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单词 ComparisonOfsinthetaAndthetaNeartheta0
释义

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.

Proof.

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.

OPQθ

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

OPQθ

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

OPQθR

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.

Proof.

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

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更新时间:2025/5/4 14:10:06