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?

Question

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.

Answer 1

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

Answer 2

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.