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


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
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
	$we solve the given equation x^2 −2xcos θ +1 = 0 for x . Discriminat Δ= 4cos^2 θ−4 = −4sin^2 θ so the roots are cos θ ± i sin θ By De Moivre′s formula x^n = cos (nθ) +i sin (nθ) x^(−n) = cos (−nθ)+i sin (−nθ) and therefore x^n +x^(−n) = {cos (nθ)+i sin (nθ)} + { cos (−nθ)+i sin (−nθ)} x^n +(1/x^n ) = 2cos (nθ) (JS ⊛ )$
	$w e s o l v e t h e g i v e n e q u a t i o n$ $x^{2} - 2 x \cos θ + 1 = 0 f o r x . D i s c r i m i n a t$ $Δ = 4 \cos^{2} θ - 4 = - 4 \sin^{2} θ$ $s o t h e r o o t s a r e \cos θ \pm i \sin θ$ $B y D e {M o i v r e}^{'} s f o r m u l a$ $x^{n} = \cos (n θ) + i \sin (n θ)$ $x^{- n} = \cos (- n θ) + i \sin (- n θ)$ $a n d t h e r e f o r e$ $x^{n} + x^{- n} = {\cos (n θ) + i \sin (n θ)} +$ ${\cos (- n θ) + i \sin (- n θ)}$ $x^{n} + \frac{1}{x^{n}} = 2 \cos (n θ)$ $(J S ⊛)$
	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

	$x + \frac{1}{x} = 2 \cos θ \Rightarrow x^{2} + 1 = 2 xcos θ \Rightarrow x^{2} - 2 xcos θ + 1 = 0$ $Δ^{^{'}} = \cos^{2} θ - 1 = - \sin^{2} θ = {(isin θ)}^{2} \Rightarrow z_{1} = \cos θ + isin θ = e^{i θ} and$ $z_{2} = \cos θ - isin θ = e^{- i θ}$ $case 1) x = z_{1} \Rightarrow x^{n} + \frac{1}{x^{n}} = z_{1}^{n} + \frac{1}{z_{1}^{n}} = \cos (n θ) + isin (n θ) + \frac{1}{\cos (n θ) + isin (n θ)}$ $= \cos (n θ) + isin (n θ) + \cos (n θ) - isin (n θ) = 2 \cos (n θ)$ $c ase 2) x = z_{2} \Rightarrow x^{n} + \frac{1}{x^{n}} = 2 \cos (n θ) because z_{2} = {\bar{z}}_{1})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum