Proof that sin(x) ≤ x for All Positive Real Numbers

# Proof that sin(x) ≤ x for All Positive Real Numbers

A very useful inequality that sometimes appears in calculus and analysis is that for any nonnegative real number $x$ we have that $\sin (x) \leq x$.

We will now prove this result using an elementary result from calculus - the Mean Value theorem. We state this result below and then prove this inequality.

 Theorem 1 (The Mean Value Theorem): If $f$ is a continuous function on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ then there exists a number $c \in (a, b)$ such that $\displaystyle{f'(c) = \frac{f(b) - f(a)}{b-a}}$.

We can now prove the inequality.

 Theorem 2: If $x$ is a nonnegative real number then $\sin (x) \leq x$.
• Proof: We break this proof into three cases.
• Case 1: Suppose that $x = 0$. Then clearly $0 =\sin (0) \leq x = 0$.
• Case 2: Suppose that $0 < x < 1$. Let $f(t) = \sin (t)$. Then $f$ is continuous and differentiable everywhere. In particular, $f$ is continuous on $[0, x]$ and differentiable on $(0, x)$. By the Mean Value theorem there exists number $c \in (0, 1)$ such that:
(1)
\begin{align} \quad f'(c) = \frac{f(x) - f(0)}{x - 0} \end{align}
• The derivative of $\sin x$ is $\cos x$. Therefore:
(2)
\begin{align} \quad \cos (c) &= \frac{\sin (x) - \sin(0)}{x} \\ \cos (c) &= \frac{\sin x}{x} \end{align}
• Note that the cosine function is bounded, that is,$-1 \leq \cos t \leq 1$ for every real number $t$. Therefore:
(3)
• Since $0 < x < 1$, we multiply both sides of the inequality above to get that $\sin x \leq x$.
• Case 3: Suppose that $1 \leq x < \infty$. We know that $\sin t$ is a bounded function and $-1 \leq \sin t \leq 1$ for every real number $t$. Thus $\sin x \leq 1 \leq x$, i.e., $\sin x \leq x$. $\blacksquare$