De Moivres formula for multiple angles. Where does the sinus go?

[mariush]

Hi!

Yesterday i read and mostly understood that (cos($\theta$) + isin($\theta$)$^{n}$ = (cos($n\theta$) + isin($n\theta$)

 

[HallsofIvy]

Okay, so what is your question? What about the multiple angles that you mention in your title?

Note that the formula you give can be derived from the formula for multiplication of complex numbers in "polar form"- $r_1(cos(\theta_1)+ i sin(\theta_1))r_2(cos(\theta_2)+ i sin(\theta_2))= (r_1r_2)(cos(\theta_1+ \theta_2)+ i sin(\theta_1+ \theta_2))$.

Also, we can use Euler's formula, $e^{ix}= cos(x)+ i sin(x)$ to write $(cos(\theta)+ i sin(\theta))^n= cos(n\theta)+ i sin(n\theta)$ in the simpler form $(e^{i\theta})^n= e^{i n\theta}$.

 

[mariush]

Sorry, I'm not best friends with the post publisher today.

My question was about the multiple angles: I understand that
$(cos x +i sinx)^{n} = cos (nx) + i*sin(nx)$. Then i read that $cos (nx) = cos^{n}(x) - (nC2)cos^{n-2}x*sin^{2}(x)- ...-(-1)^{k/2}(nCk)cos^{n-k}(k)*sin^{k}(x)$

Now, that looks a lot like what I would expect to get from $cos (nx) + i*sin(nx) = (cos x +i sinx)^{n},$ but i cannot figure out where the sin(nx) goes.

My first thought is that $cos (nx) = (cos x +i sinx)^{n} - sin(nx)$

 

[mariush]

Okay, so what is your question? What about the multiple angles that you mention in your title?

Note that the formula you give can be derived from the formula for multiplication of complex numbers in "polar form"- $r_1(cos(\theta_1)+ i sin(\theta_1))r_2(cos(\theta_2)+ i sin(\theta_2))= (r_1r_2)(cos(\theta_1+ \theta_2)+ i sin(\theta_1+ \theta_2))$.

Also, we can use Euler's formula, $e^{ix}= cos(x)+ i sin(x)$ to write $(cos(\theta)+ i sin(\theta))^n= cos(n\theta)+ i sin(n\theta)$ in the simpler form $(e^{i\theta})^n= e^{i n\theta}$.

Hi, and thanks for such a quick response! I'm sorry about the first post. Seems like I hit the submit rather than the preview button, a bit prematurely..