Let's solve the problem step by step as requested: --- ### **Step 1: Fourier Sine Series Expansion** **Given interval:** \( 0 < x < l \) **Function:** \[ f(x) = \begin{cases} x, & 0 < x < \frac{l}{2} \\ l - x, & \frac{l}{2} < x < l \end{cases} \] The Fourier sine series for \( f(x) \) on \( (0, l) \) is: \[ f(x) = \sum_{n=1}^{\infty} b_n \sin \left( \frac{n\pi x}{l} \right) \] where \[ b_n = \frac{2}{l} \int_0^l f(x) \sin \left( \frac{n\pi x}{l} \right) dx \] Split the integral according to the definition of \( f(x) \): \[ b_n = \frac{2}{l} \left[ \int_0^{\frac{l}{2}} x \sin\left(\frac{n\pi x}{l}\right) dx + \int_{\frac{l}{2}}^l (l - x) \sin\left(\frac{n\pi x}{l}\right) dx \right] \] **Explanation:** We set up the sine series and broke the integral into two parts based on the piecewise nature of \( f(x) \). --- ### **Step 2: Fourier Cosine Series Expansion** The Fourier cosine series for \( f(x) \) on \( (0, l) \) is: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos \left( \frac{n \pi x}{l} \right) \] where \[ a_0 = \frac{2}{l} \int_0^l f(x) dx \] \[ a_n = \frac{2}{l} \int_0^l f(x) \cos \left( \frac{n\pi x}{l} \right) dx \] Again, split the integrals: \[ a_0 = \frac{2}{l} \left[ \int_0^{\frac{l}{2}} x dx + \int_{\frac{l}{2}}^l (l-x) dx \right] \] \[ a_n = \frac{2}{l} \left[ \int_0^{\frac{l}{2}} x \cos\left(\frac{n\pi x}{l}\right) dx + \int_{\frac{l}{2}}^l (l-x) \cos\left(\frac{n\pi x}{l}\right) dx \right] \] **Explanation:** We set up the cosine series, again dividing the integrals according to the definition of \( f(x) \). --- ## **Final Answer (in Setup Form)** ### **Fourier Sine Series:** \[ f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{l}\right) \] with \[ b_n = \frac{2}{l} \left[ \int_0^{\frac{l}{2}} x \sin\left(\frac{n\pi x}{l}\right) dx + \int_{\frac{l}{2}}^l (l-x) \sin\left(\frac{n\pi x}{l}\right) dx \right] \] --- ### **Fourier Cosine Series:** \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi x}{l}\right) \] with \[ a_0 = \frac{2}{l} \left[ \int_0^{\frac{l}{2}} x dx + \int_{\frac{l}{2}}^l (l-x) dx \right] \] \[ a_n = \frac{2}{l} \left[ \int_0^{\frac{l}{2}} x \cos\left(\frac{n\pi x}{l}\right) dx + \int_{\frac{l}{2}}^l (l-x) \cos\left(\frac{n\pi x}{l}\right) dx \right] \] --- **You can evaluate these integrals further to get the explicit coefficients if required.** If you want the full evaluation, let me know!