Chapter 12.3, Problem 39PPS

To determine

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

Answer to Problem 39PPS

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

Explanation of Solution

Given information:

  $\cos \left(\theta \right)=-\frac{8}{17}$

Quadrant-III

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

  $\cos \left(\theta \right)=-\frac{8}{17}$

We know

  $\cos \left(\theta \right)=\frac{x}{r}$

On comparing we get

  $\begin{array}{l}x=8\\ r=17\end{array}$

Therefore,

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

Therefore,

Given

Quadrant-III

In Quadrant-III only Tan and Cot functions are Positive

On substituting values

We get

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

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

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

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

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