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I need to write a simplified formula for this:

$$A_i = n\cdot r\cdot \sin\left(90^\circ-\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\sin^2\left(90^\circ-\frac{180^\circ}{n}\right)}$$

I am not very confident that I know enough trigonometry identities to simplify this completely. Other than converting $\sin(90^\circ-x)$ to $\cos(x)$, I am not sure there isn’t anything I’m missing, with the radical sign in there and everything.

$n$ and $r$ are variables and natural numbers. $A_i$ is a value based on $n$ and $r$.

This is in degrees, if that wasn’t clear.

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  • $\begingroup$ cant you factor out $r^2$ and then it comes out of the radical as r? then there's another trig identity inside the square too: $1-sin^2$ after factoring. Doing all of this eventually removes the square root. $\endgroup$
    – user29418
    Sep 28, 2018 at 23:45
  • $\begingroup$ @user29418 yes, but when I did that I was not able to reach the answer I wanted. There is a specific equation I am deriving that I couldn’t figure out by factoring out the $r$, so I left it unsimplified in case there was a different method of rewriting that radical that led to the right answer. $\endgroup$
    – Descartes Before the Horse
    Sep 28, 2018 at 23:57

2 Answers 2

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$$n\cdot r\cdot \sin\left(90^\circ-\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\sin^2\left(90^\circ-\frac{180^\circ}{n}\right)}$$ $$=n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\cos^2\left(\frac{180^\circ}{n}\right)}$$ $$ =n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2\left(1-\cos^2\left(\frac{180^\circ}{n}\right)\right)}$$ $$ =n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2\sin^2\left(\frac{180^\circ}{n}\right)}$$ $$ =n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)r\sin\left(\frac{180^\circ}{n}\right)$$ $$ =n\cdot r^2\cdot \cos\left(\frac{180^\circ}{n}\right)\sin\left(\frac{180^\circ}{n}\right)$$ $$= \frac{1}{2}nr^2\sin\left(\frac{360}{n}\right)$$ Is this what you are looking for?

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  • $\begingroup$ Yes, that is exactly what I was looking for. Could you explain what you did in the last step; simplifying $cos(x)sin(x)$ to $0.5sin(x/2)$? I might just be forgetting about something about the trig functions, but I can't make sense of that. $\endgroup$
    – Descartes Before the Horse
    Sep 29, 2018 at 0:36
  • $\begingroup$ Sorry--Just looked at your answer again; $sin(x)cos(x) = 0.5sin(2x)$, right? So $cos(180/n)sin(180/n)$ should simplify to $sin(360/n)$, not $sin(90/n)$ as you said, correct? $\endgroup$
    – Descartes Before the Horse
    Sep 29, 2018 at 0:41
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    $\begingroup$ You are correct, $\sin(2x)=2\sin(x)\cos(x)$. I have made the correction. Basically, it is one of the trig identities. $x = 180/n$ in the problem. $\endgroup$
    – Larry
    Sep 29, 2018 at 0:44
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First of all, let's say you've already changed each $\sin(90^{\circ}-x)$ into $\cos(x)$. Then, here's a hint: $$r^2-r^2\cos^2(x)=r^2\left[1-\cos^2(x)\right]=r^2\sin^2(x),$$ and you can take the square root now.

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  • $\begingroup$ I’ve done that already; it didn’t get my anywhere as far as I could tell, so I left that portion unsimplified in case there was a different method of simplifying that got me a final answer. This is an equation for something specific that I am trying to derive, and I know I am very close, but I can’t figure out how to get the pieces to fall in the right place. I leave the actual equation I am deriving out because i believe doing a problem you already know the answer to changes the method of reaching that answer. $\endgroup$
    – Descartes Before the Horse
    Sep 28, 2018 at 23:55
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    $\begingroup$ @Matt: "I've done that already" ... In the future, please make sure that your questions include the work you have done, so that people don't waste their precious time telling you things that you already know. (When you have a particular goal in mind —as appears to be true here— you should include that, too.) $\endgroup$
    – Blue
    Sep 29, 2018 at 0:14

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