4. If \( \pi<\theta<\frac{3}{3} \) with \( \cos (\theta)=-\frac{y \sqrt{3}}{5} \), find \( \sin (\theta) \)

Given the problem, we have \(\theta\) in radians such that \(\pi < \theta < 2\). This means \(\theta\) is in the third quadrant where sine is negative and cosine is also negative. We are told that \(\cos(\theta) = -\frac{y \sqrt{3}}{5}\). Using the Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] Substituting \(\cos(\theta)\) into the equation, we get: \[ \sin^2(\theta) + \left(-\frac{y \sqrt{3}}{5}\right)^2 = 1 \] Simplifying the cosine term: \[ \sin^2(\theta) + \frac{3y^2}{25} = 1 \] Now, we can isolate \(\sin^2(\theta)\): \[ \sin^2(\theta) = 1 - \frac{3y^2}{25} \] To find \(\sin(\theta)\), we take the square root of both sides. Since we are in the third quadrant where sine is negative, we have: \[ \sin(\theta) = -\sqrt{1 - \frac{3y^2}{25}} \] This is the expression for \(\sin(\theta)\) given the constraints of the problem. To finalize, we can express it as: \[ \sin(\theta) = -\sqrt{\frac{25 - 3y^2}{25}} = -\frac{\sqrt{25 - 3y^2}}{5} \] Thus, the final result for \(\sin(\theta)\) is: \[ \sin(\theta) = -\frac{\sqrt{25 - 3y^2}}{5} \] This provides the sine of the angle \(\theta\) based on the given cosine value.