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Question Number 127206 by MathSh last updated on 27/Dec/20
Prove it:  sin(π/(2m+1))∙sin((2π)/(2m+1))∙...∙sin((mπ)/(2m+1))=((√(2m+1))/2^m )
Proveit:sinπ2m+1sin2π2m+1sinmπ2m+1=2m+12m
Answered by Olaf last updated on 28/Dec/20
  Ω = Π_(k=1) ^(k=m) sin(((kπ)/(2m+1)))  Solve (z+1)^(2m+1)  = 1 (1)  z_k  = e^((2ikπ)/(2m+1)) −1, k = 0,1,2,...,2m  z_k  = 2ie^((ikπ)/(2m+1)) sin(((kπ)/(2m+1)))  Let P(z) = Σ_(k=0) ^(2m) (z+1)^k  (2)  ⇒ P(0) = 2m+1 (3)  P(z) = 0 ⇔ Σ_(k=0) ^(2m) (z+1)^k  = 0  ⇔ (((z+1)^(2m+1) −1)/((z+1)−1)) = 0, z≠0  ⇔ z = z_k , k = 1,2,...,2m (4)  (2) and (4) :  P(z) = Π_(k=1) ^(2m) (z−z_k )  ⇒ P(0) = Π_(k=1) ^(2m) (−z_k ) = Π_(k=1) ^(2m) z_k   P(0) = Π_(k=1) ^(2m) [2ie^((ikπ)/(2m+1)) sin(((kπ)/(2m+1)))]  P(0) = 2^(2m) i^(2m) e^(((2m(2m+1))/2).((iπ)/(2m+1))) Π_(k=1) ^(2m) sin(((kπ)/(2m+1)))  P(0) = 2^(2m) (−1)^m e^(imπ) Π_(k=1) ^(2m) sin(((kπ)/(2m+1)))  P(0) = 2^(2m) Π_(k=1) ^(2m) sin(((kπ)/(2m+1))) (5)  (3) and (5) :  P(0) = 2m+1 = 2^(2m) Π_(k=1) ^(2m) sin(((kπ)/(2m+1)))  ⇒ Π_(k=1) ^(2m) sin(((kπ)/(2m+1))) = ((2m+1)/2^(2m) ) (6)  But sin(((kπ)/(2m+1))) = sin(π−((kπ)/(2m+1)))  = sin((((2m+1−k)π)/(2m+1)))  For example :  sin((π/(2m+1))) = sin(((2mπ)/(2m+1)))  ⇒ Π_(k=1) ^(2m) sin(((kπ)/(2m+1))) = Π_(k=1) ^m sin^2 (((kπ)/(2m+1)))  = (Π_(k=1) ^m sin(((kπ)/(2m+1))))^2  = Ω^2  (7)  (6) and (7) : Ω = ((√(2m+1))/2^m )
Ω=k=mk=1sin(kπ2m+1)Solve(z+1)2m+1=1(1)zk=e2ikπ2m+11,k=0,1,2,,2mzk=2ieikπ2m+1sin(kπ2m+1)LetP(z)=2mk=0(z+1)k(2)P(0)=2m+1(3)P(z)=02mk=0(z+1)k=0(z+1)2m+11(z+1)1=0,z0z=zk,k=1,2,,2m(4)(2)and(4):P(z)=2mk=1(zzk)P(0)=2mk=1(zk)=2mk=1zkP(0)=2mk=1[2ieikπ2m+1sin(kπ2m+1)]P(0)=22mi2me2m(2m+1)2.iπ2m+12mk=1sin(kπ2m+1)P(0)=22m(1)meimπ2mk=1sin(kπ2m+1)P(0)=22m2mk=1sin(kπ2m+1)(5)(3)and(5):P(0)=2m+1=22m2mk=1sin(kπ2m+1)2mk=1sin(kπ2m+1)=2m+122m(6)Butsin(kπ2m+1)=sin(πkπ2m+1)=sin((2m+1k)π2m+1)Forexample:sin(π2m+1)=sin(2mπ2m+1)2mk=1sin(kπ2m+1)=mk=1sin2(kπ2m+1)=(mk=1sin(kπ2m+1))2=Ω2(7)(6)and(7):Ω=2m+12m
Commented by MathSh last updated on 28/Dec/20
Thanks sir
Thankssir
Answered by mnjuly1970 last updated on 28/Dec/20
z^(2m+1) −1=0  , z∈C(roots::=1,z_1 ,...,z_(2m) )   z=re^(iθ)    r^(2m+1) e^(i(2m+1)θ) =1=1.e^(i2kπ)    θ_k =((2kπ)/(2m+1)) ⇒z_k =e^((i(2kπ))/(2m+1))  (k∈{0,1,2,...,2m})   (z−1)(z^(2m) +z^(2m−1) +...+z+1)=0    P(z)=z^(2m) +z^(2m−1) +...+z+1     P(z)=(z−z_1 )(z−z_2 )...(z−z_(2m) )      P(1)=(1−z_1 )(1−z_2 )...(1−z_(2m) )=2m+1     ∣1−z_1 ∣∣z−z_2 ∣....∣z−z_(2m) ∣=2m+1     ∣1−e^((i2π)/(2m+1)) ∣∣1−e^((i4π)/(2m+1)) ∣...∣1−e^((i(2(2mπ)))/(2m+1)) ∣=2m+1    2^(2m) sin((π/(2m+1)))sin(((2π)/(2m+1)))...sin(((mπ)/(2m+1)))sin((((m+1)π)/(2m+1)))...sin(((2mπ)/(2m+1)))=2m+1(∗)  note:∣sin(θ)−icos(θ)∣=1    note:1−cos(2θ)=2sin^2 (θ)          note: sin(2θ)=2sin(θ)cos(θ)       note:sin(((2mπ)/(2m+1)))=sin(π−((2mπ)/(2m+1)))=sin((π/(2m+1)))   ⋮  sin((((m+1)π)/(2m+1)))=sin(π−(((m+1)π)/(2m+1)))=sin(((mπ)/(2m+1)))    example:∣1−e^((i2π)/(2m+1)) ∣=∣1−cos(((2π)/(2m+1)))−isin(((2π)/(2m+1)))∣  =∣2sin^2 ((π/(2m+1)))−2isin((π/(2m+1)))cos((π/(2m+1)))∣  =2sin((π/(2m+1)))∣sin((π/(2m+1)))−icos((π/(2m+1)))∣=2sin((π/(2m+1)))  from (∗) and( notes)=::  2^(2m) (sin((π/(2m+1)))...sin(((mπ)/(2m+1))))^2 =2m+1    sin((π/(2m+1)))...sin(((mπ)/(2m+1)))=((√(2m+1))/2^m ) ✓
z2m+11=0,zC(roots::=1,z1,,z2m)z=reiθr2m+1ei(2m+1)θ=1=1.ei2kπθk=2kπ2m+1zk=ei(2kπ)2m+1(k{0,1,2,,2m})(z1)(z2m+z2m1++z+1)=0P(z)=z2m+z2m1++z+1P(z)=(zz1)(zz2)(zz2m)P(1)=(1z1)(1z2)(1z2m)=2m+11z1∣∣zz2.zz2m∣=2m+11ei2π2m+1∣∣1ei4π2m+11ei(2(2mπ))2m+1∣=2m+122msin(π2m+1)sin(2π2m+1)sin(mπ2m+1)sin((m+1)π2m+1)sin(2mπ2m+1)=2m+1()note:∣sin(θ)icos(θ)∣=1note:1cos(2θ)=2sin2(θ)note:sin(2θ)=2sin(θ)cos(θ)note:sin(2mπ2m+1)=sin(π2mπ2m+1)=sin(π2m+1)sin((m+1)π2m+1)=sin(π(m+1)π2m+1)=sin(mπ2m+1)example:∣1ei2π2m+1∣=∣1cos(2π2m+1)isin(2π2m+1)=∣2sin2(π2m+1)2isin(π2m+1)cos(π2m+1)=2sin(π2m+1)sin(π2m+1)icos(π2m+1)∣=2sin(π2m+1)from()and(notes)=::22m(sin(π2m+1)sin(mπ2m+1))2=2m+1sin(π2m+1)sin(mπ2m+1)=2m+12m
Commented by MathSh last updated on 28/Dec/20
Thanks sir
Thankssir
Commented by mnjuly1970 last updated on 28/Dec/20
you are welcome...
youarewelcome

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