TMUA practice: Trig Identities, problem 3

Let the regular $n$-gon have circumradius $R$ (centre to a vertex) and inradius $r$ (centre to the midpoint of a side). The inscribed circle is the one of radius $r$.

Relating $r$ and $R$. Join the centre $O$ to one vertex $V$ and to the midpoint $M$ of a side meeting at $V$. The side subtends a central angle $\dfrac{2\pi}{n}$ at $O$, and $OM$ bisects that angle, so $\angle VOM = \dfrac{\pi}{n}$. Triangle $OMV$ is right-angled at $M$, with hypotenuse $OV = R$ and the side $OM = r$ adjacent to the angle at $O$:

$r = R\cos\frac{\pi}{n}.$

Areas. The circumcircle has area $A = \pi R^2$. The inscribed circle therefore has area

$\pi r^2 = \pi R^2\cos^2\frac{\pi}{n} = A\cos^2\frac{\pi}{n}.$

Matching the options. No option is written as $A\cos^2\dfrac{\pi}{n}$ directly, so apply the double-angle identity $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$ with $\theta = \dfrac{\pi}{n}$:

Answer: B.