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I'm curious how to find out the period of an hypotrochoid.

x = (a-b) * cos(t) + h * cos( ((a-b)/b) * t )
y = (a-b) * sin(t) - h * sin( ((a-b)/b) * t )

I know that for a single cos(x) function to find the period you divide 2π by the multiplier of the x, but i have no idea with more tri function like cos(x) + cos(x1) + m1

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Could you please put a definition of a hypotrochoid into your question? I am sure that will get you answers. – Phira Dec 18 '12 at 0:06
ok added the function, thanks! – nkint Dec 18 '12 at 0:10
up vote 2 down vote accepted

The period of $\cos t$ and $\sin t$ is $2\pi$.

For any positive constant $k$, the period of $\cos kt$ and $\sin kt$ is $2\pi/k$.

(Because $\cos (k(t+p)) = \cos (kt +kp)$ so $kp$ has to be the period $2\pi$ of $\cos t$.)

So, now each of the two coordinates is a sum of a function with period $2\pi$ and a function with period $2\pi k = 2\pi \frac nd$ if we set $\frac{a-b}{b}=\frac dn$ as your book does (unless $h=0$ or $a=b$ and the function is actually simpler).

If the two function do not have the same period (i.e. $d\not=n$) we now have to find the first time when both periods have been completed an integral number of times, so we need simultaneously a multiple of $2\pi$ and a multiple of $2\pi\frac nd$.

But this happens first at $2\pi n$.

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find it out in an old math book (but i don't understand exactly the reason).

hypotrochoid function:

x = (a-b) * cos(t) + h * cos( ((a-b)/b) * t )
y = (a-b) * sin(t) - h * sin( ((a-b)/b) * t )

let say b/(a-b) = n/d, and n,d positive integer without common divisor, so period is n2π.

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What is $a$ and what is $b$? – Phira Dec 18 '12 at 0:07
added hypotrochoid function! – nkint Dec 18 '12 at 0:10

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