# Use De Moivre's Theorem to prove this:

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 :)

The first step would be to convert tanθ into $\frac{sin\theta}{cos\theta}$ and work from there.

Thanks a bunch - that helped a lot. Converted it into:
[(secθ)(cosθ+isinθ)]^n + [(secθ)(cosθ-isinθ)]^n and it was easy from there.

Cheers!

No problem. Good luck with the HSC, I just finished mine :P