1. **Problem Statement:** Determine the Fourier series of a triangular waveform $x(t)$ with amplitude $A$ and period $T=2\pi$. 2. **Fourier Series Formula:** For a periodic function with period $T$, the Fourier series is given by: $$ x(t) = a_0 + \sum_{n=1}^\infty \left(a_n \cos\left(\frac{2\pi n t}{T}\right) + b_n \sin\left(\frac{2\pi n t}{T}\right)\right) $$ where $$ a_0 = \frac{1}{T} \int_0^T x(t) dt, \quad a_n = \frac{2}{T} \int_0^T x(t) \cos\left(\frac{2\pi n t}{T}\right) dt, \quad b_n = \frac{2}{T} \int_0^T x(t) \sin\left(\frac{2\pi n t}{T}\right) dt $$ 3. **Properties of the Triangular Wave:** The triangular wave is an even function, so all $b_n = 0$ because sine terms are odd functions and their integral over symmetric intervals is zero. 4. **Calculate $a_0$:** The average value over one period is zero because the waveform is symmetric about the horizontal axis: $$ a_0 = 0 $$ 5. **Calculate $a_n$ coefficients:** The triangular wave can be expressed as a piecewise linear function. Using the known Fourier series for a triangular wave with amplitude $A$ and period $2\pi$: $$ a_n = \frac{8A}{\pi^2 n^2} (-1)^{n/2} \quad \text{for even } n, \quad a_n = 0 \text{ for odd } n $$ More precisely, the Fourier series of a triangular wave is: $$ x(t) = \frac{8A}{\pi^2} \sum_{n=1,3,5,...}^\infty \frac{(-1)^{\frac{n-1}{2}}}{n^2} \cos(nt) $$ where the sum is over odd $n$ only. 6. **Final Fourier Series:** $$ x(t) = \frac{8A}{\pi^2} \sum_{n=1,3,5,...}^\infty \frac{(-1)^{\frac{n-1}{2}}}{n^2} \cos(nt) $$ This series converges to the triangular waveform shown. **Summary:** The triangular wave has only cosine terms with coefficients decreasing as $1/n^2$ for odd harmonics.