Chapter 13.3, Problem 2CYU

To determine

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

Answer to Problem 2CYU

  $\begin{array}{l}\sin \left(\theta \right)=-\frac{15}{17}\\ \cos \left(\theta \right)=-\frac{8}{17}\\ \tan \left(\theta \right)=\frac{15}{8}\\ \cos ec\left(\theta \right)=-\frac{17}{15}\\ \sec \left(\theta \right)=-\frac{17}{8}\\ \cot \left(\theta \right)=\frac{8}{15}\end{array}$

Explanation of Solution

Given information:

  $\left(x,y\right)=\left(-8,-15\right)$

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

  $\left(x,y\right)=\left(-8,-15\right)$

Therefore,

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \Rightarrow r=\sqrt{{\left(-8\right)}^2+{\left(-15\right)}^2}\\ \Rightarrow r=\sqrt{64+225}\\ \Rightarrow r=\sqrt{289}\\ \Rightarrow r=17\end{array}$

Therefore,

On substituting values

We get

  $\begin{array}{l}\sin \left(\theta \right)=\frac{y}{r}\\ \Rightarrow \sin \left(\theta \right)=\frac{-15}{17}\end{array}$

  $\begin{array}{l}\cos \left(\theta \right)=\frac{x}{r}\\ \Rightarrow \cos \left(\theta \right)=\frac{-8}{17}\end{array}$

  $\begin{array}{l}\tan \left(\theta \right)=\frac{y}{x}\\ \Rightarrow \tan \left(\theta \right)=\frac{-15}{-8}\\ \Rightarrow \tan \left(\theta \right)=\frac{15}{8}\end{array}$

  $\begin{array}{l}\cos ec\left(\theta \right)=\frac{r}{y}\\ \Rightarrow \cos ec\left(\theta \right)=\frac{17}{-15}\\ \Rightarrow \cos ec\left(\theta \right)=-\frac{17}{15}\end{array}$

  $\begin{array}{l}\sec \left(\theta \right)=\frac{r}{x}\\ \Rightarrow \sec \left(\theta \right)=\frac{17}{-8}\\ \Rightarrow \sec \left(\theta \right)=-\frac{17}{8}\end{array}$

  $\begin{array}{l}\cot \left(\theta \right)=\frac{x}{y}\\ \Rightarrow \cot \left(\theta \right)=\frac{-8}{-15}\\ \Rightarrow \cot \left(\theta \right)=\frac{8}{15}\end{array}$