Proof. To find \(a_n\), multiply both sides of (i) by \(\cos nx\) and integrate over \([-\pi , \pi ]\). Since the series converges uniformly, we can integrate term by term. We have \[f(x) \, \cos nx = \frac {a_0}{2}\cos nx + \sum _{k = 1}^{\infty }\left [a_k \cos kx\, \cos nx + b_k \sin kx \cos nx\right ]\] \[\int _{-\pi }^{\pi }f(x)\, \cos nx\, dx = \frac {a_0}{2}\int _{-\pi }^{\pi }\cos nx \, dx + \sum _{k=1}^{\infty }a_k\int _{-\pi }^{\pi }\cos kx\, \cos nx \, dx + \sum _{k = 1}^{\infty }b_k\int _{-\pi }^{\pi }\sin kx\, \cos nx\, dx.\] But \[\int _{-\pi }^{\pi }\cos nx\, dx= 0, \hspace {0.3cm} n = 1,\, 2, \, 3, \, \cdots \cdots \] \begin {align*} \int _{-\pi }^{\pi }\sin kx \cos nx\, dx & = \frac {1}{2}\int _{-\pi }^{\pi }\left [\sin [(k+n)x] + \sin [(k -n)x]\right ]\, dx\\ \int ^{\pi }_{-\pi }\cos kx \cos nx\, dx & = \frac {1}{2}\int _{-\pi }^{\pi }\left [\cos [(k+n)x] + \cos [(k-n)x]\right ]\, dx = \begin {cases} 0, & k\neq n\\ \pi , & k = n\geq 1\\ \end {cases} \end {align*}

Thus \[a_n = \frac {1}{\pi }\int _{-\pi }^{\pi }f(x)\, \cos nx\, dx.\] For \(n=0\), we get \(\int _{-\pi }^{\pi }f(x)\, dx = 2a_0\pi \), so that \[a_0 = \frac {1}{\pi }\int _{-\pi }^{\pi } f(x)\, dx.\] □