac{2\sqrt{10}}{2}\\ \Rightarrow \cos ec\left(\theta \right)=\sqrt{10}\end{array}$

  $\begin{array}{l}\sec \left(\theta \right)=\frac{r}{x}\\ \Rightarrow \sec \left(\theta \right)=\frac{2\sqrt{10}}{-6}\\ \Rightarrow \sec \left(\theta \right)=-\frac{\sqrt{10}}{3}\end{array}$

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