Chapter 5.3, Problem 12CFU

To determine

To find:the values of the remaining five trigonometric functions for $\theta$ .

Answer to Problem 12CFU

  $\sin \theta =\frac{\sqrt{3}}{2};\tan \theta =-\sqrt{3};\csc \theta =\frac{2\sqrt{3}}{3};\cot \theta =\frac{-\sqrt{3}}{3};\text{ and }\sec \theta =-2$

Explanation of Solution

Given information:

Let $\theta$ is an angle in standard position whose terminal side lies in the given quadrant.

  $\cos \theta =-\frac{1}{2};\text{}\text{ Quadrant II}$

Calculation:

Let $\cos \theta =\frac{x}{r}$ . The value of $r=2\text{ and }x=-1$ . The graph will be similar to the one shown below

Now evaluate $y$ by using Pythagoras Theorem as shown below.

  $\begin{array}{c}{x}^2+y^2=r^2\\ {\left(-1\right)}^2+y^2=2^2\\ 1+y^2=4\\ y=\pm \sqrt{4-1}\\ y=\sqrt{3}\left[\because \text{Quadrant II}\right]\end{array}$

The values of the remaining five trigonometric functions for $\theta$ will be evaluated as follows:

  $\begin{array}{c}\sin \theta =\frac{y}{r}\\ =\frac{\sqrt{3}}{2}\left[\text{Put the value of }r\text{ and }y\right]\end{array}$

  $\begin{array}{c}\tan \theta =\frac{y}{x}\\ =\frac{\sqrt{3}}{-1}\left[\text{Put the value of }y\text{ and }x\right]\\ =-\sqrt{3}\end{array}$

  $\begin{array}{c}\csc \theta =\frac{r}{y}\\ =\frac{2}{\sqrt{3}}\left[\text{Put the value of }r\text{ and }y\right]\\ =\frac{2\sqrt{3}}{3}\left[\text{Multiply and divide by }\sqrt{3}\right]\end{array}$

  $\begin{array}{c}\cot \theta =\frac{x}{y}\\ =\frac{-1}{\sqrt{3}}\left[\text{Put the value of }y\text{ and }x\right]\\ =\frac{-\sqrt{3}}{3}\left[\text{Multiply and divide by }\sqrt{3}\right]\end{array}$

  $\begin{array}{c}\sec \theta =\frac{r}{x}\\ =\frac{2}{-1}\left[\text{Put the value of }r\text{ and }x\right]\\ =-2\end{array}$

Thus, $\sin \theta ,\tan \theta ,\csc \theta ,\cot \theta \text{ and sec}\theta$ the value of is $\frac{\sqrt{3}}{2},-\sqrt{3},\frac{2\sqrt{3}}{3},\frac{-\sqrt{3}}{3}\text{ and }-2$ respectively.