DeMoivre's theorem: cos3θ in powers of cosθ

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  • #1

Homework Statement

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

Homework Equations

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

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
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  • #2
use i^2=-1 to write (cosθ)^3 + 3[(cosθ)^2][isinθ] + 3(cosθ)[(isinθ)^2] + (isinθ)^3 in a+bi form
  • #3
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
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
Okay, thanks. I've got the answer.

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