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三角連乘公式證明

回復 2# Exponential 的帖子

回其中一個,應該都差不多
\( \displaystyle \prod\limits_{k = 1}^{n-1} {\sin (\frac{{k\pi }}{n})} \)
令\(\displaystyle \omega  = \cos \frac{{2\pi }}{{n}} + i\sin \frac{{2\pi }}{{n}}\),\( {x^{n}} - 1 = (x - 1)(1 + x + {x^2} + ...+{x^{n - 1}}) = (x - 1)(x - \omega )(x - {\omega ^2})...(x - {\omega ^{n-1}}) \)
\(1 + x + {x^2} + ...+{x^{n - 1}}=(x - \omega )(x - {\omega ^2})...(x - {\omega ^{n-1}})\)
代入\(x=1\),\(n = (1 - \omega )(1 - {\omega ^2})...(1 - {\omega ^{n - 1}})\)
\( \displaystyle 1 - {\omega ^k} = 1 - (\cos \frac{{2k\pi }}{{n}} + i\sin \frac{{2k\pi }}{{n}} )= 2\sin \frac{{k\pi }}{{n}}(\sin \frac{{k\pi }}{{n}} + i\cos \frac{{k\pi }}{{n}}) \),\( \displaystyle \left| {1 - {\omega ^k}} \right| = 2\sin \frac{{k\pi }}{{n}}\)
\(\displaystyle n = \left| {1 - \omega } \right|\left| {1 - {\omega ^2}} \right|...\left| {1 - {\omega ^{n - 1}}} \right| = {2^{n - 1}}\prod\limits_{k = 1}^{n-1} {\sin (\frac{{k\pi }}{n})} \),\( \displaystyle \prod\limits_{k = 1}^{n - 1} {\sin (\frac{{k\pi }}{n})}  = \frac{n}{{{2^{n - 1}}}}\)

話說這個有什麼比較好的背法嗎,學生看到我寫這個腦袋只會一片空白,我自己也沒有背過這個
順代一提,\(cos\)應該是代\(x=-1\)

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