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[函数] 三的幂次倍角

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hbghlyj posted 2019-6-20 23:35 |Read mode
Last edited by hbghlyj 2022-12-14 20:47sin9x/sinx=(2cos2x+1)(2cos6x+1)
cos9x/cosx=(2cos2x-1)(2cos6x-1)
这些公式都是从三倍角公式来的。
sin3x/sinx=3-4sin^2x=1+2cos2x
cos3x/cosx=4cos^2x-3=2cos2x-1
因此
$\sin3^nx/\sin x=\prod_{i=1}^{n-1}(1+2\cos2\cdot 3^ix)$
$\cos3^nx/\cos x=\prod_{i=1}^{n-1}(2\cos2\cdot 3^ix-1)$

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力工 posted 2019-6-25 09:31
回复 1# hbghlyj
我都看不下去了。码一下。
证明:(1) $\frac{\sin9x}{\sin x}=(2\cos2x+1)(2\cos6x+1)$,
         (2) $\frac{\cos9x}{\cos x}=(2\cos2x-1)(2\cos6x-1)$.

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kuing posted 2019-6-25 09:37

我都看不下去了。码一下。
力工 发表于 2019-6-25 09:31
你这码的也不见得好看,呵呵……

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