# DeMoivre's theorem: cos3θ in powers of cosθ

1. Oct 30, 2011

### ZedCar

1. The problem statement, all variables and given/known data

Use de Moivre's Theorem to express cos3θ in powers of cosθ

2. Relevant equations

z^n = [r(cosθ + isinθ)]^n = r^n (cos(nθ) + i sin(nθ))

3. The attempt at a solution

cos3θ = Re(cos3θ +isin3θ) = Re[(cosθ +isinθ)^3]

I've then expanded the brackets using binomial theorem and got;

(cosθ)^3 + 3[(cosθ)^2][isinθ] + 3(cosθ)[(isinθ)^2] + (isinθ)^3

So (cosθ)^3 is the real part and 3[(cosθ)^2][isinθ] + 3(cosθ)[(isinθ)^2] + (isinθ)^3 the imaginary part.

If anyone has any suggestions...

Thank you

Last edited: Oct 30, 2011
2. Oct 30, 2011

### lurflurf

use i^2=-1 to write (cosθ)^3 + 3[(cosθ)^2][isinθ] + 3(cosθ)[(isinθ)^2] + (isinθ)^3 in a+bi form

3. Oct 30, 2011

### ZedCar

Okay, so doing that I'm getting:

(cosθ)^3 - 3cosθ(sinθ)^2 + 3i(cosθ)^2 (sinθ) - i(sinθ)^3

So the real part is (cosθ)^3 - 3cosθ(sinθ)^2

the imaginary part 3i(cosθ)^2 (sinθ) - i(sinθ)^3

4. Oct 30, 2011

### lurflurf

Yes, though the imaginary part is usually defined to not have the i
also the question asked for an answer in cosθ so eliminate sinθ

5. Oct 30, 2011

### ZedCar

Okay, thanks. I've got the answer.