Chapter 14.1, Problem 1ACYP

To determine

To find :the value of sin $\theta$ .

Answer to Problem 1ACYP

  $\sin \theta =-2\sqrt{2}/3$

Explanation of Solution

Given information :

We know that, $\cos \theta =\frac{1}{3},270°<\theta <360°$

Calculation :

We know that

  $\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\\ \Rightarrow {\sin }^2\theta +{\left(\frac{1}{3}\right)}^2=1\\ \Rightarrow {\sin }^2\theta =1-\frac{1}{9}=\frac{8}{9}\\ \Rightarrow \sin \theta =\pm \frac{\sqrt{8}}{3}\end{array}$

But $\theta$ lies between $\text{270°}$ and $\text{360°}$

  $\Rightarrow \sin \theta =-\frac{2\sqrt{2}}{3}$