Chapter 13.3, Problem 38PPS

To determine

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

Answer to Problem 38PPS

  $\begin{array}{l}\cos \left(\theta \right)=\frac{3}{\sqrt{13}}\\ \sin \left(\theta \right)=-\frac{2}{\sqrt{13}}\\ \cos ec\left(\theta \right)=-\frac{\sqrt{13}}{2}\\ \sec \left(\theta \right)=\frac{\sqrt{13}}{3}\\ \cot \left(\theta \right)=-\frac{3}{2}\end{array}$

Explanation of Solution

Given information:

  $\tan \left(\theta \right)=-\frac{2}{3}$

Quadrant-IV

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

  $\tan \left(\theta \right)=-\frac{2}{3}$

We know

  $\tan \left(\theta \right)=\frac{y}{x}$

On comparing we get

  $\begin{array}{l}y=2\\ x=3\end{array}$

Therefore,

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \Rightarrow r=\sqrt{3^2+2^2}\\ \Rightarrow r=\sqrt{9+4}\\ \Rightarrow r=\sqrt{13}\end{array}$

Therefore,

Given

Quadrant-IV

In Quadrant-IV only Cosine and Secant functions are Positive

On substituting values

We get

  $\begin{array}{l}\cos \left(\theta \right)=\frac{x}{r}\\ \Rightarrow \cos \left(\theta \right)=\frac{3}{\sqrt{13}}\end{array}$

  $\begin{array}{l}\sin \left(\theta \right)=-\frac{y}{r}\\ \Rightarrow \sin \left(\theta \right)=-\frac{2}{\sqrt{13}}\end{array}$

  $\begin{array}{l}\cos ec\left(\theta \right)=-\frac{r}{y}\\ \Rightarrow \cos ec\left(\theta \right)=-\frac{\sqrt{13}}{2}\end{array}$

  $\begin{array}{l}\sec \left(\theta \right)=\frac{r}{x}\\ \Rightarrow \sec \left(\theta \right)=\frac{\sqrt{13}}{3}\end{array}$

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