comparison of sinθ and θ near θ=0

Theorem 1.

Let 0<θ<π2, where θ is an angle measured in radians. Then sinθ<θ.


Let O=(0,0), P=(1,0), and Q=(cosθ,sinθ). Note that the circle x2+y2=1 passes through P and Q and that the shortest arc along this circle from P to Q has length θ. Note also that the line segmentsMathworldPlanetmath OP¯ and OQ¯ are radii of the circle x2+y2=1 and therefore must each have length 1.


Draw the line segment PQ¯. Since this does not correspond to the arc, its length |PQ¯|<θ.


Drop the perpendicularMathworldPlanetmathPlanetmathPlanetmath from Q to OP¯. Let R be the point of intersectionMathworldPlanetmath. Note that |OR¯|=cosθ and |QR¯|=sinθ.


Since 0<θ<π2, 0<sinθ<1 and 0<cosθ<1. Thus, RQ and R lies strictly in between O and P. Therefore, |PR¯|>0.

By the Pythagorean theoremMathworldPlanetmathPlanetmath, |QR¯|2+|PR¯|2=|PQ¯|2. Thus, |QR¯|2<|PQ¯|2. Therefore, sinθ=|QR¯|<|PQ¯|<θ. ∎

The analogous result for θ slightly below 0 is:

Corollary 1.

Let -π2<θ<0, where θ is an angle measured in radians. Then θ<sinθ.


Since 0<-θ<π2, the previous theorem yields sin(-θ)<-θ. Since sin is an odd function, -sinθ<-θ. It follows that θ<sinθ. ∎

Title comparison of sinθ and θ near θ=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