Chapter 13.3, Problem 37PPS

To determine

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

Answer to Problem 37PPS

  $\begin{array}{l}\cos \left(\theta \right)=-\frac{3}{5}\\ \tan \left(\theta \right)=-\frac{4}{3}\\ \cos ec\left(\theta \right)=\frac{5}{4}\\ \sec \left(\theta \right)=-\frac{5}{3}\\ \cot \left(\theta \right)=-\frac{3}{4}\end{array}$

Explanation of Solution

Given information:

  $\sin \left(\theta \right)=\frac{4}{5}$

Quadrant-II

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

  $\sin \left(\theta \right)=\frac{4}{5}$

We know

  $\sin \left(\theta \right)=\frac{y}{r}$

On comparing we get

  $\begin{array}{l}y=4\\ r=5\end{array}$

Therefore,

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \Rightarrow 5=\sqrt{x^2+4^2}\\ \Rightarrow 5^2={\left(\sqrt{x^2+16}\right)}^2\\ \Rightarrow 25=x^2+16\\ \Rightarrow x^2=25-16\\ \Rightarrow x^2=9\\ \Rightarrow x=3\end{array}$

Therefore,

Given

Quadrant-II

In Quadrant-II only Sine and Cosecant functions are Positive

On substituting values

We get

  $\begin{array}{l}\cos \left(\theta \right)=-\frac{x}{r}\\ \Rightarrow \cos \left(\theta \right)=-\frac{3}{5}\end{array}$

  $\begin{array}{l}\tan \left(\theta \right)=-\frac{y}{x}\\ \Rightarrow \tan \left(\theta \right)=-\frac{4}{3}\end{array}$

  $\begin{array}{l}\cos ec\left(\theta \right)=\frac{r}{y}\\ \Rightarrow \cos ec\left(\theta \right)=\frac{5}{4}\end{array}$

  $\begin{array}{l}\sec \left(\theta \right)=-\frac{r}{x}\\ \Rightarrow \sec \left(\theta \right)=-\frac{5}{3}\end{array}$

  $\begin{array}{l}\cot \left(\theta \right)=-\frac{x}{y}\\ \Rightarrow \cot \left(\theta \right)=-\frac{3}{4}\end{array}$