Fourier series of a function

Consider $$f(t)= \begin{cases} 1 \mbox{ ; } 0<t<1\\ 2-t \mbox{ ; } 1<t<2 \end{cases}$$

Let $f_1(t)$ be the Fourier sine series and $f_2(t)$ be the Fourier cosine series of $f$, $f_1(t)=f_2(t), 0<t<2$. Write the form of the series (without computing the coefficients) and graph $f_1$ and $f_2$ on [-4,4] (including the endpoints $\pm 4$) using *'s to identify the value of the series at points of discontinuity.

I think we have:

$f_1(t)=\sum \limits_{n=1}^{\infty} b_n \sin \frac{n \pi t}{2}$
$f_2(t)=\frac{a_0}{2}+\sum \limits_{n=1}^{\infty} a_n \cos \frac{n \pi t}{2}$

I think we have $f_2=1$ and for $0<t<2, f_1=f_2=1$

Can we do anything else? Can someone help me with the end?

Thank you

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I could be mistaken, but shouldn't the angular frequency be just $n\pi$, not $n\pi / 2$, seeing as $2\pi / 2 = \pi$? –  Sam Apr 6 '13 at 16:34
I don't think so. If we have a periodic function of period $P=2L$ then in the fourier series we have $\cos \frac{n \pi t}{L}$ –  Carpediem Apr 6 '13 at 16:43
I agree with that, but isn't the function assumed to have a period $P = 2$ (that is, $L = 1$)? –  Sam Apr 6 '13 at 16:46
yeah chill i said i gonna do it ... –  Dominic Michaelis Apr 6 '13 at 17:35

Ok at first we gonna plot our function

We know that on jump discontinuities it will converge to the arithmetic mean of them, so the first approximation is just taking $\frac{1}{2}$. This gonna look like

The Cos terms gonna look like

The Sin terms are looking like

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It's only a partial answer i will add the rest after eating –  Dominic Michaelis Apr 6 '13 at 17:57
Can you edit please ? –  Carpediem Apr 6 '13 at 19:58
On the cosine graph, the horizontal line the constant function ? –  Carpediem Apr 6 '13 at 20:23
yeah it is, it should be on 1/2 but it isn't i found that strange –  Dominic Michaelis Apr 6 '13 at 20:24
So the overlaps are normal right? Since those are then other terms –  Carpediem Apr 6 '13 at 20:24