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  1. #1
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    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
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  2. #2
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    Re: cos question hard

    Quote Originally Posted by darandoma View Post
    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)
    Attached Thumbnails Attached Thumbnails cos question hard-ngonsection.jpg  
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  3. #3
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    Re: cos question hard

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

    Quote Originally Posted by darandoma View Post
    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 ...
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