Answer
See below.
Work Step by Step
1. To show that the period of $f(\theta)=sin\theta$ is $2\pi$, since we know $2\pi$ is a period, we only need to show that there is no other period less than $2\pi$.
2. Assume there is another period $p$ with $0\lt p\lt 2\pi$ and $sin(\theta+p)=sin\theta$ for all $\theta$, let $\theta=0$, we have $sin(p)=0$ which gives $p=\pi$.
3. Now let $\theta=\frac{\pi}{2}$, we have $sin(p+\frac{\pi}{2})=sin\frac{\pi}{2}=1$ which gives $p+\frac{\pi}{2}=\frac{\pi}{2}$ or $p=0$.
4. Since the results of steps 2 and 3 are contradict to each other, such $p$ value does not exist, which proves that the minimum period is $2\pi$.
