Chapter 7.6, Problem 17AYU

In Problems 9-20, use the information given about the angle $\theta$, $0\le \theta <2\pi$, to find the exact value of:

(a) $\sin \left(2\theta \right)$           (b) $cos(2\theta )$           (c) $\sin \frac{\theta }{2}$           (d) $\cos \frac{\theta }{2}$           (e) $\tan \left(2\theta \right)$           (f) $\tan \frac{\theta }{2}$

$\tan \theta =-3$, $\sin \theta \text{ < 0}$

To determine

To find: The exact value of following functions if $\text{tan}\theta =-3,\text{sin}\theta <0$.

Answer to Problem 17AYU

$\text{sin}2\theta =2.\left(\frac{-3}{\sqrt{10}}\right).\left(\frac{1}{\sqrt{10}}\right)=-\frac{6 }{10}=-\frac{3 }{5}$

$\text{cos}2\theta ={(\frac{1}{\sqrt{10}})}^2-{(\frac{-3}{\sqrt{10}})}^{2=-}\frac{8}{10}=-\frac{4}{5}$

$\text{sin}\frac{\theta }{2}=\pm {(\frac{1-(\frac{1}{\sqrt{10}}}{2})}^{\frac{1}{2}}=\pm \sqrt{\frac{\sqrt{10 }-1}{2\sqrt{10}}}$

$\text{cos}\frac{\theta }{2}=\pm {(\frac{1+(\frac{1}{\sqrt{10}}}{2})}^{\frac{1}{2}}=\pm \sqrt{\frac{\sqrt{10 }+1}{2\sqrt{10}}}$

$\text{tan}2\theta =\frac{2.\left(-3\right)}{1-{\left(-3\right)}^2}=\left(\frac{-6}{-8}\right)=\frac{3}{4}$

$\text{tan}\frac{\theta }{2}=\pm {(\frac{1-\left(\frac{1}{\sqrt{10}}\right)}{1+\left(\frac{1}{\sqrt{10}}\right)})}^{\frac{1}{2}}=\pm \sqrt{\frac{\sqrt{10 }-1}{\sqrt{10 }+1}}$

Explanation of Solution

Given:

$\text{tan}\theta =-3,\text{sin}\theta <0$

Calculation:

$\text{sinθ }=-\frac{3}{\sqrt{10}}$, $\text{cos}\theta =\frac{1}{\sqrt{10}}$

By double angle formula,

$\text{sin}2\theta =2\text{sin}\theta \text{cos}\theta$

$\text{cos}2\theta ={\text{cos}}^2\theta -{\text{sin}}^2\theta$

$\text{tan}2\theta =\frac{2\text{tan}\theta }{1-{\text{tan}}^2\theta }$

By half angle formula,

$\text{sin}\frac{\theta }{2}=\pm {(\frac{1-\text{cos}\theta }{2})}^{\frac{1}{2}}$

$\text{cos}\frac{\theta }{2}=\pm {(\frac{1+\text{cos}\theta }{2})}^{\frac{1}{2}}$

$\text{tan}\frac{\theta }{2}=\pm {(\frac{1-\text{cos}\theta }{1+\text{cos}\theta })}^{\frac{1}{2}}$

$\text{sin}2\theta =2.\left(\frac{-3}{\sqrt{10}}\right).\left(\frac{1}{\sqrt{10}}\right)=-\frac{6 }{10}=-\frac{3 }{5}$

$\text{cos}2\theta ={(\frac{1}{\sqrt{10}})}^2-{(\frac{-3}{\sqrt{10}})}^2=-\frac{8}{10}=-\frac{4}{5}$

$\text{sin}\frac{\theta }{2}=\pm {(\frac{1-(\frac{1}{\sqrt{10}}}{2})}^{\frac{1}{2}}=\pm \sqrt{\frac{\sqrt{10 }-1}{2\sqrt{10}}}$

$\text{cos}\frac{\theta }{2}=\pm {(\frac{1+(\frac{1}{\sqrt{10}}}{2})}^{\frac{1}{2}}=\pm \sqrt{\frac{\sqrt{10 }+1}{2\sqrt{10}}}$

$\text{tan}2\theta =\frac{2.\left(-3\right)}{1-{\left(-3\right)}^2}=\left(\frac{-6}{-8}\right)=\frac{3}{4}$

$\text{tan}\frac{\theta }{2}=\pm {(\frac{1-\left(\frac{1}{\sqrt{10}}\right)}{1+\left(\frac{1}{\sqrt{10}}\right)})}^{\frac{1}{2}}=\pm \sqrt{\frac{\sqrt{10 }-1}{\sqrt{10 }+1}}$