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Question Number 153009 by SANOGO last updated on 04/Sep/21
montrer que:  2cos(π/2^n )=(√(2 +(√(2+...+(√2)))))
$${montrer}\:{que}: \\ $$$$\mathrm{2}{cos}\frac{\pi}{\mathrm{2}^{{n}} }=\sqrt{\mathrm{2}\:+\sqrt{\mathrm{2}+…+\sqrt{\mathrm{2}}}} \\ $$$$ \\ $$
Commented by mr W last updated on 04/Sep/21
please make your posts better readable!  try to set the defaut font size 20.  try to avoid using extremly large or  extremly small fonts!
$${please}\:{make}\:{your}\:{posts}\:{better}\:{readable}! \\ $$$${try}\:{to}\:{set}\:{the}\:{defaut}\:{font}\:{size}\:\mathrm{20}. \\ $$$${try}\:{to}\:{avoid}\:{using}\:{extremly}\:{large}\:{or} \\ $$$${extremly}\:{small}\:{fonts}! \\ $$
Commented by SANOGO last updated on 04/Sep/21
ok thank you i understood
$${ok}\:{thank}\:{you}\:{i}\:{understood} \\ $$
Answered by puissant last updated on 04/Sep/21
montrons alors que ∀n∈N, 2cos(π/2^n )=(√(2+(√(2+....+(√2)))))  ⇒ cos(π/2^n )=(1/2)(√(2+(√(2+(√(2+..(√2)))))))  →pour n=1 la propriete^�  est vraie..  →soit n∈N, supposons la propriete^�  vraie  au rang n et prouvons au rang n+1.  Remarquons au pre^� alable que ∀x∈[0;(π/2)]  comme cos^2 x=((cos2x+1)/2)⇒cos((x/2))=(√((cosx+1)/2))  et cos((π/2^(n+1) ))=cos((1/2)((π/2^n )))  =(√((1/2)(cos((π/2^n ))+1)))  =(√((1/2)((1/2)(√(2+(√(2+...+(√2)))))+1))) on a ainsi  n radicaux d′apres l′hypothese de reccurence.  =(1/2)(√(2+(√(2+...+(√2)))))  (n+1 radicaux)  on prouve ainsi la propriete^�  par reccurence..
$${montrons}\:{alors}\:{que}\:\forall{n}\in\mathbb{N},\:\mathrm{2}{cos}\frac{\pi}{\mathrm{2}^{{n}} }=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+….+\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow\:{cos}\frac{\pi}{\mathrm{2}^{{n}} }=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+..\sqrt{\mathrm{2}}}}} \\ $$$$\rightarrow{pour}\:{n}=\mathrm{1}\:{la}\:{propriet}\acute {{e}}\:{est}\:{vraie}.. \\ $$$$\rightarrow{soit}\:{n}\in\mathbb{N},\:{supposons}\:{la}\:{propriet}\acute {{e}}\:{vraie} \\ $$$${au}\:{rang}\:{n}\:{et}\:{prouvons}\:{au}\:{rang}\:{n}+\mathrm{1}. \\ $$$${Remarquons}\:{au}\:{pr}\acute {{e}alable}\:{que}\:\forall{x}\in\left[\mathrm{0};\frac{\pi}{\mathrm{2}}\right] \\ $$$${comme}\:{cos}^{\mathrm{2}} {x}=\frac{{cos}\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}}\Rightarrow{cos}\left(\frac{{x}}{\mathrm{2}}\right)=\sqrt{\frac{{cosx}+\mathrm{1}}{\mathrm{2}}} \\ $$$${et}\:{cos}\left(\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right)={cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\right) \\ $$$$=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)+\mathrm{1}\right)} \\ $$$$=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…+\sqrt{\mathrm{2}}}}+\mathrm{1}\right)}\:{on}\:{a}\:{ainsi} \\ $$$${n}\:{radicaux}\:{d}'{apres}\:{l}'{hypothese}\:{de}\:{reccurence}. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…+\sqrt{\mathrm{2}}}}\:\:\left({n}+\mathrm{1}\:{radicaux}\right) \\ $$$${on}\:{prouve}\:{ainsi}\:{la}\:{propriet}\acute {{e}}\:{par}\:{reccurence}.. \\ $$
Commented by SANOGO last updated on 04/Sep/21
merci bien que Dieu te benisse
$${merci}\:{bien}\:{que}\:{Dieu}\:{te}\:{benisse} \\ $$
Answered by som(math1967) last updated on 04/Sep/21
cos(𝛑/4)=(1/( (√2)))  2cos(𝛑/4)=(√2) ∴2cos(𝛑/2^2 )=(√2)  cos(𝛑/8)=(√((1+cos(𝛑/4))/2))  [∵cos(𝛉/2)=(√((1+cos𝛉)/2)) ]  cos(𝛑/8)=(√(((√2)+1)/(2(√2))))=(√((2+(√2))/4))=  2cos(𝛑/2^3 )=(√(2+(√2)))  same way  2cos(𝛑/2^4 )=(√(2+(√(2+(√2)))))  2cos(𝛑/2^5 )=(√(2+(√(2+(√(2+(√2)))))))  .......2cos(𝛑/2^n )=(√(2+(√(2+(√(2+...(√2)))))))
$$\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{4}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{2}\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{4}}=\sqrt{\mathrm{2}}\:\therefore\mathrm{2}\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{8}}=\sqrt{\frac{\mathrm{1}+\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{4}}}{\mathrm{2}}}\:\:\left[\because\boldsymbol{{cos}}\frac{\boldsymbol{\theta}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}+\boldsymbol{{cos}\theta}}{\mathrm{2}}}\:\right] \\ $$$$\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{8}}=\sqrt{\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}}=\sqrt{\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}}= \\ $$$$\mathrm{2}\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{2}^{\mathrm{3}} }=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$$${same}\:\boldsymbol{{way}}\:\:\mathrm{2}\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{2}^{\mathrm{4}} }=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}} \\ $$$$\mathrm{2}\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{2}^{\mathrm{5}} }=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}} \\ $$$$…….\mathrm{2}\boldsymbol{{cos}}\frac{\boldsymbol{\pi}}{\mathrm{2}^{\boldsymbol{{n}}} }=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…\sqrt{\mathrm{2}}}}} \\ $$
Commented by SANOGO last updated on 04/Sep/21
merci beaucou le dur
$${merci}\:{beaucou}\:{le}\:{dur} \\ $$
Answered by Jonathanwaweh last updated on 05/Sep/21
montrer qu:  2cos(π/2^n )=(√(2 +(√(2+...+(√2)))))  posons U_n =(√(2+(√(2+...+(√(2 )))))) alors U_(n+1) =(√(U_n +2))  soit p(n) :∀nεN n≥2 2cos(Π/2^n )=U_n   1−initialisation(facile)  2−heredite  pour n≥2 fixe^′  supposons p(n) vraie et montrons que p(n+1) l′est aussi  ie 2cos(Π/2^(n+1) )=U_(n+1) . on a cos((Π/2^n ))=cos((Π/2^(n+1) )×2)                                                  =2cos^2 ((Π/2^(n+1) ))−1  donc 2cos((Π/2^(n+1) ))=4cos^2 ((Π/2^(n+1) ))−2  ie (√(2cos((Π/2^n ))+2))=2cos((Π/2^(n+1) ))  d^′ ou ona U_(n+1) =2cos((Π/2^(n+1) ))  conclusion:...
$${montrer}\:{qu}: \\ $$$$\mathrm{2}{cos}\frac{\pi}{\mathrm{2}^{{n}} }=\sqrt{\mathrm{2}\:+\sqrt{\mathrm{2}+…+\sqrt{\mathrm{2}}}} \\ $$$${posons}\:{U}_{{n}} =\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…+\sqrt{\mathrm{2}\:}}}\:{alors}\:{U}_{{n}+\mathrm{1}} =\sqrt{{U}_{{n}} +\mathrm{2}} \\ $$$${soit}\:{p}\left({n}\right)\::\forall{n}\epsilon{N}\:{n}\geqslant\mathrm{2}\:\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{{n}} }={U}_{{n}} \\ $$$$\mathrm{1}−{initialisation}\left({facile}\right) \\ $$$$\mathrm{2}−{heredite} \\ $$$${pour}\:{n}\geqslant\mathrm{2}\:{fixe}^{'} \:{supposons}\:{p}\left({n}\right)\:{vraie}\:{et}\:{montrons}\:{que}\:{p}\left({n}+\mathrm{1}\right)\:{l}'{est}\:{aussi} \\ $$$${ie}\:\mathrm{2}{cos}\frac{\Pi}{\mathrm{2}^{{n}+\mathrm{1}} }={U}_{{n}+\mathrm{1}} .\:{on}\:{a}\:{cos}\left(\frac{\Pi}{\mathrm{2}^{{n}} }\right)={cos}\left(\frac{\Pi}{\mathrm{2}^{{n}+\mathrm{1}} }×\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\Pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right)−\mathrm{1} \\ $$$${donc}\:\mathrm{2}{cos}\left(\frac{\Pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right)=\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\Pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right)−\mathrm{2} \\ $$$${ie}\:\sqrt{\mathrm{2}{cos}\left(\frac{\Pi}{\mathrm{2}^{{n}} }\right)+\mathrm{2}}=\mathrm{2}{cos}\left(\frac{\Pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right) \\ $$$${d}^{'} {ou}\:{ona}\:{U}_{{n}+\mathrm{1}} =\mathrm{2}{cos}\left(\frac{\Pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right) \\ $$$${conclusion}:… \\ $$
Commented by SANOGO last updated on 05/Sep/21
merci bien mon chef
$${merci}\:{bien}\:{mon}\:{chef} \\ $$

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