Showing that $\sin xLet $f(t) = \sin t$. Fix $x$ such that $0 < x < $$π\over{2}$. If you were to apply the Mean Value Theorem to $f$ for $t$ in the interval $[0, x]$:
(a) Write down precisely what the conclusion of the theorem tells you. 
(b) Explain why (a) allows you to immediately conclude that $\sin x < x$ for $x\in (0, {π\over{2}}$).
(c) Why is it the case that if $x \geq {π\over{2}}$ then $\sin x < x?$
Attempted Solutions:
(a) Let $f\in{C[0,x]}$. (Note that C is the set of continuous functions in the given interval). Assume that $f$ is differentiable on $(0,x)$. Then there exists $c\in{(0,x)}$ such that $f'(c)={f(b)-f(a)\over{b-a}}$ = $\sin x\over{x}$
(b) and (c) I am not sure of. 
 A: The theorem states that there exists $c\in (0,x)$ such that
$$\sin x-\sin 0=\sin' c \cdot (x-0)=\cos c \cdot x.$$
Since $|\cos c|<1$ for $c\in (0,\pi/2)$ we have that 
$$|\sin x|= |\sin x-\sin 0|=|\cos c||x|<x.$$
Finally, if $x\ge \pi/2$ we have that
$$|\sin x|\le 1<\pi/2 \le x.$$
A: (a) Since $f$ is continuous on $[0,x]$ and differentiable on $(0, x)$, so by the Mean Value Theorem there is a point $c \in (0, x)$ such that 
$$f(x)- f(0) = (x-0) f^\prime(c) .$$
That is, 
$$\sin x - \sin 0 = (x-0)  \cos c$$
and so
$$\sin x = x \cos c < x$$
since $0 < \cos c < 1$ for every $c$ such that $0 < c < \frac{\pi}{2}$. 
Thus we have shown that $\sin x < x$ for every $x \in \left( 0, \frac{\pi}{2} \right)$. 
A: HINT: What is the derivative $f$? Can you see that it's always less than $1$?
A: Let $g(x) = x - \sin x$. Then for given $x \in (0, \pi / 2)$, there exists $y \in (0, x)$ such that 
\begin{align*}
g'(y) & = \frac{g(x) - g(0)}{x - 0} \\
& = \frac{ x - \sin x }{x} \\
\Rightarrow x g'(y) & = x - \sin x .
\end{align*}
But $g'(y) = 1 - \cos y > 0$, since $0 < y < x < \pi / 2$, and $x > 0$, so $x - \sin x > 0 \Rightarrow x > \sin x$.
