Chapter 13, Problem 35SGR

To determine

To Prove : The given value is $\text{sin 2}\theta \text{, cos 2}\theta \text{, sin}\frac{\theta }{\text{2}},\cos \frac{\theta }{2}$

Answer to Problem 35SGR

  $\sin 2\theta =\frac{-4\sqrt{5}}{9},\cos 2\theta =-\frac{1}{9}or\frac{1}{9},\sin \frac{\theta }{2}=\sqrt{\frac{5}{6}},\cos \frac{\theta }{2}=\frac{1}{\sqrt{6}}$

Explanation of Solution

Given information :

We know that, the given equation is $\cos \theta =\frac{-2}{3},\frac{\pi }{2}<\theta <\pi$ .

Formula Used:

  $2{\cos }^2\theta -1,\sin 2\theta =2\sin \theta \cos \theta$

Calculation :

The given equation is

  $\begin{array}{l}\sin 2\theta =2\sin \theta \cos \theta \\ \sin \theta =\pm \sqrt{1-{\cos }^2\theta }=\pm \sqrt{1-\frac{4}{9}}=\pm \frac{\sqrt{5}}{3}\\ As\frac{\pi }{2}<\theta <\pi \\ \Rightarrow \sin \theta =\frac{\sqrt{5}}{3}\\ \therefore \sin 2\theta =2\times \frac{\sqrt{5}}{3}\times \frac{-2}{3}=\frac{-4\sqrt{5}}{9}\\ \cos 2\theta =\pm \sqrt{1-{\sin }^{2}2\theta }=\pm \sqrt{1-\frac{80}{81}}=\frac{-1}{9}\\ As\pi <2\theta <2\pi \\ So,\cos 2\theta =\frac{-1}{9},if\pi <2\theta <\frac{3\pi }{2}\\ and\cos 2\theta =\frac{1}{9},if\frac{3\pi }{2}<2\theta <2\pi \\ As\frac{\pi }{4}<\frac{\theta }{2}<\frac{\pi }{2}\\ \cos \theta =2{\cos }^2\frac{\theta }{2}-1\\ \Rightarrow {\cos }^2\frac{\theta }{2}=\frac{1+\cos \theta }{2}\\ \cos \frac{\theta }{2}=\pm \frac{\sqrt{1+\left(-\frac{2}{3}\right)}}{\sqrt{2}}=\pm \frac{1}{\sqrt{6}}=\frac{1}{\sqrt{6}}\\ \sin \frac{\theta }{2}=\sqrt{1-{\cos }^2\frac{\theta }{2}}=\sqrt{1-\frac{1}{6}}=\sqrt{\frac{5}{6}}\end{array}$