Question Number 43000 by abdo.msup.com last updated on 06/Sep/18
$${prove}\:{that}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+….+\sqrt{\mathrm{2}}}}\:\:=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
$${cos}\mathrm{2}\alpha=\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{1} \\ $$$${cos}^{\mathrm{2}} \alpha=\frac{\mathrm{1}+{cos}\mathrm{2}\alpha}{\mathrm{2}} \\ $$$$\mathrm{2}{cos}\alpha=\mathrm{2}\sqrt{\frac{\mathrm{1}+{cos}\mathrm{2}\alpha}{\mathrm{2}}}\: \\ $$$$\mathrm{2}{cos}\alpha=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{2}\alpha\right)}\: \\ $$$$\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\frac{\Pi}{\mathrm{2}}\right)}\:=\sqrt{\mathrm{2}}\:\: \\ $$$$\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{3}} }=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{2}} }\right)}\:=\sqrt{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)}\:=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}\:} \\ $$$$\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{4}} }=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{3}} }\right)}\:=\sqrt{\mathrm{2}\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\right)}\:=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}\:}} \\ $$$${thus}\:\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{{n}} }=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}\:}} \\ $$$${hence}\:{proved} \\ $$$${pls}\:{check} \\ $$
Commented by maxmathsup by imad last updated on 06/Sep/18
$${yes}\:{correct}\:{i}\:{can}\:{say}\:{that}\:{you}\:{have}\:{using}\:{a}\:{recurrence}\:{proof}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
$${thank}\:{you}\:{sir}… \\ $$
Answered by behi83417@gmail.com last updated on 07/Sep/18
$${y}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{……..\sqrt{\mathrm{2}}}}} \\ $$$${y}^{\mathrm{2}} =\mathrm{2}+{y}\Rightarrow{y}^{\mathrm{2}} −{y}−\mathrm{2}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{8}}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}}=\mathrm{2} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}{cos}\frac{\pi}{\mathrm{2}^{{n}} }\right)=\mathrm{2} \\ $$