Chapter 0.6, Problem 9E

To determine

To find: all the trigonometric values of $\theta$ with the given conditions.

Answer to Problem 9E

  $\sin \theta =\frac{8}{17},\cos \theta =-\frac{15}{17},\tan \theta =-\frac{8}{15},\csc \theta =\frac{17}{8},\sec \theta =-\frac{17}{15},\cot \theta =-\frac{15}{8}$ .

Explanation of Solution

Given:

  $\cos \theta =-\frac{15}{17},\sin \theta >0$ .

Calculation:

The angle $\theta$ is in the second quadrant, because the sine is positive and the cosine is negative.

The graphical representation is,

  

Now since $\text{cos}\theta \text{=}\frac{\text{base}}{\text{hypotenuse}}$

And here,

  $\begin{array}{l}\text{base=}\left|\text{-15}\right|=15\\ \text{hypotenuse}\text{=17}\end{array}$

Thus, from Pythagoras theorem

  $\begin{array}{l}{\text{p}}^{\text{2}}{\text{=h}}^{\text{2}}{\text{-b}}^{\text{2}}\\ \text{p=}\sqrt{{\text{17}}^{\text{2}}{\text{-15}}^{\text{2}}}\\ \text{p=}\sqrt{\text{289-225}}\\ \text{p=}\sqrt{\text{64}}\\ \text{p=8}\end{array}$

Therefore, from obtained value of perpendicular, it follows that

  $\begin{array}{c}\text{sin}\theta \text{=}\frac{\text{perpendicular}}{\text{hypotenuse}}\\ =\frac{\text{8}}{\text{17}}\end{array}$

Since in second quadrant $\text{sin}\theta \text{>0}$

Thus, the remain positive i.e., $\text{sin}\theta =\frac{8}{17}$

Now

  $\begin{array}{c}\text{csc}\theta \text{=}\frac{\text{hypotenuse}}{\text{perpendicular}}\\ =\frac{\text{17}}{\text{8}}\end{array}$

Since in second quadrant $\text{csc}\theta \text{>0}$

Thus, the remain positive i.e., $\text{csc}\theta =\frac{8}{17}$

For $\text{tan}\theta$

  $\begin{array}{c}\text{tan}\theta \text{=}\frac{\text{Perpendicular}}{\text{base}}\\ \text{=}\frac{\text{8}}{\text{15}}\end{array}$

Since in second quadrant $\text{tan}\theta <\text{0}$

Thus, the value of $\text{tan}\theta$ is negative i.e., $\text{tan}\theta =-\frac{8}{17}$

For $\text{cot}\theta$

  $\begin{array}{c}\cot \theta \text{=}\frac{\text{base}}{\text{perpendicular}}\\ \text{=}\frac{15}{8}\end{array}$

Since in second quadrant $\text{cot}\theta <\text{0}$

Thus, the value of $\cot \theta$ is negative i.e., $\text{cot}\theta =-\frac{15}{8}$

For $\sec \theta$

  $\begin{array}{c}\sec \theta \text{=}\frac{\text{hypotenuse}}{\text{base}}\\ \text{=}\frac{17}{15}\end{array}$

Since in second quadrant $\text{sec}\theta <\text{0}$

Thus, the value of $\sec \theta$ is negative i.e., $\text{sec}\theta =-\frac{17}{8}$

Conclusion:

  $\sin \theta =\frac{8}{17},\cos \theta =-\frac{15}{17},\tan \theta =-\frac{8}{15},\csc \theta =\frac{17}{8},\sec \theta =-\frac{17}{15},\cot \theta =-\frac{15}{8}$