If a signal is given on an interval $[0, 2\pi]$, the Fourier series can be written as $$ f(t) = c_0 + \sum_{n = 1}^\infty a_n\cos nt + b_n\sin nt $$ with coefficients $$ c_0 = \frac1{2\pi}\int_0^{2\pi}f(t)\,dt\\ a_n = \frac1{\pi}\int_0^{2\pi}f(t)\cos nt\,dt\\ b_n = \frac1{\pi}\int_0^{2\pi}f(t)\sin nt\,dt $$
Specifically, for the coefficients, where does the $1/\pi$ and $1/2\pi$ come from?
On
It's for niceness. For instance, if $f(x) = 1$, then all the $a_i$ and $b_i$ integrals turn out to be $0$, and you want to recover $c_0 = 1$. But it turns out that $\int_0^{2\pi} 1\,dt = 2\pi$ is too large, so you scale it back down.
Similarily, if $f(x) = \cos kx$, for some natural number $k>0$, then $a_k$ is the only integral that becomes non-zero, and we want to recover $a_k = 1$. However, it turns out that $\int_0^{2\pi}\cos kt\cdot \cos kt\,dt = \pi$, which is too large, so we need to scale it down. (And the argument for $b_k$ is exactly the same.)
Alternatively, we could do the definitions of $a_k, b_k$ and $c_0$ without the $\frac1{2\pi}$ and $\frac1\pi$ factors, but then we would have to make up for that someplace else, like with
$$f(x) = \frac{c_0}{2\pi} + \frac1\pi\sum_{n = 0}^\infty a_n\cos nx + b_n\sin nx$$
It's a matter of personal preference where you would best like to put these factors, but they must appear somewhere.
On
More abstractly, with an orthonormal basis $\{u_1, u_2, \dots\}$ in a (real) inner product space, we have $$ f = \sum a_n u_n $$ where the coefficients are the inner products $a_n = \langle f,u_n\rangle$. In our traditional case, if the inner product is $$ \langle f,g \rangle = \int_0^{2\pi} f(x)\;g(x)\;dx $$ we need to do integrals $$ \int_0^{2\pi} 1^2\;dx = 2\pi,\\ \int_0^{2\pi} \cos(nx)^2\;dx = \pi,\\ \int_0^{2\pi} \sin(nx)^2\;dx = \pi, $$ to find the denominators to use so that we have an orthonormal basis.
On
Please look at the derivation of these coefficients, Suppose you have the formula and you want to calculate the coefficients then for example i will tell you for $c_0$ ,
Take integration on both sides w.r.t to $t$ with limit from $0$ to $2\pi$, $$ \int_0^{2\pi}f(t)dt=\int_0^{2\pi}c_0dt+\sum_{n=1} \left( \int_0^{2\pi}a_ncos(nt)dt+\int_0^{2\pi}b_nsin(nt)dt\right) $$ All coefficients are constant w.r.t. $t$ so, $$ \int_0^{2\pi}f(t)dt=c_0\int_0^{2\pi}dt+\sum_{n=1} \left( a_n\int_0^{2\pi}cos(nt)dt+b_n\int_0^{2\pi}sin(nt)dt\right) $$ But $\int_0^{2\pi}cos(nt)dt$ and $\int_0^{2\pi}sin(nt)dt$ are $0$ so you get, $$ \int_0^{2\pi}f(t)dt=c_0\int_0^{2\pi}dt $$ Which is eqaul to, $$ \int_0^{2\pi}f(t)dt=c_02\pi $$ Thus you get, $$ c_0=\frac{1}{2\pi}\int_0^{2\pi}f(t)dt $$
They come from the periodicity of $2\pi$ (The length of the interval $[0,2\pi]$ is $2\pi-0 = 2\pi$). If the period is something else, say $T$, then the values would change.
Note that when $T = 2\pi$, then $\frac{1}{\pi} = \frac{2}{T}$ and $\frac{1}{2\pi} = \frac1T$. Similarly, the integration would be $\int_0^{2\pi} f(t) dt = \int_0^T f(t) dt.$