<s>[INST] What are the values of $a_0,a_1,...,a_{10}$ if $\cos^{10} {\theta}=\sum_{k=0}^{10}a_k\cos {k\theta }$? I was thinking of doing the following:
Let $A=a_0+a_1\cos {\theta }+a_2\cos {2\theta }+...+a_{10}\cos {10\theta }$ 
and
    $B=a_1\sin {\theta }+a_2\sin {2\theta }+...+a_{10}\sin {10\theta }$
Then, $S=A+iB=a_0+a_1(\cos {\theta}+i\sin {\theta})+a_2(\cos {2\theta}+i\sin {2\theta})+...+a_{10}(\cos {10\theta}+i\sin {10\theta})$
$=a_0+a_1e^{i\theta}+a_2e^{2i\theta}+...+a_{10}e^{10i\theta}$
If I could evaluate $S$ then my required sum would be $\Re (S)$.
 [/INST] Let $x=\cos {\theta}+i\sin {\theta}$ whence, $x+\frac{1}{x}=2\cos {\theta}$ and $x^n+\frac{1}{x^n}=2\cos {n\theta}$
$\therefore (2\cos {\theta})^{10}=(x+\frac{1}{x})^{10}$
$=(x^{10}+\frac{1}{x^{10}})+10(x^8+\frac{1}{x^8})+45(x^6+\frac{1}{x^6})+120(x^4+\frac{1}{x^4})+210(x^2+\frac{1}{x^2})+252$
$=2\cos {10\theta}+10\times 2\cos {8\theta}+45\times 2\cos {6\theta}+120\times 2\cos {4\theta}+210\times 2\cos {2\theta}+2\times 126$
$\Rightarrow \cos^{10}{\theta}=\frac{1}{2^9}\left [\cos {10\theta}+10\cos {8\theta}+45\cos {6\theta}+120\cos {4\theta}+210\cos {2\theta}+126 \right]$
So, by the problem, $a_0=\frac{126}{2^9},a_2=\frac{210}{2^9},a_4=\frac{120}{2^9},a_6=\frac{45}{2^9},a_8=\frac{10}{2^9},a_{10}=\frac{1}{2^9}$
and $a_1=a_3=a_5=a_7=a_9=0$.
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