Question

If A = ${30}^{\circ },$ verify that:

(i) $sin2A=\frac{2tanA}{1+ta{n}^{2}A}$ (ii) $cos2A=\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A}$ (iii) $tan2A=\frac{2tanA}{1-ta{n}^{2}A}$

Answer 1

(i) Given A = 30,

$sin2A=sin(30\times 2)=sin60=\frac{\sqrt{3}}{2}2tanA=2\times tan30=\frac{2}{\sqrt{3}}1+ta{n}^{2}A=1+(\frac{1}{\sqrt{3}}{)}^2=1+\frac{1}{3}=\frac{4}{3}\frac{2tanA}{1+ta{n}^{2}A}=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\frac{\sqrt{3}}{2}=sin2Asin2A=\frac{2tanA}{1+ta{n}^{2}A}$

hence proved.

(ii) Given A = 30,

$cos2A=cos(30\times 2)=cos60=\frac{1}{2}1+ta{n}^{2}A=1+(\frac{1}{\sqrt{3}}{)}^2=1+\frac{1}{3}=\frac{4}{3}1-ta{n}^{2}A=1-(\frac{1}{\sqrt{3}}{)}^2=1-\frac{1}{3}=\frac{2}{3}\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A}=\frac{\frac{2}{3}}{\frac{4}{3}}=\frac{2}{3}\times \frac{3}{4}=\frac{1}{2}=cos2Acos2A=\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A}$

(iii) Given A = 30,

$tan2A=tan(30\times 2)=tan60=\sqrt{3}2tanA=2\times tan30=\frac{2}{\sqrt{3}}1-ta{n}^{2}A=1-(\frac{1}{\sqrt{3}}{)}^2=1-\frac{1}{3}=\frac{2}{3}\frac{2tanA}{1-ta{n}^{2}A}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{2}{\sqrt{3}}\times \frac{3}{2}=\sqrt{3}=tan2Atan2A=\frac{2tanA}{1-ta{n}^{2}A}$

hence proved.