# Is $r=2\cos(\theta)$ a one-petal polar function?

I'm currently learning about polar functions and their graphs in precalculus, and one of the questions on my homework is to identify the shape of the function $r=2\cos(\theta)$. We were taught that functions in the form $r=a\cos(n\theta)$ where $a>0$ are roses. So, were $a=2$ and $n=1$, wouldn't it form this function and be rose?

I ask because the graph is a perfect circle centered at polar coordinates (1,0).

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Well, it's a flower with one perfectly circular petal. I hear that one-petal flowers can be found in nature, though I don't know how circular they get. – user31373 Jun 7 '12 at 23:34

I'd say that it's convenient to include it in the set of rose curves—it can be useful to think about $r=a\sin n\theta$ for noninteger values of $n$ (even starting at $0$), rather than just integers greater than $1$:

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Note that in polar coords

$x = r \cos \theta$

$y= r \sin \theta$

You have that $r = 2 \cos \theta$

That is

Then

$$r^2 = 2r \cos \theta$$

$$x^2+y^2 = 2x$$

$$x^2-2x+1+y^2=1$$

$$(x-1)^2+y^2=1$$

That is a circle centered at $(1,0)$ and radius $1$.

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Would it be incorrect though, to call it a rose/flower? – mowwwalker Jun 7 '12 at 23:54
Usually, a flower is of the form $\rho = a \cos( k \theta +\phi)$. You might call it a "degenerate" flower maybe. – Pedro Tamaroff Jun 7 '12 at 23:56
It is a little easier, instead of squaring both sides, to multiply both sides by $r$, and get $r^2 =2r\cos\theta$, so $x^2+y^2=2x$, etc. This avoids the problem of taking positive and negative square roots, which you seem to have ignored. – Stefan Smith Jun 8 '12 at 1:16
@user20520 For the motives you state I did disregard the $\pm$. I guess anyone would understand why it doesn't make a difference. – Pedro Tamaroff Jun 8 '12 at 4:44
It does make a difference, because $\sin\theta$ might be negative, in which you get $x^2+y^2=-2x$, which is a different circle. – Stefan Smith Jun 8 '12 at 12:18