# Thread: trig and complex numbers

1. ## trig and complex numbers

if n is an integer x is real, then show that :
$\displaystyle (1+\cos x + i\sin x )^n + (1+ \cos x-i\sin x)^n = 2^{n+1}\cos ^n {\frac{x}{2}}\cos \frac{nx}{2}$

I played around a bit with de moivre's theorem and other things...

2. ## Re: trig and complex numbers

Originally Posted by earthboy
if n is an integer x is real, then show that :
$\displaystyle (1+\cos x + i\sin x )^n + (1+ \cos x-i\sin x)^n = 2^{n+1}\cos ^n {\frac{x}{2}}\cos \frac{nx}{2}$

I played around a bit with de moivre's theorem and other things...

If you can't solve after a massive amount of calculations, there is always the mighty mathematical induction.

3. ## Re: trig and complex numbers

Originally Posted by ChessTal
If you can't solve after a massive amount of calculations, there is always the mighty mathematical induction.
yes there is.....but there got be a silly neat trick here...anyone???

4. ## Re: trig and complex numbers

Start by using the identities

$\displaystyle \cos 2A = 2\cos^{2}A - 1$ and $\displaystyle \sin 2A = 2\sin A \cos A$ so as to get everything inside the brackets in terms of half angles $\displaystyle x/2.$

Next, remove $\displaystyle 2\cos(x/2)$ as a common factor from each of the brackets.

Then, write the two complex numbers in exponential form and finally use the exponential form for $\displaystyle \cos A.$