If $x+\frac{1}{x} = 2\cos\theta$, then find the value of $x^n + \frac{1}{x^n}$.

Question

If $x+\frac{1}{x} = 2\cos\theta$, then find the value of $x^n + \frac{1}{x^n}$.

I don't even know how to start this question, pls help. Thank you :)

Answer 1

Simply consider it a quadratic equation in $x$.

$x^2 + 1 = 2x\cos\theta $

$x^2 - 2x\cos\theta + 1 = 0 $ By quadratic formula,

$$x = \frac{2\cos\theta \pm \sqrt{4\cos^2\theta - 4} }{2} = \cos\theta \pm \sqrt{\cos^2\theta - 1} = \cos\theta \pm i\sin\theta = e^{\pm i\theta }$$

$$ \frac{1}{x} = e^{\mp i\theta }$$

$$ x^n = e^{\pm in\theta } = \cos(n\theta) \pm i\sin(n\theta)$$ $$ \frac{1}{x^n} = e^{\mp in\theta } = \cos(n\theta) \mp i\sin(n\theta)$$

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

Answer 2

Hint: $2\cos\theta=e^{i\theta}+e^{-i\theta}.$

Answer 3

- 1 has to be positive for solving the quadratic equation with real roots

this gives = zero

Hence x = 1

Answer 4

$$z+\frac{1}{z}=2\cos\theta\iff z^2+1=2z\cos\theta \\ z_{1,2}=\cos\theta\pm i\sin\theta=e^{\pm i\theta}\implies \frac{1}{z_{1,2}}=\cos\theta\mp i\sin\theta=e^{\mp i\theta}$$Using De Moivre's formula, we get the following $$ z_{1,2}^n=\cos n\theta\pm i\sin n\theta \tag1$$ $$\frac{1}{z_{1,2}^n}=\cos n\theta \mp i\sin n\theta \tag2$$Adding equations $(1)$ and $(2)$, we get the desired result$$z^n+\frac{1}{z^n}=2\cos n\theta$$

Answer 5

Assume the required quantity is $2\cos n\theta$ (you can see this squaring the given identity). Then apply the induction using the identity $$x^{n+1}+\dfrac{1}{x^{n+1}}=\left(x+\dfrac1x\right)\left(x^{n}+\dfrac{1}{x^{n}}\right)-\left(x^{n-1}+\dfrac{1}{x^{n-1}}\right).$$