Question

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

Answer 1

Please see the explanation below.

Explanation:

By complex numbers,

`i^2=-1`

`cos^2theta+sin^2theta=1`

Let

`x=costheta+isintheta`

`1/x=1/(costheta+isintheta)=1/(costheta+isintheta)*(costheta-isintheta)/(costheta-isintheta)`

`=(costheta-isintheta)/(cos^2theta-i^2sin^2theta)=costheta-isintheta`

`x+1/x=2costheta`

By Demoivre's theorem

For `n in NN`

`x^n=(costheta+isintheta)^n=cos(ntheta)+isin(ntheta)`

`1/x^n=(costheta-isintheta)^n=cos(ntheta)-isin(ntheta)`

`x^n+1/x^n=2cos(ntheta)`

Answer 2

Given

`x+1/x=2costheta`

We get

`x^2-2costheta*x+1=0`

By sridharacharya's formula we get

`x=(2costhetapmsqrt((2costheta)^2-4*1*1))/2`

`=>x=(2costhetapmsqrt(-4(1-cos^2theta)))/2`

`=>x=costhetapmisintheta`

When `x=costheta+isintheta`

then `1/x=1/(costheta+isintheta)`

`=(costheta-isintheta)/((costheta-isintheta)(costheta+ isintheta))`

`=(costheta-isintheta)/(cos^2theta+sin^2theta)`

`=costheta-isintheta`

So `x^n=(costheta+isintheta)^n`

`=cosntheta+isinntheta`

[by De Moivre's Theorem.]

And similarly

`1/x^n=(costheta-isintheta)^n`

`=cosntheta-isinntheta`

Hence adding we get

`x^n+1/x^n=2cosntheta`

Similarly we get the same result when `x=costheta-isintheta`

Answer 3

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) #