Question Number 104727 by bemath last updated on 23/Jul/20

Given x + (1/x) = 2cos θ  find x^n +(1/x^n ) = ?

Commented bybemath last updated on 23/Jul/20

thank you all

Answered by mr W last updated on 23/Jul/20

for x∈R  for x>0: 2≤x+(1/x)=2 cos θ≤2  ⇒x+(1/x)=2 ⇒x=1 ⇒x^n +(1/x^n )=2    for x<0: −2≥x+(1/x)=2 cos θ≥−2  ⇒x+(1/x)=−2 ⇒x=−1 ⇒x^n +(1/x^n )=(−1)^n 2

Commented by1549442205PVT last updated on 23/Jul/20

Great Sir!

Answered by john santu last updated on 23/Jul/20

we solve the given equation   x^2 −2xcos θ +1 = 0 for x . Discriminat   Δ= 4cos ^2 θ−4 = −4sin ^2 θ  so the roots are cos θ ± i sin θ  By De Moivre′s formula  x^n  = cos (nθ) +i sin (nθ)  x^(−n)  = cos (−nθ)+i sin (−nθ)  and therefore   x^n  +x^(−n)  = {cos (nθ)+i sin (nθ)} +                         { cos (−nθ)+i sin (−nθ)}  x^n +(1/x^n ) = 2cos (nθ)   (JS ⊛ )

Answered by Dwaipayan Shikari last updated on 23/Jul/20

x^2 −2xcosθ+1=0  x=((2cosθ+(√(4cos^2 θ−4)))/2)=cosθ+2isinθ  x^n =cosnθ+2isinnθ  (De moivre′s theorem)  (1/x^n )=cosnθ−2isinnθ  x^n +(1/x^n )=2cosnθ

Answered by mathmax by abdo last updated on 24/Jul/20

x+(1/x) =2cosθ ⇒x^2  +1 =2xcosθ ⇒x^2  −2xcosθ +1 =0  Δ^′  =cos^2 θ−1 =−sin^2 θ =(isinθ)^2  ⇒z_1 =cosθ +isinθ =e^(iθ)  and  z_2 =cosθ −isinθ =e^(−iθ)   case 1 ) x =z_1  ⇒x^n  +(1/x^n ) =z_1 ^n  +(1/z_1 ^n )  = cos(nθ) +isin(nθ)+(1/(cos(nθ)+isin(nθ)))  =cos(nθ) +isin(nθ)+cos(nθ)−isin(nθ) =2cos(nθ)  case 2) x =z_2     ⇒x^(n )  +(1/x^n ) =2cos(nθ) because z_2 =z_1 ^− )