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

Consider the following figure:

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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

2 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 $P$.

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