1. **State the problem:** We need to find $\frac{dy}{dx}$ given the parametric equations $x = t \sin t$ and $y = \cos t$ at $t = \frac{\pi}{2}$. 2. **Recall the formula:** For parametric equations $x = f(t)$ and $y = g(t)$, the derivative $\frac{dy}{dx}$ is given by $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$$ 3. **Find $\frac{dx}{dt}$:** $$x = t \sin t,$$ Using the product rule, $$\frac{dx}{dt} = \sin t + t \cos t.$$ 4. **Find $\frac{dy}{dt}$:** $$y = \cos t,$$ so $$\frac{dy}{dt} = -\sin t.$$ 5. **Evaluate derivatives at $t = \frac{\pi}{2}$:** $$\frac{dx}{dt} \bigg|_{t=\frac{\pi}{2}} = \sin \frac{\pi}{2} + \frac{\pi}{2} \cos \frac{\pi}{2} = 1 + \frac{\pi}{2} \times 0 = 1,$$ $$\frac{dy}{dt} \bigg|_{t=\frac{\pi}{2}} = -\sin \frac{\pi}{2} = -1.$$ 6. **Calculate $\frac{dy}{dx}$ at $t = \frac{\pi}{2}$:** $$\frac{dy}{dx} = \frac{-1}{1} = -1.$$ **Final answer:** $$\boxed{-1}.$$