Question Number 8176 by lepan last updated on 02/Oct/16
$${Prove}\:{that}\:{cos}\mathrm{2}\theta=\frac{\mathrm{1}−{tan}^{\mathrm{2}} \theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}.{Hence}\:{deduce}\:{that}\: \\ $$$${tan}\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}=\sqrt{\mathrm{2}}−\mathrm{1}. \\ $$$$ \\ $$$$ \\ $$
Answered by prakash jain last updated on 02/Oct/16
![cos 2θ=cos^2 θ−sin^2 θ =((cos^2 θ−sin^2 θ)/(cos^2 θ+cos^2 θ)) [∵cos^2 θ+sin^2 θ=1] Divide numerator and Denominator by cos^2 θ cos 2θ=(((cos^2 θ−sin^2 θ)/(cos^2 θ))/((cos^2 θ+sin^2 θ)/(cos^2 θ)))=((1−tan^2 θ)/(1+tan^2 θ)) ________ put θ=22(1/2) cos 45°=((1−tan^2 22(1/2))/(1+tan^2 22(1/2))) x=tan 22(1/2) (1/( (√2)))=((1−x^2 )/(1+x^2 )) 1+x^2 =(√2)−(√2)x^2 (1+(√2))x^2 =(√2)−1 x^2 =(((√2)−1)/( (√2)+1))=((((√2)−1)((√2)−1))/(((√2)+1)((√2)−1)))=((((√2)−1)^2 )/(2−1))=((√2)−1)^2 ⇒x=±((√2)−1) or tan 22(1/2)=±((√2)−1) Since 22(1/2)<90, tan 22(1/2)>0 so we discard the −ve root. hence x=tan 22(1/2)=((√2)−1)](https://www.tinkutara.com/question/Q8179.png)
$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta \\ $$$$=\frac{\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta}\:\:\left[\because\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \theta=\mathrm{1}\right] \\ $$$$\mathrm{Divide}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{Denominator}\:\mathrm{by}\:\mathrm{cos}^{\mathrm{2}} \theta \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\frac{\frac{\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{cos}^{\mathrm{2}} \theta}}{\frac{\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{cos}^{\mathrm{2}} \theta}}=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta} \\ $$$$\_\_\_\_\_\_\_\_ \\ $$$$\mathrm{put}\:\theta=\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{45}°=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${x}=\mathrm{tan}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{1}+{x}^{\mathrm{2}} =\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}^{\mathrm{2}} =\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}−\mathrm{1}}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\pm\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\:\mathrm{or}\:\mathrm{tan}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}=\pm\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\mathrm{Since}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}<\mathrm{90},\:\mathrm{tan}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}>\mathrm{0}\:\mathrm{so}\:\mathrm{we}\:\mathrm{discard}\:\mathrm{the}\:−\mathrm{ve}\:\mathrm{root}. \\ $$$$\mathrm{hence}\:{x}=\mathrm{tan}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$