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