Chapter 9.7, Problem 5E

To determine

To find: To find all trigonometric functions.

Answer to Problem 5E

  $\begin{array}{l}\sin \theta =\frac{3}{\sqrt{58}},\cos \theta =-\frac{7}{\sqrt{58}},\\ \csc \theta =\frac{\sqrt{58}}{3},\sec \theta =-\frac{\sqrt{58}}{7},\cot \theta =-\frac{7}{3}\end{array}$

Explanation of Solution

Given information:

  $\tan \theta =-\frac{3}{7},$

  $\frac{\pi }{2}<\theta <\pi$

Formula used:

  $\begin{array}{l}\sin \theta =\frac{Perpendicular}{Hypotenuse},\cos \theta =\frac{Base}{Hypotenuse},\tan \theta =\frac{Perpendicular}{Base}\\ \csc \theta =\frac{Hypotenuse}{Perpendicular},\sec \theta =\frac{Hypotenuse}{Base},\cot \theta =\frac{Base}{Perpendicular}\end{array}$

  ${\left(Hypotenuse\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2$

  $\tan \theta =\frac{Perpendicular}{Base}=-\frac{3}{7}$

Negative sign will be neglected, because it only denotes that the given angle is in quadrant in which the trigonometric function is negative,

By Pythagoras theorem,

  $\begin{array}{l}{\left(Hypotenuse\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2\\ \\ {\left(Hypotenuse\right)}^2={\left(3\right)}^2+{\left(7\right)}^2\\ \\ {\left(Hypotenuse\right)}^2=9+49\\ \\ {\left(Hypotenuse\right)}^2=58\end{array}$

  $\left(Hypotenuse\right)=\sqrt{58}$

Given that $\frac{\pi }{2}<\theta <\pi$ , So $\sin \theta ,\csc \theta$ will be positive, all others will be negative

  $\begin{array}{l}\sin \theta =\frac{3}{\sqrt{58}},\cos \theta =-\frac{7}{\sqrt{58}},\\ \csc \theta =\frac{\sqrt{58}}{3},\sec \theta =-\frac{\sqrt{58}}{7},\cot \theta =-\frac{7}{3}\end{array}$