comparison of $sin\theta$ and $\theta$ near $\theta =0$

Theorem 1.

Let , where $\theta$ is an angle measured in radians. Then .

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 .

Drop the perpendicular from $Q$ to $\overline{OP}$. Let $R$ be the point of intersection. Note that $\left|\overline{OR}\right|=\cos \theta$ and $\left|\overline{QR}\right|=\sin \theta$.

Since , and . Thus, $R\ne Q$ 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, . Therefore, . ∎

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

Corollary 1.

Let , where $\theta$ is an angle measured in radians. Then .

Proof.

Since , the previous theorem yields . Since $\sin$ is an odd function, . It follows that . ∎

Title	comparison of $sin\theta$ and $\theta$ near $\theta =0$
Canonical name	ComparisonOfsinthetaAndthetaNeartheta0
Date of creation	2013-03-22 16:58:29
Last modified on	2013-03-22 16:58:29
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	7
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 26A03
Classification	msc 51N20
Classification	msc 26A06
Related topic	LimitOfDisplaystyleFracsinXxAsXApproaches0
Related topic	JordansInequality