Trigonometry problem cosine identity Let $\cos^6\theta = a_6\cos6\theta+a_5\cos5\theta+a_4\cos4\theta+a_3\cos3\theta+a_2\cos2\theta+a_1\cos\theta+a_0$. Then $a_0$ is 
(A) $0$ (B) $\frac{1}{32}$ (C) $\frac{15}{32}$ (D) $\frac{10}{32}$
Any hints on how to approach this?

Using \begin{align}&2\cos^2 x=\cos 2x +1, \\&4\cos^3x=\cos 3x+3\cos x, \\ \text{and}&2\cos a \cos b= \cos(a+b)+\cos(a-b),\end{align}
we obtain,
\begin{align}&\cos^6 x=(\cos^3 x)^2\\
\\=&\left(\dfrac{\cos 3x+3\cos x}{4}\right)^2\\
\\=&\dfrac1{16}(\cos^2 3x + 9\cos^2 x+6\cos x\cos 3x)\\
\\=&\dfrac{1}{16}\left(\dfrac{\cos 6x+1}{2}+9\dfrac{\cos 2x +1}2+3\cos 4x +3\cos 2x\right).\end{align}
It is seen that the constant term is $\dfrac{5}{16}$ or $\dfrac{10}{32}$.
(This of course, makes use of the question's assumption that the number $a_0$ is unique.) 
