1
$\begingroup$

The goal is to transform the following coordinates: $$x(t)= R(\Phi-\sin\Phi)$$ and $$z(t)=R(2 +\cos\Phi)$$

with the substitution: $u=\cos\left(\Phi/2\right)$

in order to get: $$x(t)=2R(\arccos(u)-u\sqrt{1-u^2})$$ and $$z(t)=R(1+2u^2)$$

How do I go about solving this problem? I already tried using the following addition theorems: $\cos(\Phi/2)=\sqrt{(1+\cos\Phi)/2)}$ and $\sin\Phi=\sqrt{1-\cos^2\Phi}$ and identities for $\arccos$, but somehow I failed to transform the coordinates.

$\endgroup$
4
  • 2
    $\begingroup$ Is R(x) a function or a multiplication by constant ? $\endgroup$
    – pasaba por aqui
    Nov 24, 2020 at 10:27
  • 1
    $\begingroup$ What is your source ??? Has anything to do with Brachistochrone - Tautochrone - Cycloid ??? \begin{equation} x\left(t\right) = R\Biggl[ \sqrt{\dfrac{\,g\,}{R}}\,t-\sin \left(\sqrt{\dfrac{\,g\,}{R}}\,t\right)\Biggr]=R\Bigl[\omega\,t-\sin \left(\omega\,t\right)\Bigr] \tag{01a}\label{01a} \end{equation} \begin{equation} y \left(t\right) = R\Biggl[1-\cos \left(\sqrt{\dfrac{\,g\,}{R}}\,t\right)\Biggr]=R\Bigl[1-\cos \left(\omega\,t\right)\Bigr] \tag{01b}\label{01b} \end{equation} $\endgroup$
    – Frobenius
    Nov 24, 2020 at 13:42
  • 1
    $\begingroup$ Note that \begin{equation} z(t)=R(2 +\cos\Phi)=3R-R(1 -\cos\Phi) \tag{02}\label{02} \end{equation} $\endgroup$
    – Frobenius
    Nov 24, 2020 at 13:54
  • $\begingroup$ Related : What is the position as a function of time for a mass falling down a cycloid curve?. $\endgroup$
    – Frobenius
    Nov 24, 2020 at 14:04

2 Answers 2

Reset to default
6
$\begingroup$

This is quite straightforward, I'm not sure where you're having a problem. I won't give the full answer since this is a classic "homework-style" problem, I'll just show you how to do one of them, and the other can be done in almost exactly the same way. Take the first equation, you want to go from:

$$x(t) = R(\Phi - \sin\Phi) \quad \quad \text{to} \quad \quad x(t) = R ( 2 \cos^{-1}u - 2u\sqrt{1- u^2}).$$

The most sensible substitution to start with is $$\Phi = 2 \cos^{-1}u \quad \quad \text{or}\quad \quad u = \cos\left({\frac{\Phi}{2}}\right).$$

Draw out a little triangle (or use trigonometric identities), and you should be able to show that this means that $$\sin{\left(\frac{\Phi}{2}\right)} = \sqrt{1-u^2}.$$

Then, using the simple result $$\sin\Phi = 2\sin{\left(\frac{\Phi}{2}\right)}\cos{\left(\frac{\Phi}{2}\right)},$$ you should be done. You should now be able to do the next one, after you find a relation between $\cos\Phi$ and $\cos(\Phi/2)$ and $\sin(\Phi/2)$.

$\endgroup$
0
2
$\begingroup$

As I always have problems to memorize trigonometric identities, I prefer to start by a basic identity:

$$e^{i\phi}=\left(e^{i\phi/2}\right)^2$$

an use Euler's formula for real and imaginary parts:

$$\cos(\phi)=\left(\cos(\phi/2)\right)^2-\left(\sin(\phi/2)\right)^2$$ $$\sin(\phi)=2\cos(\phi/2)\sin(\phi/2)$$

From these two trigonometric identities and introducing the definition of $u$: $$\cos(\phi)=u^2-(\sqrt{1-u^2})^2=...$$ $$\sin(\phi)=...$$

you will finish in the target equations.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .