Expectation of product of cosine and sine $\theta\sim U(-\pi,\pi)$.
When $\theta$ follows uniform distribution, what is the expected value of the producot of cosine and sine, i.e.
$$E[\sin\theta \cos\theta] = \ ?$$
 A: Hint:
$\sin(-\theta)\cos(-\theta)=-\sin\theta\cos\theta$ and $\theta$ and $-\theta$ have the same distribution.
So: $$\mathbb E\sin\theta\cos\theta=\mathbb E\sin(-\theta)\cos(-\theta)=-\mathbb E\sin\theta\cos\theta$$
A: Hint
$$E[\cos\theta \sin\theta] = \frac1{2\pi} \int_{-\pi}^\pi \sin\theta \cos\theta \mathrm d\theta$$
Note that $\sin\theta\cos\theta = \frac12 \sin(2\theta)$ is an odd function. What do you know about symmetric integrals of odd functions?
A: Since the probability density function of $\mathcal{U}\big([-\pi,\pi]\big)$ is $\displaystyle x \, \mapsto \, \frac{1}{2\pi} \mathbf{1}_{[-\pi,\pi]}(x)$, by definition, if $\theta \sim \mathcal{U}\big( [-\pi,\pi] \big)$, we have : 
$$
\begin{align*}
\mathrm{E}\big[ \cos(\theta)\sin(\theta) \big] &= {} \int \cos(x)\sin(x) \frac{1}{2\pi} \mathbf{1}_{[-\pi,\pi]}(x) \, dx \\[2mm]
 &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \cos(x)\sin(x) \, dx \\[2mm]
 &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1}{2} \sin(2x) \, dx \\[2mm]
 &= \frac{1}{4\pi} \Big[ -\frac{1}{2} \cos(2x) \Big]_{-\pi}^{\pi} \\[2mm]
 &= 0.
\end{align*} $$
A: $$\begin{align}
\mathsf E[\cos\theta\sin\theta]
 &= \frac 1{2\pi}\int_{-\pi}^\pi \cos \theta\sin \theta\operatorname d \theta
\\ &= \frac 1{4\pi}\int_{-\pi}^\pi \sin (2\theta)\operatorname d \theta 
\end{align}$$
