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I'd like to work out $\operatorname{Cov}(\cos(2U), \cos(3U))$ where $U$ is uniformly distributed on $[0, \pi]$.

I believe this involves computing $\mathbb{E}[\cos(2U)\cos(3U)]$. If so, then I first need the joint density, which I'm stuck on.

I can work out the density of $\cos(3U)$; this is similar to this question. $f_{\cos(3U)}(x) = \frac{1}{\pi\sqrt{1 - x^2}}$, $x \in (-1, 1)$. But that and $f_{\cos(2U)}(x)$ are only (directly) useful for individual expected values.

How do I work out the joint density?

(In this answer, the joint density seems to be $1$. Don't know why though.)

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    $\begingroup$ You don't need the joint density to find this expectation, thanks to en.wikipedia.org/wiki/Law_of_the_unconscious_statistician. $\endgroup$
    – StubbornAtom
    Commented Dec 30, 2023 at 7:06
  • $\begingroup$ Much easier now. $\endgroup$
    – johnsmith
    Commented Dec 30, 2023 at 9:39
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    $\begingroup$ If you consider that for a given value of $X=\cos(2U)$, then either (i) $Y=\cos(3U)$ takes one of two possible values, equal in magnitude and opposite in sign (one from $U<\pi/2$, one from $U>\pi/2$), or (ii) $Y$ is $0$; then it's immediately clear that the pair of possible values are equally probable and hence that their corresponding contributions to the covariance cancel. Given $X$ and $Y$ are both bounded, immediately $E(XY)=0$. $\endgroup$
    – Glen_b
    Commented Dec 31, 2023 at 2:50

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