Question 2.10.5: Proving an Inequality for sin x Prove that |sin a| ≤ |a| for...

First, note that f (x) = sin x is continuous and differentiable on any interval and that for any a,

since sin 0 = 0. From the Mean Value Theorem, we have that (for a≠0)

\frac{\sin a-\sin 0}{a-0}=f^{\prime}(c)=\cos c, (10.6)

for some number c between a and 0. Notice that if we multiply both sides of (10.6) by a and take absolute values, we get

|\sin a|=|\sin a-\sin 0|=|\cos c||a-0|=|\cos c||a| . (10.7)

But, |cos c| ≤ 1, for all real numbers c and so, from (10.6), we have

as desired.