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

Answer:

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) #