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Question Number 43000 by abdo.msup.com last updated on 06/Sep/18

prove that (√(2+(√(2+....+(√2))))) =2cos((π/2^n ))

$${prove}\:{that}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+....+\sqrt{\mathrm{2}}}}\:\:=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right) \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18

cos2α=2cos^2 α−1 cos^2 α=((1+cos2α)/2) 2cosα=2(√((1+cos2α)/2)) 2cosα=(√(2(1+cos2α))) 2cos(Π/2^2 )=(√(2(1+cos(Π/2)))) =(√2) 2cos(Π/2^3 )=(√(2(1+cos(Π/2^2 )))) =(√(2(1+(1/(√2))))) =(√(2+(√2) )) 2cos(Π/2^4 )=(√(2(1+cos(Π/2^3 )))) =(√(2(1+((√(2+(√2)))/2)))) =(√(2+(√(2+(√2) )))) thus 2cos(Π/2^n )=(√(2+(√(2+(√(2+...)) )))) hence proved pls check

$${cos}\mathrm{2}\alpha=\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{1} \\ $$$${cos}^{\mathrm{2}} \alpha=\frac{\mathrm{1}+{cos}\mathrm{2}\alpha}{\mathrm{2}} \\ $$$$\mathrm{2}{cos}\alpha=\mathrm{2}\sqrt{\frac{\mathrm{1}+{cos}\mathrm{2}\alpha}{\mathrm{2}}}\: \\ $$$$\mathrm{2}{cos}\alpha=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{2}\alpha\right)}\: \\ $$$$\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\frac{\Pi}{\mathrm{2}}\right)}\:=\sqrt{\mathrm{2}}\:\: \\ $$$$\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{3}} }=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{2}} }\right)}\:=\sqrt{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)}\:=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}\:} \\ $$$$\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{4}} }=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\frac{\Pi}{\mathrm{2}^{\mathrm{3}} }\right)}\:=\sqrt{\mathrm{2}\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\right)}\:=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}\:}} \\ $$$${thus}\:\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{{n}} }=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+...}\:}} \\ $$$${hence}\:{proved} \\ $$$${pls}\:{check} \\ $$

Commented by maxmathsup by imad last updated on 06/Sep/18

yes correct i can say that you have using a recurrence proof...

$${yes}\:{correct}\:{i}\:{can}\:{say}\:{that}\:{you}\:{have}\:{using}\:{a}\:{recurrence}\:{proof}... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18

thank you sir...

$${thank}\:{you}\:{sir}... \\ $$

Answered by behi83417@gmail.com last updated on 07/Sep/18

y=(√(2+(√(2+(√(........(√2))))))) y^2 =2+y⇒y^2 −y−2=0 y=((1±(√(1+8)))/2)=((1+3)/2)=2 lim_(n→∞) (2cos(π/2^n ))=2

$${y}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{........\sqrt{\mathrm{2}}}}} \\ $$$${y}^{\mathrm{2}} =\mathrm{2}+{y}\Rightarrow{y}^{\mathrm{2}} −{y}−\mathrm{2}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{8}}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}}=\mathrm{2} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}{cos}\frac{\pi}{\mathrm{2}^{{n}} }\right)=\mathrm{2} \\ $$

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