Question

If A = 30°, verify that:
(i) $\sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}$
(ii) $\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$
(iii) $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$

Answer 1

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

​(i) sin 2A = sin 60^o = $\frac{\sqrt{3}}{2}$
$\frac{2\tan A}{1+{\tan }^{2}A}=\frac{2\tan {30}^o}{1+{\tan }^2{30}^o}=\frac{2\times \left({\displaystyle \frac{1}{\sqrt{3}}}\right)}{1+{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2}=\frac{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}{1+{\displaystyle \frac{1}{3}}}=\frac{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}{{\displaystyle \frac{4}{3}}}=\left({\displaystyle \frac{2}{\sqrt{3}}}\right)\times \frac{3}{4}=\frac{\sqrt{3}}{2}$
∴ $\sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}$

(ii) cos 2A = cos 60^o = $\frac{1}{2}$

$\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}=\frac{1-{\tan }^2{30}^o}{1+{\tan }^2{30}^o}=\frac{1-{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2}{1+{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2}=\frac{\left(1-{\displaystyle \frac{1}{3}}\right)}{\left(1+{\displaystyle \frac{1}{3}}\right)}=\frac{\left({\displaystyle \frac{2}{3}}\right)}{{\displaystyle \frac{4}{3}}}=\left({\displaystyle \frac{2}{3}}\right)\times \frac{3}{4}=\frac{1}{2}$
∴ $\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$

(iii) tan 2A = tan 60^o = $\sqrt{3}$
$\frac{2\tan A}{1-{\tan }^{2}A}=\frac{2\tan {30}^o}{1-{\tan }^2{30}^o}=\frac{2\times \left({\displaystyle \frac{1}{\sqrt{3}}}\right)}{1-{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2}=\frac{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}{1-{\displaystyle \frac{1}{3}}}=\frac{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}{{\displaystyle \frac{2}{3}}}=\left({\displaystyle \frac{2}{\sqrt{3}}}\right)\times \frac{3}{2}=\sqrt{3}$
∴ $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$=2tanA1+tan2A