1. **State the problem:** Evaluate the expression $\cos^2(10^\circ) + \cos^2(50^\circ) - \sin(40^\circ) \sin(80^\circ)$.\n\n2. **Recall relevant formulas:** Use the Pythagorean identity and product-to-sum formulas.\n- $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$\n- $\sin A \sin B = \frac{\cos(A-B) - \cos(A+B)}{2}$\n\n3. **Apply the formulas:**\n\nConvert each term:\n$$\cos^2(10^\circ) = \frac{1 + \cos(20^\circ)}{2}$$\n$$\cos^2(50^\circ) = \frac{1 + \cos(100^\circ)}{2}$$\n$$\sin(40^\circ) \sin(80^\circ) = \frac{\cos(40^\circ - 80^\circ) - \cos(40^\circ + 80^\circ)}{2} = \frac{\cos(-40^\circ) - \cos(120^\circ)}{2}$$\n\n4. **Simplify the expression:**\n$$\cos^2(10^\circ) + \cos^2(50^\circ) - \sin(40^\circ) \sin(80^\circ) = \frac{1 + \cos(20^\circ)}{2} + \frac{1 + \cos(100^\circ)}{2} - \frac{\cos(-40^\circ) - \cos(120^\circ)}{2}$$\nCombine terms over common denominator 2:\n$$= \frac{1 + \cos(20^\circ) + 1 + \cos(100^\circ) - \cos(-40^\circ) + \cos(120^\circ)}{2}$$\n\n5. **Use even property of cosine:** $\cos(-\theta) = \cos \theta$, so $\cos(-40^\circ) = \cos(40^\circ)$.\n\nRewrite numerator:\n$$2 + \cos(20^\circ) + \cos(100^\circ) - \cos(40^\circ) + \cos(120^\circ)$$\n\n6. **Evaluate cosines (in degrees):**\n- $\cos(20^\circ) \approx 0.9397$\n- $\cos(100^\circ) \approx -0.1736$\n- $\cos(40^\circ) \approx 0.7660$\n- $\cos(120^\circ) \approx -0.5$\n\nSum numerator:\n$$2 + 0.9397 - 0.1736 - 0.7660 - 0.5 = 2 + 0.9397 - 0.1736 - 0.7660 - 0.5 = 1.5$$\n\n7. **Divide by 2:**\n$$\frac{1.5}{2} = 0.75$$\n\n**Final answer:**\n$$\cos^2(10^\circ) + \cos^2(50^\circ) - \sin(40^\circ) \sin(80^\circ) = 0.75$$