How is this function even? $f(t)=(\pi-t)^2$, for $0\le t \le2\pi$ How is the following function even?
$$f(t)=(\pi-t)^2, \quad 0\le t \le2\pi$$
Since $$f(-\pi)= \text{not defined} \ne 0 = f(\pi)$$
It is only even in the case that $t=0$

On the topic of Fourier Series now, lecturer claims that since the function is even, the $b_n$ coefficients will all vanish.
 A: This function is, of course, not even as a function $\mathbb R\rightarrow\mathbb R$ because of the issue you point out. However, it is common in Fourier analysis to use an abuse of notation that justify the manipulation. Perhaps the professor forgot to mention this (or, worse, just looked at the Fourier coefficients and said something without thinking too hard about what it meant).
The typical view in Fourier analysis is that a function $f:[0,2\pi)\rightarrow\mathbb R$ is really meant to be thought as a function that takes an angle as input and outputs a real number; so it's a function whose domain is a circle. Then, the quantities $\pi$ and $-\pi$ both refer to the same angle on a circle, so both would evaluate to $0$ under this function. Essentially, to evaluate the function at an angle, we first shift the angle into the interval $[0,2\pi)$ and then apply the function. The more elegant way to do this is to write $f$ as a function from $S^1$ (the unit circle) to $\mathbb R$, but it's often easier to work in terms of $[0,2\pi)$ in order to define a function, especially since different authors define the unit circle in different ways and since "take the angle from the interval $[0,2\pi)$ and square its difference from $\pi$" is hard to express naturally in terms of a circle. Geometrically, "even" then means that the function is the same evaluated at a counterclockwise rotation of $\theta$ and a clockwise rotation of $\theta$.
This is also sometimes expressed as saying that we are really studying functions with period $2\pi$ - that is function for which
$$f(x)=f(x+2\pi).$$
Then, given a definition of $f$ on the interval $[0,2\pi)$, we can figure out a function $\mathbb R\rightarrow\mathbb R$ by using the above property. Applying this to the definition you were given does indeed yield an even function.
