Question

If $\theta ={30}^{\circ }$, verify that :
(i) $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

(ii) $\tan 2\theta =\frac{4\tan \theta }{1+{\tan }^2\theta }$

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

(iv) $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

Answer 1

Given $\theta ={30}^{\circ }$

To verify,

$\left(i\right)\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

$\tan 2\left({30}^{\circ }\right)=\frac{2\tan {30}^{\circ }}{1-{\tan }^2{30}^{\circ }}$

$\tan \left({60}^{\circ }\right)=\frac{2\tan {30}^{\circ }}{1-{\tan }^2{30}^{\circ }}$

$\sqrt{3}=\frac{2\times \frac{1}{\sqrt{3}}}{1-{\left(\frac{1}{\sqrt{3}}\right)}^2}$

$\sqrt{3}=\frac{\frac{2}{\sqrt{3}}}{1-\left(\frac{1}{3}\right)}$

$\sqrt{3}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$

$\sqrt{3}=\sqrt{3}$

$\left(ii\right)\tan 2\theta =\frac{4\tan \theta }{1+{\tan }^2\theta }$

We have, $1+{\tan }^2\theta ={\sec }^2\theta$

$\therefore \tan 2\theta =\frac{4\tan \theta }{{\sec }^2\theta }$

$\tan 2\left({30}^{\circ }\right)=\frac{4\tan {30}^{\circ }}{{\sec }^2{30}^{\circ }}$

$\tan \left({60}^{\circ }\right)=\frac{4\tan {30}^{\circ }}{{\sec }^2{30}^{\circ }}$

$\sqrt{3}=\frac{4\frac{1}{\sqrt{3}}}{{\left(\frac{2}{\sqrt{3}}\right)}^2}$

$\sqrt{3}=\frac{\frac{4}{\sqrt{3}}}{\frac{4}{3}}$

$\sqrt{3}=\frac{3}{\sqrt{3}}$

$\sqrt{3}=\sqrt{3}$

$\left(iii\right)\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$\cos 2\left({30}^{\circ }\right)=\frac{1-{\tan }^2{30}^{\circ }}{1+{\tan }^2{30}^{\circ }}$

$\cos {60}^{\circ }=\frac{1-{\tan }^2{30}^{\circ }}{1+{\tan }^2{30}^{\circ }}$

$\frac{1}{2}=\frac{1-{\left(\frac{1}{\sqrt{3}}\right)}^2}{1+{\left(\frac{1}{\sqrt{3}}\right)}^2}$

$\frac{1}{2}=\frac{3-1}{3+1}$

$\frac{1}{2}=\frac{2}{4}$

$\frac{1}{2}=\frac{1}{2}$

$\left(iv\right)\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

$\cos 3\left({30}^{\circ }\right)=4{\cos }^3{30}^{\circ }-3\cos {30}^{\circ }$

$\cos \left({90}^{\circ }\right)=4\left({\cos }^3{30}^{\circ }\right)-3\cos {30}^{\circ }$

$0=4{\left(\frac{\sqrt{3}}{2}\right)}^3-3\left(\frac{\sqrt{3}}{2}\right)$

$0=4\left(\frac{3\sqrt{3}}{8}\right)-3\left(\frac{\sqrt{3}}{2}\right)$

$0=3\left(\frac{\sqrt{3}}{2}\right)-3\left(\frac{\sqrt{3}}{2}\right)$

$0=0$