How to analytically prove this result?

Question

Let x and y coordinates be

$$x = a_1 cos(\theta_1)+a_2cos(\theta_1)cos(\theta_2) - a_2sin(\theta_1)sin(\theta_2)$$

$$y = a_1 sin(\theta_1)+a_2cos(\theta_1)sin(\theta_2) + a_2cos(\theta_2)sin(\theta_1)$$

How can I analytically prove that if $a_1>a_2$ the result will be the shape below for $\theta_1$ and $\theta_2$ ranging from 0 to 360.

@Atbey Yes. I have chosen $a_1$ and $a_2$ arbitrarily given that $a_1>a_2$. — csg, Mar 08 '18 at 13:49

score 0 · Accepted Answer · 2018-03-08T19:58:01.720

0

You can write your equations as $$x = a_1 \cos(\theta_1)+a_2\cos(\theta_1+\theta_2)$$ and $$y = a_1 \sin(\theta_1)+a_2\sin(\theta_1+\theta_2)$$ This implies \begin{align}x^2+y^2&=a^2_1+a^2_2+2a_1a_2(\cos(\theta_1)\cos(\theta_1+\theta_2)+\sin(\theta_1)\sin(\theta_1+\theta_2))\\&=a^2_1+a^2_2+2a_1a_2\cos(\theta_2).\end{align} Since $-1\le\cos(\theta_2)\le1$, we have $$(a_1-a_2)^2\le x^2+y^2\le(a_1+a_2)^2.$$ And that's what we see in your picture: all points lie in the annulus between the circles with radii $a_1-a_2$ and $a_1+a_2$.

edited Mar 08 '18 at 19:58

answered Mar 08 '18 at 08:14

Thank you for your answer. Can you fix the typo in your second equation? – csg Mar 08 '18 at 19:50
@canberk_gurel Yes, of course, thanks for pointing that out. – Mar 08 '18 at 19:59

How to analytically prove this result?

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