Let's solve the problem step-by-step: --- ### **Step 1: Identify the Function and Its Symmetry** The function is given as: \[ f(x) = \begin{cases} -\frac{\pi}{4}, & -\pi < x < 0 \\ \frac{\pi}{4}, & 0 < x < \pi \end{cases} \] and is periodic with period \(2\pi\). Notice that \(f(x)\) is an **odd function** since \(f(-x) = -f(x)\). Therefore, the Fourier series will only have **sine terms** (i.e., only \(b_n\) coefficients). **Explanation:** Recognizing the symmetry simplifies our calculations, as we only need to compute the sine coefficients \(b_n\). --- ### **Step 2: Calculate Fourier Sine Coefficients \(b_n\)** The general formula for the sine coefficients for a function of period \(2\pi\) is: \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \] Since \(f(x)\) is odd, and \(\sin(nx)\) is also odd, their product is **even**, so we can write: \[ b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin(nx) \, dx \] But for \(0 < x < \pi\), \(f(x) = \frac{\pi}{4}\). So, \[ b_n = \frac{2}{\pi} \int_{0}^{\pi} \frac{\pi}{4} \sin(nx) \, dx = \frac{2}{\pi} \cdot \frac{\pi}{4} \int_{0}^{\pi} \sin(nx) \, dx = \frac{1}{2} \int_{0}^{\pi} \sin(nx) \, dx \] Now, compute the integral: \[ \int_{0}^{\pi} \sin(nx) \, dx = \left[ -\frac{1}{n} \cos(nx) \right]_{0}^{\pi} = -\frac{1}{n} (\cos(n\pi) - \cos(0)) = -\frac{1}{n} ((-1)^n - 1) \] So, \[ b_n = \frac{1}{2} \left(-\frac{1}{n} ((-1)^n - 1)\right) = -\frac{1}{2n} ((-1)^n - 1) \] For **odd \(n\)**: \((-1)^n - 1 = -2\), For **even \(n\)**: \((-1)^n - 1 = 0\). Thus, \[ b_n = \begin{cases} \frac{1}{n}, & n \text{ odd} \\ 0, & n \text{ even} \end{cases} \] **Explanation:** We simplified the calculation by using the function's symmetry and the properties of sine integrals, resulting in nonzero sine coefficients only for odd \(n\). --- ### **Final Answer: Fourier Series** \[ \boxed{ f(x) = \sum_{n=1,\, n\,\text{odd}}^{\infty} \frac{1}{n} \sin(nx) = \sin(x) + \frac{1}{3} \sin(3x) + \frac{1}{5} \sin(5x) + \cdots } \] This is the Fourier series representation of the given function.