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

3 Answers
Aug 1, 2018

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)

Aug 1, 2018

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

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)