# Can I simplify $n\cdot r\cdot \sin(90^\circ-\frac{180^\circ}{n})\sqrt{r^2-r^2\sin^2(90^\circ-\frac{180^\circ}{n})}$ further?

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.

• 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. – user29418 Sep 28 '18 at 23:45
• @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. – nine-hundred Sep 28 '18 at 23:57

$$n\cdot r\cdot \sin\left(90^\circ-\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\sin^2\left(90^\circ-\frac{180^\circ}{n}\right)}$$ $$=n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2-r^2\cos^2\left(\frac{180^\circ}{n}\right)}$$ $$=n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2\left(1-\cos^2\left(\frac{180^\circ}{n}\right)\right)}$$ $$=n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)\sqrt{r^2\sin^2\left(\frac{180^\circ}{n}\right)}$$ $$=n\cdot r\cdot \cos\left(\frac{180^\circ}{n}\right)r\sin\left(\frac{180^\circ}{n}\right)$$ $$=n\cdot r^2\cdot \cos\left(\frac{180^\circ}{n}\right)\sin\left(\frac{180^\circ}{n}\right)$$ $$= \frac{1}{2}nr^2\sin\left(\frac{360}{n}\right)$$ Is this what you are looking for?
• 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. – nine-hundred Sep 29 '18 at 0:36
• 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? – nine-hundred Sep 29 '18 at 0:41
• 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. – Larry Sep 29 '18 at 0:44
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.