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comparison of $\sin þeta$ and $þeta$ near $þeta = 0$ (Theorem)
Theorem   Let % latex2html id marker 344 $ \displaystyle 0< \theta < \frac{\pi}{2}$, where % latex2html id marker 346 $ \theta$ is an angle measured in radians. Then % latex2html id marker 348 $ \sin \theta<\theta$.
Proof. Let $ O=(0,0)$, $ P=(1,0)$, and % latex2html id marker 357 $ Q=(\cos \theta, \sin \theta)$. Note that the circle $ x^2+y^2=1$ passes through $ P$ and $ Q$ and that the shortest arc along this circle from $ P$ to $ Q$ has length % latex2html id marker 369 $ \theta$. Note also that the line segments $ \overline{OP}$ and $ \overline{OQ}$ are radii of the circle $ x^2+y^2=1$ and therefore must each have length 1.

\begin{pspicture} % latex2html id marker 68 (-1,1)(-2,2) \psdots(0,0)(2,0)(1.2,1... ....3}{0}{55.5} \rput[l](0.3,0.2){$\theta$} \psarc(0,0){2}{0}{55.5} \end{pspicture}

Draw the line segment $ \overline{PQ}$. Since this does not correspond to the arc, its length % latex2html id marker 379 $ \left\vert \overline{PQ} \right\vert<\theta$.


\begin{pspicture} % latex2html id marker 84 (-1,1)(-2,2) \psdots(0,0)(2,0)(1.2,1... ....3}{0}{55.5} \rput[l](0.3,0.2){$\theta$} \psarc(0,0){2}{0}{55.5} \end{pspicture}

Drop the perpendicular from $ Q$ to $ \overline{OP}$. Let $ R$ be the point of intersection. Note that % latex2html id marker 387 $ \left\vert \overline{OR} \right\vert=\cos \theta$ and % latex2html id marker 389 $ \left\vert \overline{QR} \right\vert=\sin \theta$.


\begin{pspicture} % latex2html id marker 107 (-1,1)(-2,2) \psdots(0,0)(2,0)(1.2,... ...0,0){2}{0}{55.5} \psline(1.2,0)(1.2,1.6) \rput[b](1.2,-0.5){$R$} \end{pspicture}

Since % latex2html id marker 391 $ \displaystyle 0< \theta <\frac{\pi}{2}$, % latex2html id marker 393 $ 0< \sin \theta <1$ and % latex2html id marker 395 $ 0< \cos \theta <1$. Thus, $ R \neq Q$ and $ R$ lies strictly in between $ O$ and $ P$. Therefore, $ \left\vert \overline{PR} \right\vert>0$.

By the Pythagorean theorem, $ \left\vert \overline{QR} \right\vert^2+\left\vert \overline{PR} \right\vert^2=\left\vert \overline{PQ} \right\vert^2$. Thus, $ \left\vert \overline{QR} \right\vert^2< \left\vert \overline{PQ} \right\vert^2$. Therefore, % latex2html id marker 411 $ \sin \theta = \left\vert \overline{QR} \right\vert< \left\vert \overline{PQ} \right\vert < \theta$. $ \qedsymbol$

The analogous result for % latex2html id marker 413 $ \theta$ slightly below 0 is:

Corollary   Let % latex2html id marker 419 $ \displaystyle \frac{-\pi}{2}< \theta <0$, where % latex2html id marker 421 $ \theta$ is an angle measured in radians. Then % latex2html id marker 423 $ \theta < \sin \theta$.
Proof. Since % latex2html id marker 428 $ \displaystyle 0< -\theta < \frac{\pi}{2}$, the previous theorem yields % latex2html id marker 430 $ \sin (-\theta)<-\theta$. Since $ \sin$ is an odd function, % latex2html id marker 434 $ -\sin \theta < -\theta$. It follows that % latex2html id marker 436 $ \theta < \sin \theta$. $ \qedsymbol$



"comparison of $\sin þeta$ and $þeta$ near $þeta = 0$" is owned by Wkbj79.
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See Also: limit of $\displaystyle \frac{\sin x}{x}$ as $x$ approaches 0, Jordan's inequality

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Cross-references: odd function, Pythagorean theorem, strictly, intersection, point, drop the perpendicular, radii, line segments, length, arc, passes through, circle, radians, angle
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This is version 4 of comparison of $\sin þeta$ and $þeta$ near $þeta = 0$, born on 2007-04-24, modified 2007-04-24.
Object id is 9251, canonical name is ComparisonOfSinThetaAndThetaNearTheta0.
Accessed 590 times total.

Classification:
AMS MSC51N20 (Geometry :: Analytic and descriptive geometry :: Euclidean analytic geometry)
 26A03 (Real functions :: Functions of one variable :: Foundations: limits and generalizations, elementary topology of the line)
 26A06 (Real functions :: Functions of one variable :: One-variable calculus)

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