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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 ))
provethat2+2+.+2=2cos(π2n)
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
cos2α=2cos2α1cos2α=1+cos2α22cosα=21+cos2α22cosα=2(1+cos2α)2cosΠ22=2(1+cosΠ2)=22cosΠ23=2(1+cosΠ22)=2(1+12)=2+22cosΠ24=2(1+cosΠ23)=2(1+2+22)=2+2+2thus2cosΠ2n=2+2+2+henceprovedplscheck
Commented by maxmathsup by imad last updated on 06/Sep/18
yes correct i can say that you have using a recurrence proof...
yescorrecticansaythatyouhaveusingarecurrenceproof
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
thank you sir...
thankyousir
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=2+2+..2y2=2+yy2y2=0y=1±1+82=1+32=2limn(2cosπ2n)=2

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