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Use De Moivre's Theorem to prove this:

  1. Jan 10, 2012 #1
    Use De Moivre's Theorem to show that for any n greater that equal to 1

    (1+itanθ)n + (1-itanθ)n =2cosnθ/cosnθ

    where cosθ ≠ 0

    I tried to approach this by converting into modulus argument form but wasn't really sure if that was correct. It's a common New South Wales HSC question but I couldn't find a solution anywhere. Help would be greatly appreciated :)
  2. jcsd
  3. Jan 10, 2012 #2
    The first step would be to convert tanθ into [itex]\frac{sin\theta}{cos\theta}[/itex] and work from there.
  4. Jan 10, 2012 #3
    Thanks a bunch - that helped a lot. Converted it into:
    [(secθ)(cosθ+isinθ)]^n + [(secθ)(cosθ-isinθ)]^n and it was easy from there.

  5. Jan 10, 2012 #4
    No problem. Good luck with the HSC, I just finished mine :P
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