1. **Problem statement:** Use the Mean Value Theorem (MVT) to show that $$\cos b - \cos a \leq b - a$$ for all real numbers $$a,b$$ such that $$a < b$$. 2. **Recall the Mean Value Theorem:** If a function $$f$$ is continuous on $$[a,b]$$ and differentiable on $$(a,b)$$, then there exists some $$c \in (a,b)$$ such that $$$f'(c) = \frac{f(b) - f(a)}{b - a}.$$$ 3. **Apply MVT to $$f(x) = \cos x$$:** - $$f$$ is continuous and differentiable everywhere. - So, there exists $$c \in (a,b)$$ such that $$$-\sin c = \frac{\cos b - \cos a}{b - a}.$$$ 4. **Analyze the inequality:** - Since $$-1 \leq \sin c \leq 1$$, we have $$-1 \leq \sin c \leq 1$$. - Therefore, $$-\sin c \leq 1$$. - Multiply both sides by $$b - a > 0$$ (since $$a < b$$), preserving inequality: $$$\cos b - \cos a = (b - a)(-\sin c) \leq (b - a) \cdot 1 = b - a.$$$ 5. **Conclusion:** $$$\cos b - \cos a \leq b - a,$$$ which is what we wanted to prove using the Mean Value Theorem.