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Give a geometric proof that sin x < x for 0 < x < π/2

by RoRi

Consider the following figure:

Rendered by QuickLaTeX.com

Compare the area of the triangle OAP with the area of the sector OAP to prove

    \[ \sin x < x \qquad \text{for} \qquad 0 < x < \frac{\pi}{2}. \]

Further, prove,

    \[ | \sin x | < |x| \qquad \text{if} \qquad 0 < |x| < \frac{\pi}{2}. \]

Proof. Assume the radius is r = 1. For 0 < x < \frac{\pi}{2} we have the triangle OAP_t has base length a = 1 and height b = \sin x the area is

    \[ \text{Area}(OAP_t) = \frac{1}{2} ab = \frac{\sin x}{2}. \]

The area of the circular sector OAP_s is

    \[ \text{Area}(OAP_s) = \frac{x}{2 \pi} \pi r^2 = \frac{x}{2}. \]

Since the area of the triangle OAP_t is less than the area of the sector OAP_s we have,

    \begin{align*} \frac{\sin x }{2} < \frac{x}{2} && \implies && \sin x &< x. \end{align*}

Then, since \sin (-x) = - \sin x we have for 0 < x < \frac{\pi}{2}

    \[ \sin x < x \quad \implies \quad -\sin x > -x \quad \implies \quad | \sin x | < |x| \]

for 0 < |x| < \frac{\pi}{2}. \qquad \blacksquare

4 comments

    • RoRi says:

      Hi, thanks! Yes, that was definitely incorrect. I put up a fix now, and also corrected the typo in the diagram that had the origin labelled with $P$.

      • Daniel says:

        Your implication is false. x < pi/2 doesn't imply that pi/2 < x < pi, as you are implicitly stating by holding sin(x) < x for higher values of x.

        It's like if I say "0 + x < 10 for 0 < x < 9" and you answer with "but this would imply that 0 + x < 10 for 0 < x < 548392 which is not true". Review your logic.

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