# Thread: cos question hard

1. ## cos question hard

show that the area of a regular n-gon inscribed in a circle to the area of a regular n-gon circumscribing the same circle is cos^2(pi/n) : 1

2. ## Re: cos question hard

Originally Posted by darandoma
show that the area of a regular n-gon inscribed in a circle to the area of a regular n-gon circumscribing the same circle is cos^2(pi/n) : 1
area of triangular sections from each n-gon ...

red area = $\frac{1}{2}r^2 \sin\left(\frac{2\pi}{n}\right)$

red area + blue area = $r^2 \tan\left(\frac{\pi}{n}\right)$

ratio ...

$\frac{\sin\left(\frac{2\pi}{n}\right)}{\tan \left(\frac{\pi}{n}\right)}$

$\frac{2\sin\left(\frac{\pi}{n}\right) \cos\left(\frac{\pi}{n}\right)}{2\tan \left(\frac{\pi}{n}\right)}$

$\sin\left(\frac{\pi}{n}\right)\cos\left(\frac{\pi} {n}\right) \cdot \frac{\cos\left(\frac{\pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)}$

$\cos^2\left(\frac{\pi}{n}\right)$

3. ## Re: cos question hard

thanks for commenting but why does red area + blue area = r^2tan(pi/n) ?

4. ## Re: cos question hard

Originally Posted by darandoma
thanks for commenting but why does red area + blue area = r^2tan(pi/n) ?
look at the upper large right triangle (half of the entire large triangular section of the circumscribed n-gon)...

angle = $\frac{\pi}{n}$

adjacent side = $r$

opposite side = $r \tan\left(\frac{\pi}{n}\right)$

area of upper right triangle = $\frac{1}{2} \cdot r \cdot r \tan\left(\frac{\pi}{n}\right)$

double that ...