Conceptual understanding of the effects of time shift on the coefficients of a Fourier Series. I have a function of a triangle wave that goes from a minimum value of 0 volts to a maximum value of V. It has a period of T, and is a maximum at t = 0. I'm trying to understand how a time shift $\tau$ would affect the coefficients $a_n$ and $b_n$ of the Fourier series.
Without a time shift I can write:
$$a_n = \frac 2T \int_{-\frac T2}^0 f(t) \cos(\frac {2\pi nt}{T})\,dt\ + \frac 2T \int_{0}^{\frac T2}g(t) \cos(\frac {2\pi nt}{T})\,dt$$
But once a time shift is introduced I have: 
$$a_n(\tau) = \frac 2T \int_{-\frac T2}^0 f(t +\tau) \cos(\frac {2\pi nt}{T})\,dt\ + \frac 2T \int_{0}^{\frac T2} g(t+ \tau) \cos(\frac {2\pi nt}{T})\,dt$$
My assumption is that this adds an extra few terms to the oringinal $a_n$ such that $a_n(\tau) = a_n + C$, where C depends on n. Is this statement accurate?
Thinking of the coefficients as a projection of $f(t)$ onto the vector $\cos(n\omega t)$, I don't understand why shifting the graph would change the value of this projection for a given point on the function. Why is this? 
Maybe I'm thinking of this all wrong or making wrong assumptions, any help would be appreciated!
 A: Suppose that $f(t)$ is a periodic function with period $T$.
The fourier series of $f$ is,
$$ f(t) = a_0 + \sum_{n=1}^\infty \Big[a_n \cos(n\omega t) + b_n \sin(n\omega t)\Big]$$ 
Where 
$$ a_0 = \frac1T \int_0^T f(t) \mathrm{d}t$$
$$ a_n = \frac2T \int_0^T f(t) \cos(n\omega t)\mathrm{d}t$$
$$ b_n = \frac2T \int_0^T f(t) \sin(n\omega t)\mathrm{d}t$$
$$ \omega = \frac{2\pi}{T}$$

Now lets study the effect of a shift in $t$ by an amount $\tau$ on the Fourier coefficients of $f$. We could study this by analyzing the integral expressions for the coefficients, but it is easier to study it by manipulating the Fourier series term by term. 
$$ f(t+\tau) = a_0 + \sum_{n=1}^\infty \Big[a_n \cos\Big(n\omega (t+\tau)\Big) + b_n \sin\Big(n\omega (t+\tau)\Big)\Big]$$
$$ = a_0 + \sum_{n=1}^\infty \Big[a_n \cos(n\omega t+n\omega\tau) + b_n \sin(n\omega t+ n\omega\tau)\Big]$$
$$ = a_0 + \sum_{n=1}^\infty \Big[a_n \Big(\cos(n\omega t)\cos(n\omega\tau) - \sin(n\omega t)\sin(n \omega \tau) \Big) + b_n \Big( \sin(n\omega t)\cos(n\omega\tau) + \cos(n\omega t) \sin(n\omega\tau)\Big)\Big]$$
Collecting the coefficients of $\cos(n\omega t)$ and $\sin(n\omega t)$ together we get,
$$ = a_0 + \sum_{n=1}^\infty \Big[\Big(\color{blue}{ a_n \cos(n\omega\tau) + b_n \sin(n\omega\tau)} \Big) \cos(n\omega t) + \Big(\color{blue}{b_n \cos(n\omega\tau)-a_n \sin(n\omega\tau) }\Big) \sin(n\omega t) \Big]$$
The blue expressions are the new coefficients, lets call them $a_n'$ and $b_n'$
$$ = a_0 + \sum_{n=1}^\infty \Big[\color{blue}{ a_n'}  \cos(n\omega t) + \color{blue}{b_n'}\sin(n\omega t) \Big]$$
Note that the coefficient $a_0$ remains unchanged. 
The transformation rule is, 
$$ a_0 \rightarrow a_0$$
$$ a_n \rightarrow a_n \cos(n\omega\tau) + b_n \sin(n\omega\tau) \qquad (n\geq 1)$$
$$ b_n \rightarrow b_n \cos(n\omega\tau)-a_n \sin(n\omega\tau)$$

Note that these coefficients transform exactly as if they were the $x$ and $y$ components of a vector being rotated in the $xy$ plane by an angle $n\omega\tau$. 
