Chapter 7.1, Problem 28E

To determine

To determine the exact trigonometric value.

Answer to Problem 28E

  $\sin \theta =\frac{\sqrt{5}}{3}$

Explanation of Solution

Given information :

  $\cos \theta =-\frac{2}{3},{90}^{\circ }<\theta <{180}^{\circ }$

Now we know the identity, ${\sin }^2\theta +{\cos }^2\theta =1$

Hence we get,

  $\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\\ {\sin }^2\theta +{\left(-\frac{2}{3}\right)}^2=1\\ {\sin }^2\theta =1-\frac{4}{9}\\ {\sin }^2\theta =\frac{5}{9}\\ \sin \theta =\pm \frac{\sqrt{5}}{3}\end{array}$

Given that $\theta$ is in II quadrant, sine is positive in II quadrant therefore $\sin \theta =\frac{\sqrt{5}}{3}$