# Reparametrization of angles

Why is it true that $(\cos (a \theta +b), \sin (a \theta +b), (c \theta +d))$ for $\theta \in [\theta_1,\theta_2]$ can always be written as $(\cos \alpha \theta' , \sin \alpha \theta', \beta \theta')$ for a suitable choice of $\theta'\in [\theta'_1,\theta'_2]$?

Update: The parametrization describes a spiral of constant speed.

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It's not true. Take $a = 1$, $b = 0$, $c = 0$, $d = 1$, so that the first expression is $(\cos \theta, \sin \theta, 1)$. To have $\beta \theta' = 1$ you have to have $\theta'$ constant, but then you can't get $\cos \alpha\theta' = \cos \theta$ for all $\theta \in [\theta_1, \theta_2]$. Is there something you've omitted in the question? –  Rahul Mar 11 '12 at 11:52
@RahulNarain: Thank you for your reply! I have added some info. Would it be true now? –  Mister Mar 11 '12 at 11:56
No, the mathematical content of the question is exactly the same. The latter parametrization describes a spiral that passes through $(1,0,0)$, but the former does not, so they cannot be equivalent. –  Rahul Mar 11 '12 at 11:58
@RahulNarain: Thanks, but I don't think the second parametrization necessarily passes through $(1,0,0)$ since $\theta$ does not necessarily start from $0$... –  Mister Mar 11 '12 at 12:09
It's a spiral that if extended sufficiently (i.e. enough to meet the plane $z=0$) would necessarily pass through $(1,0,0)$. Anyway, Neal has given you the formal answer. –  Rahul Mar 11 '12 at 12:15
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I don't think it's true. For the arguments of cos and sin, you must have that $\theta' = \frac{1}{\alpha}(a\theta + b)$, but then this means that $\theta = \frac{1}{a}(\alpha\theta' - b)$. Then the third coordinate must be $\frac{c}{a}(\alpha\theta' - b) + d = \frac{c\alpha}{a}\theta' + (d - \frac{bc}{d})$. Now you pick $\beta = \frac{c\alpha}{a}$, but it's not so clear why $d - \frac{bc}{a}$ must vanish; you don't get to choose $a,b,c,$ or $d$ so you don't have any control over whether or not they vanish.