<s>[INST] Fourier Series for $\cos^3(x)$ Consider the familiar trigonometric identity: $\cos^3(x) = \frac{3}{4} \cos(x) + \frac{1}{4} \cos(3x)$
Show that the identity above can be interpreted as Fourier series expansion.
so we know that cos is periodic between $\pi$ and $-\pi$ and $\cos$ is an even function, therefore, $\cos^3$ is even. 
so we need to compute $a_0$ ( the integral of $f(x)$ and it will equal $0$) and $a_n$ ( the integral from $\pi$ to $-\pi$ of $\cos^3(x) \cos(nx)$ )
how to compute $a_0$
thanks 
 [/INST] Recall: from a familiar trigonometric identity,
$$\cos^3(x) = \cos(x)\color{blue}{\cos^2(x)} = \cos(x)\color{blue}{(1 - \sin^2(x))}$$
Thus,
$$\int \cos^3(x)dx = \int (1-\sin^2(x))\cos(x)dx$$
Make the $u$-substitution $u = \sin(x), du = \cos(x)dx$ and you should be able to find the result easily. You should get
$$\int \cos^3(x)dx = \sin(x) - \frac 1 3 \sin^3(x) + C$$
Apply the fundamental theorem of calculus on the bounds $0,\pi$. Since $\sin(n\pi) \equiv 0 \; \forall n \in \Bbb Z$,
$$\int_0^\pi \cos^3(x)dx = \sin(0) - \frac 1 3 \sin^3(0) - \left( \sin(\pi) - \frac 1 3 \sin^3(\pi) \right) = 0$$
and thus, $a_0 = 0$.
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