# comparison of $\mathop{sin}\nolimits\theta$ and $\theta$ near $\theta=0$

###### Theorem 1.

Let $\displaystyle 0<\theta<\frac{\pi}{2}$, where $\theta$ is an angle measured in radians. Then $\sin\theta<\theta$.

###### Proof.

Let $O=(0,0)$, $P=(1,0)$, and $Q=(\cos\theta,\sin\theta)$. 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 $\theta$. 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.

Draw the line segment $\overline{PQ}$. Since this does not correspond to the arc, its length $\left|\overline{PQ}\right|<\theta$.

Since $\displaystyle 0<\theta<\frac{\pi}{2}$, $0<\sin\theta<1$ and $0<\cos\theta<1$. Thus, $R\neq Q$ and $R$ lies strictly in between $O$ and $P$. Therefore, $\left|\overline{PR}\right|>0$.

By the Pythagorean theorem   , $\left|\overline{QR}\right|^{2}+\left|\overline{PR}\right|^{2}=\left|\overline{% PQ}\right|^{2}$. Thus, $\left|\overline{QR}\right|^{2}<\left|\overline{PQ}\right|^{2}$. Therefore, $\sin\theta=\left|\overline{QR}\right|<\left|\overline{PQ}\right|<\theta$. ∎

The analogous result for $\theta$ slightly below 0 is:

###### Corollary 1.

Let $\displaystyle\frac{-\pi}{2}<\theta<0$, where $\theta$ is an angle measured in radians. Then $\theta<\sin\theta$.

###### Proof.

Since $\displaystyle 0<-\theta<\frac{\pi}{2}$, the previous theorem yields $\sin(-\theta)<-\theta$. Since $\sin$ is an odd function, $-\sin\theta<-\theta$. It follows that $\theta<\sin\theta$. ∎

Title comparison of $\mathop{sin}\nolimits\theta$ and $\theta$ near $\theta=0$ ComparisonOfsinthetaAndthetaNeartheta0 2013-03-22 16:58:29 2013-03-22 16:58:29 Wkbj79 (1863) Wkbj79 (1863) 7 Wkbj79 (1863) Theorem msc 26A03 msc 51N20 msc 26A06 LimitOfDisplaystyleFracsinXxAsXApproaches0 JordansInequality