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
 A: According to the Power-reduction formula, one can interchange between $\cos(x)^n$ and $\cos(nx)$ like the following:
$$
\cos^n\theta = \frac{2}{2^n} \sum_{k=0}^{\frac{n-1}{2}} \binom{n}{k} \cos{((n-2k)\theta)} \tag{odd}\\
$$
$$
\cos^n\theta = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} \binom{n}{k} \cos{((n-2k)\theta)}. \tag{even}
$$
Now combine $\sum_m c_m \cos^m\theta$, such that only $\cos(nx)$ is left.
A: There were two corrections in the posting ($1102$ should be $1120$ and $\cos x$ term should be $\cos^2 x$ term:
$$\cos 10x = 512 (\cos x)^{10} -1280 (\cos x)^8 +1120 (\cos x)^6 -400(\cos x)^4+50(\cos x)^2-1$$
Follow Arturo's answer you should get the following
$$
\begin{align*}
\cos 2x &= 2 \cos^2 x -1 \\
\cos 3x &= 4 \cos^3 x - 3 \cos x\\
\cos 4x &= 8 \cos^4 x - 8 \cos^2 x +1 \\
\cos 5x &= 16 \cos^5 x - 20 \cos^3 x + 5 \cos x\\
\cos 6x &= 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x -1\\
\cos 8x &= 128 \cos^8 x - 256 \cos^6 x + 160 \cos^4 x -32 \cos^2 x +1\\
\cos 10x &= 512 \cos^{10} x -1280 \cos^8 x +1120 \cos^6 x -400 \cos^4 x +50 \cos^2 x-1\\
\end{align*}
$$
(Corrected $-1120 $ to $1120$ )
A: These are usually denoted $T_n$ and called Chebyshev polynomials of the first kind. For every $n\geqslant0$, $\cos(nu)=T_n(\cos(u))$ with
$$
T_n(x)= \sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} (x^2-1)^k x^{n-2k}.
$$
For example,
$$
T_8(x) = 128x^8 - 256x^6 + 160x^4 - 32x^2 + 1,
$$
and
$$
T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x.
$$
Likewise,
$$
T_{10}(x)=512x^{10} − 1280x^8 + 1120x^6 − 400x^4 + 50x^2 − 1,
$$
$$
T_{11}(x)=1024x^{11} − 2816x^9 + 2816x^7 − 1232x^5 + 220x^3 − 11x,
$$
and
$$
T_{12}(x)=2048x^{12} − 6144x^{10} + 6912x^8 − 3584x^6 + 840x^4 − 72x^2 + 1.
$$
A: \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}
A: Here is a neat way to derive what the answer will be, using Euler's formula $e^{ix}=\cos x + i \sin x$.
We have that $\cos nx$ is the real part of $e^{i(nx)}=\left(e^{ix}\right)^n=(\cos x + i \sin x)^n$
By the binomial formula, $(\cos x + i \sin x)^n=\displaystyle\sum_{k=0}^n i^k \binom{n}{k} \sin^k(x) \cos^{n-k}(x)$.  Since $i^k$ is real if and only if $k$ is even, we therefore have (replacing $k$ with a new indexing variable $2\ell$)
$$\cos nx= \sum_{\ell=0}^{\lfloor n/2\rfloor} (-1)^{\ell} \binom{n}{2\ell} (\sin^2(x))^{\ell} \cos^{n-2\ell}(x).$$
Finally, we use the Pythagorean identity $\sin^2=1-\cos^2$ to rewrite
$$\cos nx= \sum_{\ell=0}^{\lfloor n/2\rfloor} (-1)^{\ell} \binom{n}{2\ell} (1-\cos^2(x))^{\ell} \cos^{n-2\ell}(x)=\sum_{\ell=0}^{\lfloor n/2\rfloor} \binom{n}{2\ell} (\cos^2(x)-1)^{\ell} \cos^{n-2\ell}(x).$$
This agrees with Didier's answer.
A: You can always repeatedly use
$$\begin{align*}
\cos(a\pm b) &= \cos a\cos b \mp \sin a\sin b\\
\sin(a\pm b) &= \sin a\cos b \pm \cos a\sin b\\
\sin^2(r) &= 1-\cos^2(r).
\end{align*}$$
For example,
$$\begin{align*}
\cos(4x) &= \cos(2x+2x)\\
&= \cos(2x)^2 - \sin^2(2x)\\
&= \cos(2x)^2 - (1-\cos^2(2x))\\
&= 2\cos(2x)^2 - 1\\
&= 2(\cos(x+x))^2 - 1\\
&= 2(\cos x\cos x - \sin x\sin x)^2 - 1\\
&= 2(\cos^2 x - \sin^2 x)^2 - 1\\
&= 2(\cos^2x - (1-\cos^2 x))^2 - 1\\
&= 2(2\cos^2x - 1)^2 - 1\\
&= 2(4\cos^4 x - 4\cos^2 x + 1) - 1\\
&= 8\cos^4 x - 8\cos^2 x + 1.
\end{align*}$$
