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| &= \frac12 \quad \quad \quad \text{Q.E.D.} | | &= \frac12 \quad \quad \quad \text{Q.E.D.} |
| \end{align}</math> | | \end{align}</math> |
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− | To better see why the equation is true, it is better to go to the complex plane. cos(2k pi/7) <!--<math>\textstyle{ \cos \frac{2k\pi}{7} }</math>--> is the real part of the k-th 7-th root of unity, exp(2 k i pi/7)<!--<math>\textstyle{ \exp \frac{2 k i\pi}{7} }</math>-->. The seven 7-th roots of unity (for 0 <= k <= 6) sum up to zero, hence so do their real parts:
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− | <!--:<math>0 = \cos \frac{0\pi}{7} + \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} + \cos \frac{8\pi}{7} + \cos \frac{10\pi}{7} + \cos \frac{12\pi}{7} </math>-->
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− | :0 = cos(0 pi/7) + cos(2 pi/7) + cos(4 pi/7) + cos(6 pi/7) + cos(8 pi/7) + cos(10 pi/7) + cos(12 pi/7)
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− | But one of these roots is just 1, and all other root go by pairs of conjugate roots, which have the same real part (alternatively, consider that cos(x) = cos(2 pi - x)):
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− | <!--:<math>0 = 1 + 2 ( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} ) </math>-->
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− | :0 = 1 + 2 (cos(2 pi/7) + cos(4 pi/7) + cos(6 pi/7))
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− | Hence
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− | <!--:<math>\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = - 1/2 </math>-->
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− | :cos(2 pi/7) + cos(4 pi/7) + cos(6 pi/7) = - 1/2
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− | which, because cos(x) = cos(pi - x),<!--<math>\cos (x) = - \cos(\pi - x)</math>,--> can be rewritten as
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− | <!--:<math>\cos \frac{5\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{pi}{7} = 1/2 </math>-->
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− | :cos(5 pi/7) + cos(3 pi/7) + cos(pi/7) = 1/2
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− | Q.E.D.
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| ==Transcript== | | ==Transcript== |