# If 2costheta=x+1/x then prove that the value of the x^n+1/(x^n), ninN is 2cosntheta?

Aug 1, 2018

#### Explanation:

By complex numbers,

${i}^{2} = - 1$

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

Let

$x = \cos \theta + i \sin \theta$

$\frac{1}{x} = \frac{1}{\cos \theta + i \sin \theta} = \frac{1}{\cos \theta + i \sin \theta} \cdot \frac{\cos \theta - i \sin \theta}{\cos \theta - i \sin \theta}$

$= \frac{\cos \theta - i \sin \theta}{{\cos}^{2} \theta - {i}^{2} {\sin}^{2} \theta} = \cos \theta - i \sin \theta$

$x + \frac{1}{x} = 2 \cos \theta$

By Demoivre's theorem

For $n \in \mathbb{N}$

${x}^{n} = {\left(\cos \theta + i \sin \theta\right)}^{n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right)$

$\frac{1}{x} ^ n = {\left(\cos \theta - i \sin \theta\right)}^{n} = \cos \left(n \theta\right) - i \sin \left(n \theta\right)$

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

Aug 1, 2018

Given

$x + \frac{1}{x} = 2 \cos \theta$

We get

${x}^{2} - 2 \cos \theta \cdot x + 1 = 0$

By sridharacharya's formula we get

$x = \frac{2 \cos \theta \pm \sqrt{{\left(2 \cos \theta\right)}^{2} - 4 \cdot 1 \cdot 1}}{2}$

$\implies x = \frac{2 \cos \theta \pm \sqrt{- 4 \left(1 - {\cos}^{2} \theta\right)}}{2}$

$\implies x = \cos \theta \pm i \sin \theta$

When $x = \cos \theta + i \sin \theta$

then $\frac{1}{x} = \frac{1}{\cos \theta + i \sin \theta}$

$= \frac{\cos \theta - i \sin \theta}{\left(\cos \theta - i \sin \theta\right) \left(\cos \theta + i \sin \theta\right)}$

$= \frac{\cos \theta - i \sin \theta}{{\cos}^{2} \theta + {\sin}^{2} \theta}$

$= \cos \theta - i \sin \theta$

So ${x}^{n} = {\left(\cos \theta + i \sin \theta\right)}^{n}$

$= \cos n \theta + i \sin n \theta$

[by De Moivre's Theorem.]

And similarly

$\frac{1}{x} ^ n = {\left(\cos \theta - i \sin \theta\right)}^{n}$

$= \cos n \theta - i \sin n \theta$

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

Similarly we get the same result when $x = \cos \theta - i \sin \theta$

Aug 1, 2018

By de Moivre's identity we know

 \cos \theta = (e^(i \theta)+e^(-i\theta))/2

then

 2\cos \theta = e^(i \theta)+e^(-i\theta)

then

 e^(i n\theta)+e^(-i n\theta) = 2\cos(n\theta)