Thread: 2 triangles
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Old November 10th, 2009, 07:37 PM
ukorov ukorov is offline
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Quote:
Originally Posted by Grandad View Post
Solving the quadratic, taking the positive root gives
\cos(\tfrac{2\pi}{5})=\frac{\sqrt5-1}{4}

\Rightarrow 2\cos^2(\tfrac{\pi}{5}) -1 = \frac{\sqrt5-1}{4}
why \cos(\tfrac{2\pi}{5}) = 2\cos^2(\tfrac{\pi}{5})-1 ??
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