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

G i v e n x + \frac{1}{x} = 2 \cos θ

f i n d x^{n} + \frac{1}{x^{n}} = ?

Commented by bemath last updated on 23/Jul/20

t h a n k y o u a l l

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

f o r x \in R

f o r x > 0 : 2 ⩽ x + \frac{1}{x} = 2 \cos θ ⩽ 2

\Rightarrow x + \frac{1}{x} = 2 \Rightarrow x = 1 \Rightarrow x^{n} + \frac{1}{x^{n}} = 2

f o r x < 0 : - 2 ⩾ x + \frac{1}{x} = 2 \cos θ ⩾ - 2

\Rightarrow x + \frac{1}{x} = - 2 \Rightarrow x = - 1 \Rightarrow x^{n} + \frac{1}{x^{n}} = {(- 1)}^{n} 2

Commented by 1549442205PVT last updated on 23/Jul/20

Great Sir!

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

w e s o l v e t h e g i v e n e q u a t i o n

x^{2} - 2 x \cos θ + 1 = 0 f o r x . D i s c r i m i n a t

Δ = 4 \cos^{2} θ - 4 = - 4 \sin^{2} θ

s o t h e r o o t s a r e \cos θ \pm i \sin θ

B y D e {M o i v r e}^{'} s f o r m u l a

x^{n} = \cos (n θ) + i \sin (n θ)

x^{- n} = \cos (- n θ) + i \sin (- n θ)

a n d t h e r e f o r e

x^{n} + x^{- n} = {\cos (n θ) + i \sin (n θ)} +

{\cos (- n θ) + i \sin (- n θ)}

x^{n} + \frac{1}{x^{n}} = 2 \cos (n θ)

(J S ⊛)

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

x^{2} - 2 x c o s θ + 1 = 0

x = \frac{2 c o s θ + \sqrt{4 {c o s}^{2} θ - 4}}{2} = c o s θ + 2 i s i n θ

x^{n} = c o s n θ + 2 i s i n n θ (D e {m o i v r e}^{'} s t h e o r e m)

\frac{1}{x^{n}} = c o s n θ - 2 i s i n n θ

x^{n} + \frac{1}{x^{n}} = 2 c o s n θ

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

x + \frac{1}{x} = 2 \cos θ \Rightarrow x^{2} + 1 = 2 xcos θ \Rightarrow x^{2} - 2 xcos θ + 1 = 0

Δ^{^{'}} = \cos^{2} θ - 1 = - \sin^{2} θ = {(isin θ)}^{2} \Rightarrow z_{1} = \cos θ + isin θ = e^{i θ} and

z_{2} = \cos θ - isin θ = e^{- i θ}

case 1) x = z_{1} \Rightarrow x^{n} + \frac{1}{x^{n}} = z_{1}^{n} + \frac{1}{z_{1}^{n}} = \cos (n θ) + isin (n θ) + \frac{1}{\cos (n θ) + isin (n θ)}

= \cos (n θ) + isin (n θ) + \cos (n θ) - isin (n θ) = 2 \cos (n θ)

c ase 2) x = z_{2} \Rightarrow x^{n} + \frac{1}{x^{n}} = 2 \cos (n θ) because z_{2} = {\bar{z}}_{1})