<s>[INST] How to expand $\cos nx$ with $\cos x$? Multiple Angle Identities:
How to expand  $\cos nx$  with  $\cos x$, such as 
$$\cos10x=512(\cos x)^{10}-1280(\cos x)^8+1120(\cos x)^6-400(\cos x)^4+50(\cos x)^2-1$$
See a list of trigonometric identities in english/ chinese
 [/INST] \begin{equation}
\text{You can use De Moivre's identity:}
\end{equation}
\begin{equation} 
\text{Let's Call:}\\\\
\end{equation}
\begin{equation}
\mathrm{z=\cos x+i \sin x}\\
\mathrm{\frac{1}{z}=\cos x-i \sin x}\\
\end{equation}
\begin{equation}
\text{Now, addind both equations together, we get:}\\
\end{equation}
\begin{equation}
\mathrm{2\cos x=z+\frac{1}{z}}\\
\text{And we know that:}\\
\end{equation}
\begin{equation}
\mathrm{z^n=(cis x)^n=cis~nx}\\
\end{equation}
\begin{equation}
\text{So:}\\
\mathrm{2\cos x=z+\frac{1}{z}}\Rightarrow\\
\mathrm{1024\cos^{10} x=\left (z+\frac{1}{z}  \right )^{10}}\\
\end{equation}
\begin{equation}
\text{Expanding the RHS:}\\
\end{equation}
\begin{equation}
\mathrm{1024\cos^{10} x=z^{10}+\frac{1}{z^{10}}+10\left (z^{8}+\frac{1}{z^{8}} \right )+45\left (z^{6}+\frac{1}{z^{6}} \right )+120\left (z^{4}+\frac{1}{z^{4}} \right )+210\left (z^{2}+\frac{1}{z^{2}} \right )+252}\\ 
\end{equation}
\begin{equation}
\mathrm{1024\cos^{10} x=2\cos 10x+20\cos 8x+90\cos 6x+240\cos 4x+420\cos 2x+252}\\
\end{equation}
\begin{equation}
\boxed{\boxed{\mathrm{\therefore\cos^{10} x=\frac{1}{512}\cos 10x+\frac{5}{256}\cos 8x+\frac{45}{512}\cos 6x+\frac{15}{64}\cos 4x+\frac{105}{256}\cos 2x+\frac{63}{256}}}}
\end{equation}
</s>