Expectation of 2 functions with one random variable This may be a trivial question but I want to consult with you all. 
Let U be a continuous random variable taking values int he interval [0,2pi]. Let X = cos(U), Y = sin(U). Determine the Pearson correlation between X and Y.
How to calculate the E[XY]?
 A: Using the identity $\sin(2U) = 2\sin(U)\cos(U)$, we get:
$\displaystyle\mathbb{E}[XY] = \mathbb{E}[\sin(U)\cos(U)] =   \mathbb{E}\left[\frac{\sin(2U)}{2}\right] = \frac{1}{2}\mathbb{E}\left[\sin(2U)\right] = \frac{1}{2}\int_0^{2\pi}\frac{1}{2\pi}\sin(2\theta)d\theta = \frac{1}{8\pi}\int_0^{4\pi}\sin(x)dx = 0 $
A: Solution by regression
This argument requires some geometric intuition and knowing that the Pearson correlation is proportional to the linear regression coefficient of $Y$ against $X$.
As $U$ ranges from $0$ to $2\pi$, $(X,Y)$ describes a circle, covering the upper half of the circle in the same way it covers the lower half. Thus the conditional distribution of $Y$ given $X$ is symmetric about $0$, forcing the regression line to be $Y=0$, implying the correlation is zero, whence the covariance is zero, too.  Since (by circular symmetry) both $X$ and $Y$ have zero expectations, the covariance is merely $\mathbb{E}(XY)$ and we're done.

Solution by circular symmetry
That had a hand-waving intuitive flavor to it.  Maybe we can do a little better.  Here's a more elaborate version of the same idea.  It doesn't require knowing about regression.
Changing $U$ to $-U$ does not change the distribution of $(X,Y)$: it only causes $(X,Y)$ to traverse the circle clockwise rather than counterclockwise and the circle is still covered uniformly because the distribution of $-U$ is uniform between $-2\pi$ and $0$.  But changing $U$ to $-U$ leaves $X$ alone and converts $Y$ to $-Y$.  Because the distribution is unchanged, the expectations remain the same, too: $$\mathbb{E}(XY)=\mathbb{E}(X(-Y)) = -\mathbb{E}(XY).$$  Since $|XY|$ is bounded, the expectation of $XY$ exists and is finite, whence the only possible value of its expectation is zero.

Solution by algebra and (very simple) trigonometry
OK, maybe that doesn't seem sufficiently rigorous.  Here's another approach.
Cosine and sine depend only on $U$ up to a multiple of $2\pi$.  Therefore you may allow $U$ to range uniformly over any interval of length $2\pi$, such as from $\pi/2$ to $\pi/2+2\pi$.  That's equivalent to adding the start of the interval, $\pi/2$, to $U$. In light of this, notice that
$$X = \cos(U)\text{ becomes } \cos(U+\pi/2) = \sin(U) = -Y$$
and
$$Y = \sin(U)\text{ becomes } \sin(U+\pi/2) = \cos(U) = X$$
The expectation of $(X+Y)^2$ thereby becomes
$$\eqalign{
\mathbb{E}(X^2+Y^2) + \mathbb{E}(2XY) &= \mathbb{E}((X+Y)^2) \\
&= \mathbb{E}((-Y+X)^2) = \mathbb{E}(X^2+Y^2) + \mathbb{E}(-2XY).
} $$
(These equations are the consequences of straightforward algebraic relationships $(X\pm Y)^2 = X^2+Y^2\pm 2XY$ and the linearity of expectation.)
Subtracting the right from the left and dividing that by $4$ yields
$$\mathbb{E}(XY) = 0,$$
QED.
