# Prove using de moivre's theorem of I=cos^2n theta

## Homework Statement

Use de Moivre's Theorem and the result that integral (with limits pi, 0) of cos[nx]dx = 0 for all non-zero integers of n. when n does equal 0 the answer is 1.

[this i proved in the first section of the qu using logic and limits(x->0).

so we need to prove that

integral[with limits pi, 0] [(cos(x))2n]dx = [pi / (2)2n] * [(2n)! / (n!)2]

## Homework Equations

de moivre -> (cos(nx) + i*sin(nx)) = (cos(x) + i*sin(x))n

eulers -> (cos(x) + i*sin(x)) = eix

## The Attempt at a Solution

okay ive tried not using de moivres and trying to find a pattern by doing the integral a few times, and ive also thought that using eulers formula might help, but i seriously just cant get started...i know that i need to use the real part of the formula but i dont know what to do first.

this is a past exam paper by the way and my uni doesnt issue mark schemes, plus it being bank hols weekend no one will be there until wednesday

i appreciate any help or even just ideas on how to get started
thanks

cos(x) = (exp(ix) + exp(-ix))/2

Take both sides to the power 2n

Use the binomial theorem for the right-hand side

Integrate

well ive just done that between posting the qu and reading your reply and this is what ive done

cos(x)2n = 1 / 2^(2n) * (values of cos + (2n)! / ((n!)2))

the values of cos were found with setting n to various numbers and expanding
ie-
if n=2 :-values of cos were cos4x + 4cos2x
if n=3 :-values of cos were cos6x + 6cos4x + 15cos2x

but i dont know if i can do this without expanding and leaving n in there.

so with this equ integrated the cos becomes sin and then fall to 0 because integar numbers of pi is 0, and (2n)! / (n!)2 becomes (2n)! / ((n!)2) *x and the times this by the 1 / 2^(2n) gives the answer.

but once again i dont know whether i should know how to represent the 'values of cos' relating it to n [the main prob is the coefficients], cause i wont have this kind of time in the exam

Last edited:
tiny-tim
Homework Helper
Hi indie452! (have a pi: π )
Use de Moivre's Theorem and the result that integral (with limits pi, 0) of cos[nx]dx = 0 for all non-zero integers of n. when n does equal 0 the answer is 1.

[this i proved in the first section of the qu using logic and limits(x->0).

so we need to prove that

integral[with limits pi, 0] [(cos(x))2n]dx = [pi / (2)2n] * [(2n)! / (n!)2]
Use nealjking's two hints: cos(x) = (eix + e-ix)/2,

and (a + b)2n = ∑akb2n-k 2nCk

which coefficient(s) are the only ones with nonzero integral? 