Establish the identity\\n
1+z+z²+......+zⁿ=(1-zⁿ⁺¹)/(1-z))/(1-z);
(z ≠ 1) and then use it to derive Lagrange's trigonometric identity: 1 + cosθ + cos2θ + ··· + cosnθ = 1/2 + (sin[(2n + 1)θ/2])/2sin(0/2); (0 < θ < 2π). Suggestion:
As for the first identity, write
S= 1+z+z²+......+zⁿ
and consider the difference S - zS. To derive the second identity, write
z=eᶦθ
in the first one.
