$\cos^5 (x)$ as a sum How can I write the $$ \cos^{5}(x) $$ as a sum of first order terms? I have started noticing that:
$$ \cos^{5}(x) = \cos^{4}(x) \cos(x),$$ and then: $$ \cos^{4}(x)=\cos^{2}(x)\cos^{2}(x). $$ Finally:  $$ \cos^{2}(x)=\frac{1}{2} (1+\cos(2x)). $$  
But I would like to find an easier formula, like the one for $$\cos^{2}(x).$$
 A: This is probably a little advanced, but a nice trick...
$\cos x = \frac 12 (e^{ix} + e^{-ix})$
$\cos^5 x = $$(\frac 12)^5 (e^{ix} + e^{-ix})^5\\
(\frac 12)^5 (e^{5ix} + 5e^{4ix}e^{-ix} + 10e^{3ix}e^{-2ix} + 10e^{2ix}e^{-3ix} + 5e^{ix}e^{-4ix} + e^{-5ix})$
Simplify
$\cos^5 x = $$(\frac 12)^5 (e^{5ix} + e^{-5ix} + 5(e^{3ix}+ e^{-3ix}) + 10(e^{ix}+ e^{-ix}))\\
(\frac 12)^4(\cos 5x + 5\cos 3x + 10\cos x)$
A: The following Power reduction formula holds
$$\cos^5 \theta=\frac{10\cos \theta+5 \cos 3\theta+\cos 5\theta}{16}$$
A: The Chebyshev polynomials of the first kind allow you to express $\cos(nx)$ as a polynomial of degree $n$ in $\cos(x)$.
You can use them in reverse to write, say $\cos^5(x)$ as a multiple of $\cos(5x)$ plus a fourth-degree polynomial in $\cos(x)$. Then use the same technique to get rid of the $\cos^4(x)$ term (actually, as it turns out, there won't be any even-degree terms), and then $\cos^3(x)$ and $\cos^2(x)$.
At the end you get the finite Fourier expansion
$$ \cos^5(x) = a_5\cos(5x)+a_4\cos(4x) + a_3\cos(3x)+a_2\cos(2x)+a_1\cos(x)$$
for some $a_i$s that you should be able to work out (though I don't care to).
A: The general answer is yes:  any power $\cos^nx$ and $\sin^nx$ can be linearised.
The simplest way (except for small $n$s, say $n<4$) is to use complex numbers.
Here, for instance, as $\cos x=\frac12(\mathrm e^{ix}+ \mathrm e^{-ix})$, we have
$$\cos^5 x=\frac1{2^5}\bigl(\mathrm e^{ix}+ \mathrm e^{-ix}\bigr)^5.$$
We'll expand it with Newton's formula. 
To simplify the calculation, we'll set $u=\mathrm e^{ix}$, so $\bar u=\mathrm e^{-ix}$ and $u\,\bar u=1$:
\begin{align}
\cos^5 x&=\frac1{32}\bigl(u+\bar u\bigr)^5=\frac1{32}\bigl(u^5+5u^4\bar u+10u^3\bar u^2 +10u^2\bar u^3+5u\bar u^4+ \bar u^5 \bigr)\\
&=\frac1{32}\bigl(u^5+ \bar u^5 +5(u^3+\bar u^3)+10(u+ \bar u)\bigr).
\end{align}
Now note that $\;u^n+\bar u^{\mkern 1mun}=2\cos nx$, thus
\begin{align}
\cos^5 x&=\frac1{16}\bigl(2\cos5x +5\cos 3x+10\cos x\bigr).
\end{align}
A: I think your way is very easy:
$$\cos^5x=\frac{\cos{x}(1+\cos2x)^2}{4}=\frac{\cos{x}+2\cos{x}\cos2x+\cos{x}\cos^22x}{4}=$$
$$=\frac{\cos{x}+\cos3x+\cos{x}+\frac{\cos{x}+\cos{x}\cos4x}{2}}{4}=\frac{5\cos{x}+2\cos3x+\frac{\cos5x+\cos3x}{2}}{8}=$$
$$=\frac{10\cos{x}+5\cos3x+\cos5x}{16}.$$
