# cos question hard

• Sep 18th 2011, 11:07 PM
darandoma
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
• Sep 19th 2011, 06:13 PM
skeeter
Re: cos question hard
Quote:

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

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

ratio ...

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

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

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

$\displaystyle \cos^2\left(\frac{\pi}{n}\right)$
• Sep 20th 2011, 08:35 AM
darandoma
Re: cos question hard
thanks for commenting but why does red area + blue area = r^2tan(pi/n) ?
• Sep 20th 2011, 01:37 PM
skeeter
Re: cos question hard
Quote:

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

adjacent side = $\displaystyle r$

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

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

double that ...