1. **State the problem:** Find the derivative of the function $$y = \sin^2(\pi t - 2)$$ with respect to $$t$$. 2. **Recall the formula:** To differentiate $$y = [f(t)]^2$$, use the chain rule: $$\frac{dy}{dt} = 2 f(t) \cdot f'(t)$$. 3. **Identify inner function:** Here, $$f(t) = \sin(\pi t - 2)$$. 4. **Differentiate inner function:** $$f'(t) = \cos(\pi t - 2) \cdot \frac{d}{dt}(\pi t - 2) = \cos(\pi t - 2) \cdot \pi$$. 5. **Apply chain rule:** $$\frac{dy}{dt} = 2 \sin(\pi t - 2) \cdot \pi \cos(\pi t - 2)$$ 6. **Simplify using double-angle identity:** Recall $$\sin(2x) = 2 \sin x \cos x$$, so $$\frac{dy}{dt} = \pi \sin(2(\pi t - 2)) = \pi \sin(2\pi t - 4)$$. **Final answer:** $$\boxed{\frac{dy}{dt} = \pi \sin(2\pi t - 4)}$$