How do these equations get derived?

Question

Please look at this webpage for reference: http://mathworld.wolfram.com/Hypocycloid.html

Go to line 7 to 15 on that webpage. Line 7 and 8 show 2 parametric equations. These have been rewritten as in line 12 and 13. I understand how this has been done. But I want to know how these parametric equations (line 12 and 13) have been used to get the arc length and the area equations (line 14 and 15 respectively).

I have tried using the arc length formula:

I rearranged to get the radius in terms of S and theta. Then I substitute that into the area equation:

But that gives me the area in terms of theta. However, on the webpage, there is no angle or theta.

How should I do this? If there is something I am mission or these is another, better way to do this, please tell me.

Sorry for my incompetence and thanks in advance.

EDIT

By "LINE" I mean the numbers that are on the right side of the page in brackets

Answer 1

Note that hypocycloid is simply connected when $\displaystyle \frac{a}{b}=n=3,4,5, \ldots$

Now \begin{align*} x &= \frac{a}{n}[(n-1)\cos t+\cos (n-1)t] \\ y &= \frac{a}{n}[(n-1)\sin t-\sin (n-1)t] \\ x' &= \frac{a(n-1)}{n} [-\sin t-\sin (n-1)t] \\ y' &= \frac{a(n-1)}{n} [\cos t-\cos (n-1)t] \\ x'^2+y'^2 &= \frac{a^2(n-1)^2}{n^2} (2-2\cos nt)] \\ &= \frac{4a^2(n-1)^2}{n^2} \sin^2 \frac{nt}{2} \\ ds &= \frac{2a(n-1)}{n} \left| \sin \frac{nt}{2} \right| dt \\ P &= n\int_{0}^{2\pi/n} \frac{2a(n-1)}{n} \sin \frac{nt}{2} \, dt \\ &= 2a(n-1) \left[ -\frac{2}{n} \cos \frac{nt}{2} \right]_{0}^{2\pi/n} \\ &= \frac{8a(n-1)}{n} \\ A &= \oint_{C} x\, dy \\ &= \frac{a^2(n-1)}{n^2} \times \pi [(n-1)-1] \quad \quad \text{(see the note below)} \\ &= \frac{\pi a^2(n-1)(n-2)}{n^2} \end{align*}

Note that $$x\, dy = \frac{a^2(n-1)}{n^2} [(n-1)\cos^2 t+(2-n)\cos t \cos (n-1)t-\cos^2 (n-1)t] \, dt$$

By $$\int_{0}^{2\pi} \cos nt \, \cos mt \, dt = \pi \delta_{n,m}$$ only the square terms in $x\, dy$ survive after integration.

For the area of hypocycloid with other values of $n$, see another post here.