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My A-Level algebra is failing me. Can someone please tell me how to rearrange this formula to give $n$ when you know $R$ and $r$.

$R \sin(180^\circ/n)/(1 - \sin(180^\circ/n)) = r$

This formula is devised here, and is a formula to find the radius 'r' of several smaller circles, that will fit around a larger circle of radius 'R'. I want to find out the number of circles that would fit if you know both the radii for a project I am working on. I know that there might be an issue here with fractions of a circle but that's OK for now.

I'm struggling with how to extract the 'n' out of the 'sin' function. Help much appreciated.

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Thanks for the formatting Steven. –  JimmyDeemo Nov 12 '12 at 20:52
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3 Answers

up vote 4 down vote accepted

First of all, get all the stuff you know on one side, and all the stuff you don't on the other: $\dfrac{\sin(180^\circ/n)}{1-\sin(180^\circ/n)} = \dfrac{r}{R}$

Now, for convenience I'm going to introduce a new variable, $t$, just to hold on to $\sin(180^\circ/n)$ - once we've got the expression written down in terms of $t$ we can 'unpack' it again. This gives us $\dfrac{t}{1-t} = \dfrac{r}{R}$. We can multiply both sides by $(1-t)$ and expand: $t = \dfrac{r}{R}(1-t) = \dfrac{r}{R}-t\dfrac{r}{R}$. Now, add $t\dfrac{r}{R}$ to both sides, and factor out the factor of $t$ on the left: $t+t\dfrac{r}{R} = \dfrac{r}{R}$, or $t(1+\dfrac{r}{R}) = \dfrac{r}{R}$. Divide out by the factor on the left, and then finally multiply numberator and denominator on the right by R to clear the fractions: $t=\dfrac{\frac{r}{R}}{1+\frac{r}{R}} = \dfrac{r}{R+r}$. Now that we have an expression in terms of $t$, we can reintroduce our $n$: $\sin(180^\circ/n) = \dfrac{r}{R+r}$. Now just apply arcsin to both sides and divide $180^\circ$ into both sides: $\dfrac{180^\circ}{n} = \arcsin\left(\dfrac{r}{R+r}\right)$, or finally:

$$n=\dfrac{180^\circ}{\arcsin\left(\dfrac{r}{R+r}\right)}$$

Also, one thing I strongly encourage doing whenever you're solving a problem like this: once you have what you think is the final formula, test it! In this case, we know that six equally-sized circles will fit around a circle, so if we plug in $R=r$ we should find $n=6$: $$\begin{align} n &= \dfrac{180^\circ}{\arcsin\left(\dfrac{r}{R+r}\right)} \\ &= \dfrac{180^\circ}{\arcsin\left(\dfrac{r}{r+r}\right)} \\ &= \dfrac{180^\circ}{\arcsin\left(\dfrac{r}{2r}\right)} \\ &= \dfrac{180^\circ}{\arcsin(\frac{1}{2})} \\ &= \dfrac{180^\circ}{30^\circ} \\ &= 6 \end{align} $$

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Thank you to all replies but this one broke it down nicely. Cheers! –  JimmyDeemo Nov 12 '12 at 21:07
    
Yes. It is the best answer. –  Simon Hayward Nov 12 '12 at 21:09
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$$R \sin(180^\circ/n)/(1 - \sin(180^\circ/n)) = r$$

$$R/r = (1 - \sin(180^\circ/n)) / \sin(180^\circ/n)$$

$$R/r = 1 / \sin(180^\circ/n) -1$$

$$1 / \sin(180^\circ/n) = (r + R)/r $$

$$\sin(180^\circ/n) = r/(r + R) $$

$$ \text{etc.}$$

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Shouldn't the third line read $\frac{R}{r}=\frac{1}{\sin(180/n)}-1$ or am I being a boob? –  Simon Hayward Nov 12 '12 at 21:02
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Yeah, the passage from the second to the third line is incorrect, and that throws off the arithmetic on the rest of it. –  Steven Stadnicki Nov 12 '12 at 21:04
    
It should indeed - now edited –  Henry Nov 12 '12 at 22:10
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Rearranging gives $$\frac{r}{R}=\frac{\sin(180/n)}{1-\sin(180/n)}\\ \Rightarrow \frac{r(1-\sin(180/n)}{R}=\sin(180/n)\\ \Rightarrow \frac{r}{R}-\frac{r\sin(180/n)}{R}=\sin(180/n)\\ \Rightarrow \frac{r}{R}= \left(1+\frac{r}{R}\right)(\sin(180/n)\\ \Rightarrow \sin^{-1}\left(\frac{r}{R}\left(1+\frac{r}{R}\right)^{-1}\right)=(180/n)\\ \Rightarrow \frac{180}{\sin^{-1}\left(\frac{r}{R}\left(1+\frac{r}{R}\right)^{-1}\right)}^{-1}=n\\ \Rightarrow \frac{180}{\sin^{-1}\left(\frac{r}{R + r}\right)}=n$$

I hope that helps. You extract the value using the arcsine (inverse sine) function, which is on almost all scientific calculators.

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Oh lord. Inverse sine. Of course. Thank you. –  JimmyDeemo Nov 12 '12 at 21:01
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This answer isn't quite correct: you failed to distribute the factor of $r$ over $(1-\sin(180/n))$ when passing from the second equation to the third. –  Steven Stadnicki Nov 12 '12 at 21:02
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And your last line is missing the arcsin, which has to stay around. –  Steven Stadnicki Nov 12 '12 at 21:03
    
You are absolutely correct, will amend. >.< –  Simon Hayward Nov 12 '12 at 21:03
    
This still has a calculation error - your passage from the first line with arcsin to the second isn't right (it would work if the RHS were n/180, but not for 180/n). –  Steven Stadnicki Nov 12 '12 at 21:06
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