Answer
See explanations.
Work Step by Step
Step 1. We know $2\pi$ is a period of $f(\theta)=sin\theta$. But we need to show that it is the smallest period.
Step 2. Assume there is a smaller period $0\lt p\lt 2\pi$ such that $sin(\theta+p)=sin\theta$. Let $\theta=0$, we have $sin(p)=0\longrightarrow p=\pi$
Step 3. Let $\theta=\frac{\pi}{2}$, we have $sin(\frac{\pi}{2}+p)=1\longrightarrow \frac{\pi}{2}+p=\frac{\pi}{2}\longrightarrow p=0$. This contradicts to the assumption of $0\lt p\lt 2\pi$ as well as the result from Step 2.
Step 4. Thus the smallest period of $f(\theta)=sin\theta$ is $2\pi$.
