I'm having trouble figuring out the Fourier series of $\operatorname{sgn}(\sin(x))+\operatorname{sgn}(\cos(x))$
from $−\pi$ to $\pi$. How to determine if it's odd or even function? And how to determine period? This is the first time I'm seeing sgn function under integral and I don't know how to calculate anything I need for Fourier series.
[Math] Fourier series for $\operatorname{sgn}(\sin(x))+\operatorname{sgn}(\cos(x))$
calculusfourier analysisfourier seriessequences-and-series
Related Solutions
Although $ \int_0^\pi \cos(x)\,dx = 0$, $a_0\ne 0$ because $$\int_0^{\pi/2} |\cos(x)|\,dx=\int_{\pi/2}^{\pi} |\cos(x)|\,dx. $$
We can evaluate it as follows, as can be seen in the plot below
$$a_0 = \frac 1 \pi \int_{-\pi}^\pi |\cos(x)|\,dx=\frac 2 \pi \int_0^\pi |\cos(x)|\,dx=\frac 4 \pi \int_0^{\pi/2} |\cos(x)|\,dx = \frac 4 \pi \int_0^{\pi/2} \cos(x)\,dx=\frac 4 \pi.$$ $$\tag{1}$$
Plot of $\cos x$ (doted line) and $|\cos x|$ (solid line) in the interval $[-\pi,\pi]$.
The coefficients $b_n=0$ as you concluded. As for the $a_n$ coefficients only the odd ones are equal to $0$ (see below). The functions $\cos(x)$ and $\cos(nx)$ are orthogonal in the interval $[-\pi,\pi]$, but $|\cos(x)|$ and $\cos(nx)$ are not. Since
\begin{equation*} \left\vert \cos (x)\right\vert =\left\{ \begin{array}{c} \cos (x) \\ -\cos (x) \end{array} \begin{array}{c} \text{if} \\ \text{if} \end{array} \begin{array}{c} 0\leq x\leq \pi /2 \\ \pi /2\leq x\leq \pi, \end{array} \right. \tag{2} \end{equation*}
we have that
\begin{eqnarray*} a_{n} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }\left\vert \cos (x)\right\vert\cos (nx)\,dx=\frac{2}{\pi }\int_{0}^{\pi }\left\vert \cos (x)\right\vert \cos (nx)\,dx \\ &=&\frac{2}{\pi }\int_{0}^{\pi /2}\left\vert \cos (x)\right\vert \cos (nx)\,dx+\frac{2}{\pi }\int_{\pi /2}^{\pi }\left\vert \cos (x)\right\vert \cos (nx)\,dx \\ &=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos (x)\cos (nx)\,dx-\frac{2}{\pi } \int_{\pi /2}^{\pi }\cos (x)\cos (nx)\,dx. \\ a_{1} &=&\frac{2}{\pi }\int_{0}^{\pi /2}\cos ^{2}(x)\,dx-\frac{2}{\pi }\int_{\pi /2}^{\pi }\cos ^{2}(x)\,dx=0. \end{eqnarray*}
Using the following trigonometric identity, with $a=x,b=nx$, \begin{equation*} \cos (a)\cos (b)=\frac{\cos (a+b)+\cos (a-b)}{2},\tag{3} \end{equation*}
we find
\begin{eqnarray*} a_{2m} &=&\frac{4}{\pi \left( 1-4m^{2}\right) }\cos (\frac{2m\pi }{2})=\frac{ 4}{\pi \left( 1-4m^{2}\right) }(-1)^{m} \\ a_{2m+1} &=&\frac{4}{\pi ( 1-4(2m+1)^{2}) }\cos (\frac{(2m+1)\pi }{2})=0,\qquad m=1,2,3,\ldots.\tag{4} \end{eqnarray*}
The expansion of $\left\vert \cos (x)\right\vert $ into a trigonometric Fourier series in the interval $[-\pi ,\pi ]$ is thus
\begin{equation*} \left\vert \cos x\right\vert =\frac{a_{0}}{2}+\sum_{n=1}^{\infty }\left( a_{n}\cos (nx)+b_{n}\sin (nx)\right) =\frac{2}{\pi }+\frac{4}{\pi } \sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx)\tag{5} \end{equation*}
$$|\sin(x)|\ \text{(blue) and the partial sum }\frac{2}{\pi }+\frac{4}{\pi } \sum_{m=1}^{5 }\frac{(-1)^{m}}{1-4m^{2}}\cos (2mx) \ \text{(red) in }[-\pi,\pi]$$
Setting $x=0$ in $(5)$, we obtain
\begin{equation*} 1=\frac{2}{\pi }+\frac{4}{\pi }\sum_{m=1}^{\infty }\frac{(-1)^{m}}{1-4m^{2}}=\frac{2}{\pi }-\frac{4}{\pi }\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}.\tag{6} \end{equation*}
Hence
\begin{equation*} \sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{1-4n^{2}}=\frac{1}{2}-\frac{\pi }{4}.\tag{7} \end{equation*}
$ y(t)= sgn(\sin{\frac{2 \pi t}{T_0}})$ equals $1$ between $t=0$ and $t= $$T_0 \over 2$, and equals $-1$ between $t=$ $T_0 \over 2$ and $t = T_0$.
The fourier coefficients are: $c_k=\frac{1}{T_0}\int_{0}^{T_0}y(t)e^{-2 \pi i k t \over T_0 }dt=\frac{1}{T_0}\int_{0}^{T_0 \over 2}y(t)e^{-2 \pi i k t \over T_0 }+\frac{1}{T_0}\int_{T_0 \over 2}^{T_0 }y(t)e^{-2 \pi i k t \over T_0 }= $
$\frac{1}{T_0}\int_{0}^{T_0 \over 2}1\cdot e^{-2 \pi i k t \over T_0 }+\frac{1}{T_0}\int_{T_0 \over 2}^{T_0}(-1)\cdot e^{-2 \pi i k t \over T_0 }dt = $
$\frac{1}{-2\pi i k}(-1)^k + \frac{1}{2\pi i k} + \frac{1}{2\pi i k}+ \frac{1}{-2\pi i k}(-1)^k = $
$\frac{1}{\pi i k}+\frac{1}{\pi i k}(-1)^{k+1} = 0$ for $k$ even and $\frac{2}{\pi i k}$ for $k$ odd.
And $c_0= \frac{1}{T_0}\int_{0}^{T_0}y(t)dt = \frac{1}{T_0}\int_{0}^{T_0 \over 2}1 dt+\frac{1}{T_0}\int_{T_0 \over 2}^{T_0 }(-1)dt = \frac{1}{2} + 0 - 1 + \frac{1}{2} = 0$
Best Answer
HINT
Rewrite your function as a branch function. Note that $$\operatorname{sgn}(\sin(x)) = \begin{cases} 1, & x \in (0, \pi] \\ -1, & x \in [-\pi,0) \end{cases}$$ and you can do the same to the other term and then add them together. Once you have the branch function completely, it will be easy both to integrate and determine exact properties.