Why do you do what you do with $\displaystyle a_0$?? Do you hate yourself? Much easier to split the integral into a sum:
$\displaystyle \displaystyle{\int\limits^1_{-1}(1-|t|)dt=\int\limits^0_{-1} (1+t)dt+\int\limits^1_0(1-t)dt=1}$ , and then do the same for $\displaystyle a_n\,,\,b_n$
If this exercise is what I think it is then it's a very nice and beautiful one. You'll get both
the sums of $\displaystyle \displaystyle{\sum\limits^\infty_{n=1}\frac{1}{(2n-1)^2}\,\,and\,\,\sum\limits^\infty_{n=1}\frac{1}{n ^2}}$
Tonio
Sketch a sine curve and observe its behaviour, if you actually try it is not difficult to answer your own question.
Also if you do not know the value of the sine and cosine at interger multiples of pi you are not ready to study Fourier series and need to repeat trig.
CB
It is a little weird, imho, to make this kind of trivial questions in very basic trigonometry
if you're dealing with Fourier series...even if you don't remember. and this is already weird,
why don't you check your high school books or, at least, google it?
The sine of any multiple of pi is zero...
Tonio
i think my question was not clear, my question was based on what does $\displaystyle sin(1-n)$ equal to.
We know that if $\displaystyle cos(1-n)\pi = (-1)^{1-n}$ so would this apply to $\displaystyle sin(1-n)\pi$ as well.