91Ó°ÊÓ

A Fourier series of a periodic function with period \( 2\pi \) is given by: \[ f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nt) + b_n \sin(nt) \right). \]Where, \[ a_0 = \frac{1}{2\pi}\int_0^{2\pi} f(t) \, dt\]\[ a_n = \frac{1}{\pi}\int_0^{2\pi} f(t) \cos(nt) \, dt\]\[ b_n = \frac{1}{\pi}\int_0^{2\pi} f(t) \sin(nt) \, dt. \]For this function, \(f(t) = 5 \sin(t)\) for \(0 \leq t \leq \pi \) and \(f(t) = 0\) for \(\pi \leq t \leq 2\pi \). First, we calculate \( a_0 \):\[ a_0 = \frac{1}{2\pi} \left( \int_0^{\pi} 5 \sin(t) \, dt + \int_{\pi}^{2\pi} 0 \, dt \right). \]Calculating this gives:\[ a_0 = \frac{5}{2\pi} \left( -\cos(t) \right)_0^{\pi} = 0. \]

\( a_n \) is given by:\[ a_n = \frac{1}{\pi} \left( \int_0^{\pi} 5 \sin(t) \cos(nt) \, dt + \int_{\pi}^{2\pi} 0 \cdot \cos(nt) \, dt \right). \]The integral in the first term is specifically:\[ a_n = \frac{5}{\pi} \int_0^{\pi} \sin(t) \cos(nt) \, dt. \]Using integration by parts or known results, this evaluates to zero due to the orthogonality of \( \sin(t) \) and \( \cos(nt) \) over \([0,\pi]\). Thus, \( a_n = 0 \) for all \( n \geq 1 \).

\( b_n \) is given by:\[ b_n = \frac{1}{\pi} \left( \int_0^{\pi} 5 \sin(t) \sin(nt) \, dt + \int_{\pi}^{2\pi} 0 \cdot \sin(nt) \, dt \right). \]The integral in the first term is particularly:\[ b_n = \frac{5}{\pi} \int_0^{\pi} \sin(t) \sin(nt) \, dt. \]Using the identity \( \sin A \sin B = \frac{1}{2} \left[ \cos(A-B) - \cos(A+B) \right] \), this becomes:\[ b_n = \frac{5}{\pi} \times \frac{1}{2} \left[ \int_0^{\pi} \cos((1-n)t) \, dt - \int_0^{\pi} \cos((1+n)t) \, dt \right]. \]Evaluating these integrals provides:- When \( n = 1 \), the integral over \( \cos(0 \cdot t)\) results in \(\pi\), so \( b_1 = \frac{5}{2} \).- When \( n eq 1 \), both terms result in zero because the integrals of cosine functions over their periods are zero.Thus, \( b_n = 0 \) for \( n eq 1 \) and \( b_1 = \frac{5}{2} \).

We now construct the Fourier series for \( f(t) \) using the coefficients:\[ f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nt) + b_n \sin(nt) \right). \]Thus, with \( a_0 = 0 \), \( a_n = 0 \) for \( n \geq 1 \), and \( b_n = 0 \) except \( b_1 = \frac{5}{2} \), the Fourier series becomes:\[ f(t) = \frac{5}{2} \sin(t). \]