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$$

f(\theta) = \begin{cases}

\theta, & 0\leq \theta \leq\pi\\

\pi + \theta, & -\pi\leq \theta < 0

\end{cases}.

$$

$$

a_0 = \frac{1}{\pi}\int_0^{\pi}\theta d\theta + \frac{1}{\pi}\int_{-\pi}^0\theta d\theta + \int_{-\pi}^0 d\theta

$$

The first and second integral together are 0 so the $a_0 = \pi$

$$

a_n = \frac{1}{\pi}\int_0^{\pi}\theta\cos n\theta d\theta + \frac{1}{\pi}\int_{-\pi}^0\theta\cos n\theta d\theta + \int_{-\pi}^0 \cos n\theta d\theta

$$

The first and second integral together are 0 so the $a_n = 0$

$$

b_n = \frac{1}{\pi}\int_0^{\pi}\theta\sin n\theta d\theta + \frac{1}{\pi}\int_{-\pi}^0\theta\sin n\theta d\theta + \int_{-\pi}^0\sin n\theta d\theta

$$

The first and second integral together are 0 so the $b_n = \begin{cases}\frac{-2}{n}, & \text{if n is odd}\\0, & \text{if n is even}\end{cases}$

So

$$

\frac{\pi}{2} - 2\sum_{n = 1}^{\infty}\frac{1}{2n-1}\sin(2n - 1)\theta

$$

Correct?