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Shape of the Sine Curve vs. Circle


Date: 07/18/2001 at 01:11:24
From: Aaron
Subject: Shape of the Sine Curve vs. Circle

I have a question regarding the shape of the sine curve. 

A sine curve appears almost linear between its high points. In other 
words, if the graph y = 5sinx is graphed, the curve looks almost 
linear except near the points where y = 5. I wonder why the sine curve 
looks this way instead of completely curvy like a circle, since the 
sine curve originates from the circle. 

To compare the two, I graphed y = sqrt(1-(1-x)^2), which is the top 
half of a circle of radius 1, translated to the right by 1, and I 
graphed the equation y = sin((pi/2)x). This way the circle and the 
first "hill" of the sine curve intersect the x-axis at the same places 
and reach the same high point. Clearly, the circle is much more 
curve-like, whereas the sine curve approaches its peak in an almost 
linear fashion. I was wondering why this is. Could you please 
explain why?


Date: 07/18/2001 at 10:42:05
From: Doctor Rick
Subject: Re: Shape of the Sine Curve vs. Circle

Hi, Aaron.

I think you'll see things differently if you make a different 
comparison. You are choosing your sine curve so that its zeros match 
those of the semicircle. Instead, try graphing these:

  y = sin(x)

  y = sqrt(1-(x-pi/2))^2)

Tell me what you see. I'd glad to discuss the reasons for these 
observations, but I'll let you make the observations first.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 07/18/2001 at 17:12:00
From: Aaron
Subject: Re: Shape of the Sine Curve vs. Circle

Well, this is also very interesting. It now appears that with the 
equations

  y = sin(x)
  y = sqrt(1-(x-pi/2))^2)

the circle seems to fit inside the top of the sine curve. Then, the 
sine curve seems to linearize until it reaches y = -1, where 
apparently another circle could fit. But why is this so? Why is the 
sine curve like a circle but then linear?


Date: 07/19/2001 at 09:02:07
From: Doctor Rick
Subject: Re: Shape of the Sine Curve vs. Circle

Hi again, Aaron.

You observed exactly what I hoped you'd observe. Good work! Now you 
see that the sine curve looked "pointy" at first only because you were 
looking at it with the x axis scrunched up. When it's spread out in 
the natural way (by using radians as the angle measure), it looks much 
"rounder": a circle fits neatly into the sine curve near the peak.

You have observed:

(1) The sine curve is approximated by a circle near x = pi/2 and 
    x = -pi/2.

(2) The sine curve is approximated by a straight line near x = 0 and 
    x = pi (in fact, the slopes of these lines are 1 and -1 
    respectively).

Why??

Since you know that the sine curve is derived from a circle, I assume 
you are familiar with the unit circle:

               |
            ---+--- B
          /    |  /|\
         /    1| / | \
        |      |/  |  |
--------+------+---+--+------------
       D|     O|   C  |A
         \     |     /
          \    |    /
            ---+---
               |

The radius of the circle, OA, is 1. The sine of the angle AOB (which 
I'll call theta) is BC (the y coordinate of point B). The cosine of 
theta is OC (the x coordinate of point B).

The radian measure of an angle is defined as the arc length subtended 
by the angle in a unit circle. Thus, the radian measure of theta is 
the arc length AB. For small angles, C is close to A, and the y 
coordinate BC is approximately equal to the arc AB. Thus the sine of 
theta is approximately equal to theta itself:

  sin(theta) ~= theta          (theta near 0)

For angles near 180 degrees (pi radians), the sine BC is approximately 
equal to the arc BD, which is the radian measure of angle DOB. This is 
the supplement of theta. Thus

  sin(theta) ~= pi - theta 

What happens at angles near pi/2? Now it is the x coordinate (the 
cosine) that is approximately equal to an arc, namely the complement 
of theta. The coordinates (x,y) are on the circle, so

  y = sqrt(1 - x^2)

Since x is approximately equal to pi/2 - theta, we have

  sin(theta) ~= sqrt(1 - (pi/2-theta)^2)

A similar result holds for theta near -pi/2.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Equations, Graphs, Translations

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