Correlation between sine and cosine

Question

Suppose $X$ is uniformly distributed on $[0, 2\pi]$. Let $Y = \sin X$ and $Z = \cos X$. Show that the correlation between $Y$ and $Z$ is zero.

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It seems I would need to know the standard deviation of the sine and cosine, and their covariance. How can I calculate these?

I think I need to assume $X$ has uniform distribution, and the look at the transformed variables $Y=\sin(X)$ and $Z=\cos(X)$. Then the law of the unconscious statistician would give the expected value

$$E[Y] = \frac{1}{b-a}\int_{-\infty}^{\infty} \sin(x)dx$$ and $$E[Z] = \frac{1}{b-a}\int_{-\infty}^{\infty} \cos(x)dx$$

(the density is constant since it is a uniform distribution, and can thus be moved out of the integral).

However, those integrals are not defined (but have Cauchy principal values of zero I think).

How could I solve this problem? I think I know the solution (correlation is zero because sine and cosine have opposite phases) but I cannot find how to derive it.

Answer 1

Since

$$\begin{align} \operatorname{Cov}(Y, Z) &= E[(Y - E[Y])(Z - E[Z])] \\ &= E[(Y - {\textstyle \int}_0^{2\pi} \sin x \;dx)(Z - {\textstyle \int}_0^{2\pi} \cos x \;dx)] \\ &= E[(Y - 0)(Z - 0)] \\ &= E[YZ] \\ &= \int_0^{2\pi} \sin x \cos x \;dx \\ &= 0 , \end{align}$$

the correlation must also be 0.

Answer 2

I really like @whuber's argument from symmetry and don't want it to be lost as a comment, so here's a bit of elaboration.

Consider the random vector $(X, Y)$, where $X = \cos(U)$, and $Y = \sin(U)$, for $U \sim U(0, 2 \pi)$. Then, because $\theta \mapsto (\cos(\theta), \sin(\theta))$ parameterizes the unit circle by arc length, $(X, Y)$ is distributed uniformly on the unit circle. In particular, the distribution of $(-X, Y)$ is the same as the distribution of $(X, Y)$. But then

$$ - \text{Cov} (X, Y) = \text{Cov} (-X, Y) = \text{Cov} (X, Y) $$

so it must be that $\text{Cov} (X, Y) = 0$.

Just a beautiful geometric argument.