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[函数] 三角恒等式

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等待hxh posted 2014-4-30 22:11 |Read mode
Last edited by hbghlyj 2025-3-13 04:05\[
\prod_{k=1}^n \sin \frac{k \pi}{2 n+1}=\frac{\sqrt{2 n+1}}{2^n}\qquad \prod_{k=1}^n \sin \frac{k \pi}{2 n}=\frac{\sqrt{n}}{2^{n-1}}
\]不用复数怎么证呀(此题只能用复数证明吗,最近想收集一些三角恒等式,不知有这样的帖子不?)

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其妙 posted 2014-4-30 23:04
回复 1# 等待hxh
能复数证明而且好用,为何不用?正如一条好路不走,还要去创新?

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青青子衿 posted 2019-8-16 13:35
Last edited by hbghlyj 2025-5-27 18:35回复 1# 等待hxh
zhihu.com/question/336984088/answer/782336316

3. 证明: $\prod_{k=1}^{m-1} \cos \frac{k \pi}{2 m}=\frac{\sqrt{m}}{2^{m-1}}$
4. 证明: $\prod_{k=1}^{m} \sin \frac{(2 k-1) \pi}{2(2 m+1)}=\frac{1}{2^{m}}$
5. 证明: $\prod_{k=1}^{m} \sin \frac{k \pi}{2 m+1}=\frac{\sqrt{2 m+1}}{2^{m}}$
6. 证明: $\prod_{k=1}^{m} \cos \frac{(2 k-1) \pi}{2(2 m+1)}=\frac{\sqrt{2 m+1}}{2^{m}}$
7. 证明: $\prod_{k=1}^{m} \sin \frac{(2 k-1) \pi}{4 m}=\frac{\sqrt{2}}{2^{m}}$
8. 证明: $\prod_{k=1}^{m} \cos \frac{(2 k-1) \pi}{4 m}=\frac{\sqrt{2}}{2^{m}}$

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kuing posted 2019-8-16 13:54
回复 3# 青青子衿

zhihu竟然没有防盗图?那以后大图是不是可以借那边的地来发了

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lemondian posted 2019-11-12 21:13
回复 1# 等待hxh
用复数证明如何证明?

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kuing posted 2019-11-13 15:35

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lemondian posted 2019-11-14 14:24
回复 6# kuing
3#的那些呢?如何搞?

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hbghlyj posted 2025-5-27 18:35
回复 6# kuing
3#的那些呢?如何搞?

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