Shape of the Sine Curve vs. CircleDate: 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/ |
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