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Consider a sine wave having $4$ cycles wrapped around a circle of radius 1 unit (its center needs not be the origin of a Cartesian coordinate system). Assume that the length of axis of the sine wave is as same as the circumference of the circle.

The circumference of circle is assumed to be mapped to $2\pi\ \mathrm{rad}$. Therefore, the sine wave represents the equation along the $x$-axis:

$$ y = \sin(4x) $$

To find the equation of the sine wave with the circumference of circle acting as the $x$-axis, one approach is to consider the sine wave along a rotated line like aligned $\frac\pi4\ \mathrm{rad}$ to $x$-axis. But it doesn't suffice for the circular path. This is where the problem of finding the equation is stuck. A hint/help taking to a right answer would be appreciated.

For more clarity, here is a rough image. In the image, four lines are drawn to clearly distinguish between crests and troughs.

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    Let $P(x,y)$ be a point on your desired wave, and let $M$ be its midpoint. Write the desired function of the distance $|PM|$.2012-10-30
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    Do you mean something like the curve given by the polar equation $r=1+a \sin \theta$?2012-10-30
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    @Matthew: assuming we're interpreting the quedtion correctly, wouldn't it be $\sin 4\theta$?2012-10-30
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    @JavierBadia Oops. Yeah, $r=1+a \sin 4 \theta$. Thanks for catching that.2012-10-31
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    @all, please read the updated question again. Sorry for not-so-clear-question earlier.2012-10-31
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    @Neeraj: From what I can tell, people have understood your question correctly, but it is you who have not understood their comments and answers. Instead of asking them to re-read your question, I think you would do better to mention what it is you find lacking in their responses.2012-10-31
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    @RahulNarain, a polar equation is given above as $r=1+a\sinθ$. Taking a=1, its polar graph is shown in [graph1](http://imgur.com/rRWMN) for $ \theta = [0, 2\pi] $. But, this graph (being polar) doesn't match with the graph (being cartesian) shown in [graph2](http://imgur.com/l0cY9) (of course, circle is merely shown as the basis for drawing the sine graph). It makes sense to compare cartesian to cartesian graphs, not polar to cartesian.2012-10-31
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    @NeerajTuteja: Try a smaller $a$, such as $a=0.25$: http://www.wolframalpha.com/input/?i=polar+plot+r+%3D+1+%2B+0.25+sin%284t%292012-10-31
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    I don't see the distinction you're making; polar and Cartesian coordinates are simply different ways of expressing the same graph. For example, you can substitute $r = \sqrt{x^2+y^2}$ and $\sin 4\theta=4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta=(4x^3y-4xy^3)/r^4$ into the polar equation to get $4axy(x^2-y^2)=(x^2+y^2)^2(\sqrt{x^2+y^2}-1)$, which is [the same graph](http://www.wolframalpha.com/input/?i=plot%204%20%281/4%29%20x%20y%20%28x%5E2%20-%20y%5E2%29%20=%20%28x%5E2%20%2b%20y%5E2%29%5E2%20%28sqrt%28x%5E2%20%2b%20y%5E2%29%20-%201%29) in Cartesian form.2012-11-01

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Do it first for the circle centered at the origin in polar coordinates.

Then switch do Cartesian coordinates, then shift to the actual center of the circle.

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    Apology for counter-commenting. Irrespective of the location of center, your solution is not pointing to a right answer, does it?2012-10-30
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    @NeerajTuteja Your comment is not telling me where your are stuck, is it? Maybe you are not familiar with polar coordinates, maybe you do not really know what you mean by "wrapping around the circle", maybe you have trouble switching to Cartesian coordinates, how should I know when you just tell me my answer must be wrong?2012-10-30
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it should be, in cartesian coordinates

x = (R + a · sin(n·θ)) · cos(θ) + xc

y = (R + a · sin(n·θ)) · sin(θ) + yc

where

R is circle's radius

a is sinusoid amplitude

θ is the parameter (angle), from 0 to 2π

xc,yc is circle's center point

n is number of sinusoids on circle

you can also get a pure cartesian equation (non-parametric) on x/y, but just for half circle, solving second for sin(θ) and replacing it on first one.