I try to determine if $X$ and $Y$ are (un)correlated and (in)dependent, when $X = \sin(\theta)$, $Y = \cos(\theta)$ and $\theta \sim N(0,1)$.
I know about the following formula:
$corr(X,Y) = \frac{cov(X,Y)}{\sqrt{var(X)var(Y)}}$
$cov(X,Y) = E(XY ) − E(X)E(Y)$
so I need to determine the following:
$E(X) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \sin(x) \cdot e^{\frac{-x^2}{2}} dx $
$E(Y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \cos(x) \cdot e^{\frac{-x^2}{2}} dx $
$E(XY) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \sin(x) \cdot \cos(x) \cdot e^{\frac{-x^2}{2}} dx$
There should be some way to solve these integrals by symmetry, considering the fact that $sin(x)$ is an odd and $cos(x)$ is an even function. How do I deal with the exponential term $e^{\frac{-x^2}{2}}$?
Integral of an odd function over $\mathbb R$ is $0$. This makes $EX=0$. Also, $\sin x \cos x=\frac 1 2 \sin (2x)$ (an odd function) and so $EXY=0$. Hence, $X$ and $Y$ are uncorrelated.
$X^{2}=1-Y^{2}$. If $X$ and $Y$ are independent then $X^{2}$ would be independent of $1-Y^{2}=X^{2}$ and $X^{2}$ would be a constant. I will let you see why this is not true.