Help me derive Lagrange's Trigonometric Identity

[Ed Quanta]

Using the fact that z=e^ibeta

and the identity 1 + z + z^2...+z^n=1-z^(n+1)/1-z

Help me derive

1 + cosbeta +cos2beta + ... +cosnbeta= 1/2 + sin((2n+1)beta/2)/2sin(beta/2) where beta is between 0 and 2pi

 

[HallsofIvy]

z= e^iβ= cos(β)+ i sin(β)
z²= e^2iβ= cos(2β)+ i sin(2β)
.
.
.
zⁿ= e^niβ= cos(nβ)+ i sin(nβ

so the left hand side is the real part of 1+ z+ z²+...+zⁿ.


(By the way: the right hand side of the first sum should be (1- z^(n+1))/(1-z).
The parentheses are necessary!)