Question

Hi, I'm having trouble understanding a step in the solution for z^(n)+(1/(z^n))=2cos(nθ). I will write the given solution below:

Answer 1

Using de Moivre's theorem:
\[z^{n}+\frac{1}{z^{n}}
=(\cos\theta+\imath\sin\theta)^{n}+(\cos\theta+\imath\sin\theta)^{-n}\]
\[=\cos n\theta +\imath\sin n\theta +\cos (-n \theta) + \imath\sin(-n \theta)\]
\[=\cos n\theta +\imath\sin n\theta +\cos (n \theta) - \imath\sin(n \theta)\]
\[=2\cos n\theta\]

But I am confused about how to get from the second to the third line. Specifically how does cos(-nθ)+isin(-nθ) become cos(nθ)-isin(nθ)?

Answer 2

I also need to show that z^n - 1/(z^n) = 2isin(nθ) but hopefully it will be clear once I understand the first problem. Thanks.

Answer 3

Because cos(-a)=cos(a), for every a, And because sin(-a)=-sin(a) for every a, you can replace them

Answer 4

(Answering from my mobile phone, typing not easy, srry)

Answer 5

WOW, thank you so much, not sure why I didn't realize that before but it's so obvious when I think about it. Just goes to show you really need to master the fundamentals before attempting more challenging things. Thanks again.

Answer 6

Yw!