Maybe it's a dumb question but it's new for me and I want to understant it clearly.
Why or which method is used there, I have formula
$$ \sum_{n=0}^{\infty} \cos^{2n+1}x = \frac{1}{2^{2n}} \left\{\cos((2n+1)x) + {{2n+1}\choose 1 }\cos((2n-1)x)+ \ldots+ {{2n+1}\choose n}\cos(x) \right\}$$
I need to know why I can obtain that series. Never seen writen the odd power of $ \cos $ like that.
Thank you for anwsering and have a nice day.
Let $$z=cos(x)+i \sin(x)$$ Then $$\frac{1}{z}=\cos(x)-i \sin(x) \\ z+\frac{1}{z}=2 \cos(x) \\ \cos(x)= \frac{1}{2}(z+\frac{1}{z}) \\ \cos{2n+1}(x)= \frac{1}{2^{2n+1}}(z+\frac{1}{z})^{2n+1} $$
Now, by the Binomial theorem, and using $\binom{2n+1}{k}=\binom{2n+1}{2n+1-k}$ you get $$\cos{2n+1}(x)= \frac{1}{2^{2n+1}}(\sum_{k=1}^n \binom{2n+1}{k} (z^k+\frac{1}{z^k})$$
Use the Demoivre formula to see that $$z^k+\frac{1}{z^k}=2 \cos((2k+1)x)$$