I was given an assignment by my instructor where i had to write the function
$$ f(t) = \begin{cases} 1-t & 0\leq t < 1 \\ t-1 & 1 \leq t < 2 \end{cases}\\ f(t + 2) = f(t) $$
as a complete Fourier series with the hint (you should only get cosine terms)
Now i realized this function was just a triangle wave and the even extension of 1-t on the interval 0 to 1 would be the given function, so i decided to expand it by simply writing it as the Fourier cosine series of 1-t from 0 to 1
would i have gotten an equivalent Fourier series if i expanded the piece-wise function above from 0 to 2 ? I get awfully confused about the limits and normalization factors, but from what i understand the even extension of the function i expanded should represent exactly the same periodic function?
If the function is periodic to begin with, then expanding it on any interval equal to an integer multiple of its periods will give you the same Fourier series.
For example, if the function is defined by the rules
$$ f(t) = \begin{cases} 1 & 0\leq t < 1 \\ 0 & 1 \leq t < 2 \end{cases}\\ f(t + 2) = f(t) $$
then you will get the same Fourier series whether you expand on $[0,2]$ or $[0,4]$ or $[0,6]$, etc. If you expand the series over a different interval (e.g. $[0,1]$ or $[0,2.7]$ or whatever), then you will get a different result. Try thinking about drawing the appropriate extensions, and this will be easier to see.
In your case, extension after expanding on $[0,1]$ will be a "sawtooth" because you are only capturing and extending the behavior between $0$ and $1$.
To get a triangular wave, you also have to use an interval that includes the part where the function "goes up again" (i.e. from $1$ to $2$)