Show that, (1+cos theta + i*sin theta)^n + (1+cos theta - i*sin theta)^n = 2^(n+1) * (cos theta/2)^n * cos (n*theta/2) ?

1 Answer
Jan 27, 2018

Please see below.

Explanation:

Let 1+costheta+isintheta=r(cosalpha+isinalpha), here r=sqrt((1+costheta)^2+sin^2theta)=sqrt(2+2costheta)
= sqrt(2+4cos^2(theta/2)-2)=2cos(theta/2)

and tanalpha=sintheta/(1+costheta)==(2sin(theta/2)cos(theta/2))/(2cos^2(theta/2))=tan(theta/2) or alpha=theta/2

then 1+costheta-isintheta=r(cos(-alpha)+isin(-alpha))=r(cosalpha-isinalpha)

and we can write (1+costheta+isintheta)^n+(1+costheta-isintheta)^n using DE MOivre's theorem as

r^n(cosnalpha+isinnalpha+cosnalpha-isinnalpha)

= 2r^ncosnalpha

= 2*2^ncos^n(theta/2)cos((ntheta)/2)

= 2^(n+1)cos^n(theta/2)cos((ntheta)/2)