Fourier cosine series of $\sin ax$

Let $f(x)=\sin ax, \; x \in (0, \pi) $ and let $a \in \mathbb{Z}$ . Prove that $f$ can be expanded into Fourier cosine series and that:

$$\sin ax \sim \left\{\begin{matrix} \displaystyle \dfrac{4a}{\pi}\sum_{n=0}^{\infty}\frac{\cos (2n+1)x}{a^2-(2n+1)^2} &,\;\;\;  a {\rm\; is \; even}\\  \displaystyle \frac{4a}{\pi}\left [ \frac{1}{2a^2}+\sum_{n=1}^{\infty}\frac{\cos 2nx}{a^2-4n^2} \right ] &,\;\;\;  a {\rm\; is \; odd} \end{matrix}\right.$$

Solution:



We extend the function \( f \) to a \( 2 \pi \) periodic function $$ F(x)=\left\{\begin{matrix}
f(x) &,x \in (0, \pi) \\
 f(-x)&, x \in (-\pi, 0) \\
 0&, x \in \left \{ -\pi, 0, \pi \right \}
\end{matrix}\right. $$

that is continuous and even. \( F \) can be expanded in a Fourier series of the form:

$$F(x) \sim \frac{a_0}{2}+\sum_{n=1}^{\infty} a_n \cos n x, \; x \in (-\pi, \pi) \setminus \{0 \}$$

We are now evaluating the coefficients:

\( \color{grey} \blacksquare \)  \( \displaystyle a_0= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin ax \, {\rm d}x= \frac{2\left ( 1-\cos a \pi \right )}{a\pi} \)
\( \color{grey} \blacksquare \)  \( \displaystyle  a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin ax \cos nx \, {\rm d}x= \frac{-2\left ( n\sin n\pi \sin a \pi + a \cos a\pi  \cos n\pi-a \right )}{a^2-n^2} \)

We distinguish cases for \(a \) using at the same time the identity: $$ \cos n \pi = (-1)^n , \; n \in \mathbb{N}$$

\( \color{red}  \blacksquare\) If \(a \) is even a straightforward calculation shows that:

$$\sin ax \sim \dfrac{4a}{\pi}\sum_{n=0}^{\infty}\frac{\cos (2n+1)x}{a^2-(2n+1)^2}$$

while ,

 \( \color{red}  \blacksquare\) If \(a \) is odd we have that:

$$ \sin ax \sim \frac{4a}{\pi}\left [ \frac{1}{2a^2}+\sum_{n=1}^{\infty}\frac{\cos 2nx}{a^2-4n^2} \right ]$$

Source: mathimatikoi.org