Step 1: Compute the marginal expectations of X and Y.
Since \(\Theta\) is uniform on \((0, 2\pi)\), its density is \(f(\theta) = \frac{1}{2\pi}\) for \(\theta \in (0, 2\pi)\). \[ E[X] = \int_0^{2\pi} \cos\theta \cdot \frac{1}{2\pi}\, d\theta = \frac{1}{2\pi}\Big[\sin\theta\Big]_0^{2\pi} = 0 \] \[ E[Y] = \int_0^{2\pi} \sin\theta \cdot \frac{1}{2\pi}\, d\theta = \frac{1}{2\pi}\Big[-\cos\theta\Big]_0^{2\pi} = 0 \]
Step 2: Compute \(E[XY]\) and hence \(cov(X,Y)\).
\[ E[XY] = E[\cos\theta \sin\theta] = \int_0^{2\pi} \cos\theta \sin\theta \cdot \frac{1}{2\pi}\, d\theta = \int_0^{2\pi} \frac{\sin 2\theta}{2} \cdot \frac{1}{2\pi}\, d\theta \] The integral of \(\sin 2\theta\) over a full period \((0, 2\pi)\) is zero, so \(E[XY] = 0\). Using the hint, \[ cov(X,Y) = E[(X-E(X))(Y-E(Y))] = E[XY] - E[X]E[Y] = 0 - (0)(0) = 0 \] Since the covariance is zero, X and Y are uncorrelated.
Step 3: Test independence directly, not just correlation.
Independence requires the joint density to factor as the product of the marginals for every value, a much stronger condition than zero covariance. Here X and Y satisfy the deterministic identity \[ X^2 + Y^2 = \cos^2\theta + \sin^2\theta = 1 \] So once X is known, Y is restricted to at most two possible values, \(Y = \pm\sqrt{1-X^2}\); Y is not free to take any value independent of X. This deterministic constraint means X and Y are dependent, even though they are uncorrelated.
Step 4: Match to the options.
The conclusion is that X and Y are dependent and uncorrelated, which is option (A). Option (B) is wrong because independence fails, Step 3. Options (C) and (D) are wrong because the covariance was shown to be exactly zero in Step 2, so X and Y cannot be correlated.
\[ \boxed{\text{Option (A): } X \text{ and } Y \text{ are dependent and uncorrelated}} \]