1. **Problem:** Evaluate $\sin^2(60^\circ) + \cos^2(60^\circ)$. 2. **Formula and rule:** The Pythagorean identity states that for any angle $\theta$, $$\sin^2(\theta) + \cos^2(\theta) = 1.$$ This is a fundamental trigonometric identity valid for all angles. 3. **Intermediate work:** Substitute $\theta = 60^\circ$: $$\sin^2(60^\circ) + \cos^2(60^\circ) = 1.$$ We can verify by calculating values: $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, so $\sin^2(60^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$. $\cos(60^\circ) = \frac{1}{2}$, so $\cos^2(60^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$. Sum: $$\frac{3}{4} + \frac{1}{4} = 1.$$ 4. **Explanation:** This confirms the identity and shows the sum is exactly 1. **Final answer:** $$\sin^2(60^\circ) + \cos^2(60^\circ) = 1.$$