The question:
On the interval [−π, π] consider the function $$f( x) \ =\ \begin{cases} 0 & if\ | x| \ >\ \delta \\ 1-| x| /\delta & if\ | x| \ \leqslant \ \delta \end{cases}$$ Thus the graph of f has the shape of a triangular tent. Show that $$f( x ) =\frac{\delta }{2\pi } +2\sum _{n=1}^{\infty }\frac{1\ -\ cos( n\delta )}{\pi n^{2} \delta } \ cos( x )$$
My attempts:
Since the function in question is even, I thought that this would be sufficient to find only cos coefficients. I decided to obtain the coefficients integrating the interval [-,] as follows
$B_{n} \ =\ \frac{1}{\pi }\int _{( -\pi )}^{\pi } f( x) \ cos( nx) \ dx$
giving that ||<
$$ \begin{array}{l} \frac{1}{\pi }\int _{( -\pi )}^{\pi } f( x) \ cos( nx) \ dx\ \\ =\frac{1}{\pi }\left( \ \int _{( -\delta )}^{0}( 1+x/\delta ) cos( nx) \ dx+\int _{0}^{\delta }( 1-x/\delta ) cos( nx) \ dx\ \right)\\ \\ =\frac{1\ -\ cos( n\delta )}{\pi n^{2} \delta } \end{array}$$
Then I thought that maybe the sine coefficients would eventually turn into a vertical constant offset. but they end up being zero, which is pretty obvious since the function is even. How can I get this constant $\frac{\delta }{2\pi }$?