SOLUTION 12: Consider the function $ f(x) = \sin x$ on the interval $ [0, 2\pi] $, and assume that $ 0 \le z < w \le 2\pi $. This function is continuous (well-known) on the closed interval $ [z, w]$. The derivative of $f$ is $$ f'(x) = \cos x $$ We can now see that $f$ is differentiable on the open interval $ (z, w) $. The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem on the closed interval $ [z, w] $. Then there is some number $c$, $ z < c < w$, so that $$ f'(c) = \displaystyle{ f(w)-f(z) \over w-z } \ \ \ \ \longrightarrow $$ $$ \cos c = \displaystyle{ \sin w - \sin z \over w-z } \ \ \ \ \longrightarrow $$ $$ \Bigg| \displaystyle{ \sin w - \sin z \over w-z } \Bigg| = | \cos c | \le 1 \ \ \ \ \longrightarrow $$ $\Big($ since $ -1 \le \cos c \le 1 $ $\Big)$ $$ \displaystyle{ | \sin w - \sin z | \over | w-z | } \le 1 \ \ \ \ \longrightarrow $$ $$ | \sin w - \sin z | \le | w-z | $$

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