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I am not sure if the question is a duplicate. The conclusion from the title is clear, for example by viewing the wolframalpha graph at here.

But on the other hand I feel intuitively there should be only 4, since a line intersect with $y=\cos[\theta]$ at 2 points, thus $y=1/2$ intersect with $y=\cos[2\theta]$ at 4 points in $[0,2\pi]$. Since $\cos[2\theta]=1/2$ implies a solution, the number of solutions of $$y=\cos[2x]-1/2,x\in [0,2\pi]$$ and $$r=\cos[2\theta],\theta\in \mathbb{S}^{1}$$ must be the same. So why there is such a difference?

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You also get intersections when $\cos 2\theta=-1/2$. Polar coordinates are weird like that. :) –  Micah Aug 26 '12 at 3:25
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More formally: The curves are $(x,y)=(\cos2\theta\cos\theta,\cos2\theta\sin\theta)$ and $x^2+y^2=1/4$, which intersect when $\cos^22\theta=1/4$, or $\cos2\theta=\pm1/2$. –  Rahul Aug 26 '12 at 3:33
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up vote 3 down vote accepted

The problem is that there are two widely used conventions for interpreting $r=f(\theta)$ when $f(\theta)\lt 0$.

One convention is to say that if $f(\theta)\lt 0$, then the curve is undefined. Makes sense, $r$ is a distance.

Another convention is that if $f(\theta)\lt 0$, we plot $(|f(\theta)|,\theta)$ as usual, and then reflect the result across the origin, or equivalently rotate through $180^\circ$.

That second convention will give you $8$ points.

Remark: I slightly prefer the second convention, it gives prettier curves.

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This is surprising to me. I never expect this kind of twisting logic. –  Bombyx mori Aug 26 '12 at 7:49
    
@user32240: The logic is not actually twisted! Imagine a ray from the origin at an angle $\theta$. If $r$ is positive, the point is at a distance $r$ along the ray. If $r$ is negative, it's just at a distance $|r|$ in the opposite direction. You could imagine (as I prefer) extending the ray backwards to mark a point at a negative distance, like a rotated version of the real number line. Or you could take the point at $|r|$ along $\theta$ and then turn it $180^\circ$, which ends up being the same thing. –  Rahul Aug 27 '12 at 8:53
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