Understanding the Sin Function

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Homework Help Overview

The discussion revolves around the properties of the sine function, specifically the inequality \(\sin \theta \leq \theta\) for angles in the range \(0 \leq \theta \leq \pi/2\). Participants are seeking deeper understanding and justification for this statement as presented in a complex analysis context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods to prove the inequality, including geometric interpretations, the use of derivatives, and series definitions. Questions arise about the validity of these approaches and the assumptions underlying them.

Discussion Status

The discussion is active, with multiple perspectives being explored. Some participants provide insights into geometric proofs and series expansions, while others express confusion about certain assumptions and seek clarification on specific points.

Contextual Notes

There is an emphasis on the requirement that angles be measured in radians for the discussed properties to hold. Additionally, some participants note the need for further justification regarding the behavior of series terms in relation to the sine function.

ehrenfest
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[SOLVED] sin function

Homework Statement


My complex analysis book says, "from calculus, [itex]\sin \theta \leq \theta[/itex] for [itex]0 \leq \theta \leq \pi/2[/itex]." Could someone please give me a better reason why that is true?

Homework Equations


The Attempt at a Solution

 
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well, the only way i know how to prove that [tex]|sin(x)|<|x|[/tex] is with the aid of geometry. YOu need to construct a circle with radius R=1, and denote wit OAB a triangle, and with AB the arch. Denote also by x the angle that the two sides, the radiuses of the circle enclose. And you will see the validity of this. This is true only when the angle is measured in radians. Because also remember that a portion of the arch of the circle, call it acrAB is actually the measure of the angle in radians, if we draw two lines that pass through the two points in the circle A,B.
 
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You could also consider the function f(x)=x-sin(x) and it's derivative.
 
morphism said:
You could also consider the function f(x)=x-sin(x) and it's derivative.

Very nice.
 
It should also be clear if you use the series definition for sine.
 
Drawing their graphs will also make it clear.
 
… chord-length < arc-length …

Hi ehrenfest! :smile:

The intuitive explanation is: 2sinx = chord-length (a straight line :smile: ), which is less than 2x = arc-length (not a straight line :frown: ), for 2x < π.

(This is sutupidmath's explanation, of course, with the details left out.)
 
Gib Z said:
It should also be clear if you use the series definition for sine.

Its not clear. For that to be true, you need to know that the sum of all the terms after the first is negative and has magnitude less than the first term in the series. Why is that true?
 
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ehrenfest said:
Its not clear. For that to be true, you need to know that the sum of all the terms after the first is negative and has magnitude less than the first term in the series. Why is that true?

Neither to me...
 
  • #10
ehrenfest said:
Its not clear. For that to be true, you need to know that the sum of all the terms after the first is negative and has magnitude less than the first term in the series. Why is that true?

My bad. The things you stated there are sometimes used when deriving the series in the first place and showing the Taylor Remainder term goes to zero, but not always. I wrongly assumed you learned it the same way I did.
 

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