Theorem (0.4 in notes)
\(\sin(x) \leq x\) for all \(x \geq 0\).
Proof
Take \(f(x) = \sin(x) - x\). Then
$$
\begin{align*}
f'(x) = \cos(x) - 1 \leq 0
\end{align*}
$$
Hence, \(f\) is decreasing for all \(x \geq 0\) (By Theorem 0.4). But now observe that \(f(0) = 0\). Hence
$$
\begin{align*}
f(x) \leq f(0) = 0
\end{align*}
$$
Therefore
$$
\begin{align*}
\sin(x) - x &\leq 0 \\
\sin(x) &\leq x. \ \blacksquare
\end{align*}
$$
References
- Introduction to Analysis, An, 4th edition by William Wade
- Lecture Notes by Professor Chun Kit Lai