Expressing $\cos(x)^6$ as a linear combination of $\cos(kx)$'s Let $$(\cos^6(x)) = m\cos(6x)+n\cos(5x)+o\cos(4x)+p\cos(3x)+q\cos(2x)+r\cos(x)+a.$$ 
What is the value of $a$?
 A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\cos^{6}\pars{x}&=\pars{\expo{\ic x} + \expo{-\ic x} \over 2}^{6}=
{1 \over 2^{6}}\sum_{\ell = 0}^{6}{6 \choose \ell}\expo{\ic\pars{6 - 2\ell}x}
\\[3mm]&=
{1 \over 2^{6}}\sum_{\ell = 0}^{2}{6 \choose \ell}\expo{\ic\pars{6 - 2\ell}x}
+{1 \over 2^{6}} \overbrace{6 \choose 3}^{\ds{20}}
+
{1 \over 2^{6}}
\color{#f00}{\sum_{\ell = 4}^{6}{6 \choose \ell}\expo{\ic\pars{6 - 2\ell}x}}
\tag{1}
\end{align}

\begin{align}
\color{#f00}{\sum_{\ell = 4}^{6}{6 \choose \ell}\expo{\ic\pars{6 - 2\ell}x}}&=
\sum_{\ell = -2}^{0}{6 \choose \ell + 6}\expo{\ic\pars{-6 - 2\ell}x}
=
\sum_{\ell = 2}^{0}\overbrace{6 \choose -\ell + 6}^{\ds{6 \choose \ell}}\
\expo{\ic\pars{-6 + 2\ell}x}
\\[3mm]&=
\color{#f00}{\sum_{\ell = 0}^{2}{6 \choose \ell}\expo{\ic\pars{-6 + 2\ell}x}}
\end{align}

We'll replace this result in $\pars{1}$:
\begin{align}
\cos^{6}\pars{x}&={5 \over 16}
+ {1 \over 32}\sum_{\ell = 0}^{2}{6 \choose \ell}\cos\pars{\bracks{6 - 2\ell}x}
={5 \over 16} + {\cos\pars{6x} + 6\cos\pars{4x} + 15\cos\pars{2x}\over 32}
\end{align}

$$
\color{#00f}{\large\cos^{6}\pars{x}
={5 \over 16} + {15 \over 32}\,\cos\pars{2x} + {3 \over 16}\,\cos\pars{4x}
+ {1 \over 32}\,\cos\pars{6x}}
$$

A: Method $1:$
Use $\displaystyle\cos3A=4\cos^3A-3\cos A\iff \cos^3A=\cdots$ for $\displaystyle\cos^6x=(\cos^3x)^2$
and $\displaystyle\cos2B=2\cos^2B-1\iff \cos^2B=\cdots$
Method $2:$
Using Euler Formula, $\displaystyle\cos y=\frac{e^{iy}+e^{-iy}}2$
A: If:
$$\cos^6(x)=\sum_{k=1}^6 a_k \cos(kx)+a_0$$ 
Then we have:
$$\cos^6(0)+\cos^6(\pi)=2=a_2+ a_4+a_6+2a_0$$ 
$$\cos^6(3\pi/4)+\cos^6(\pi/4)=1/4=-2a_4+2a_0$$
$$\cos^6(3\pi/2)+\cos^6(\pi/2)=0=-2a_2+2a_4-2a_6+2a_0$$
Add them all up to find $a_0$.
