余弦函数n倍角的那些高级公式

2022-11-27 14:16 作者:fangquping 0人读过 | 我要投稿

对于三角函数的n倍角的公式，高中生一般都学过，太俗的今天就不谈了，和大家探讨下那些很少见的余弦n倍角公式

首先要说一下棣莫弗公式，即

$%5Ccos%20nx%3D%5Ccos%5Enx-C_n%5E2%5Ccos%5E%7Bn-2%7Dx%20%5Csin%5E2%20x%2BC_n%5E4%5Ccos%5E%7Bn-4%7Dx%5Csin%5E4%20x-%5Ccdots$

这个比较重要，今天的公式大多由它推导而得，这个公式是怎么来的呢？要借助欧拉公式

$e%5E%7Bix%7D%3D%5Ccos%20x%2Bi%5Csin%20x$

$e%5E%7Binx%7D%3D(%5Ccos%20x%2Bi%5Csin%20x)%5En%3D%5Ccos%7Bnx%7D%2Bi%5Csin%20%7Bnx%7D$

中间部分展开，得

$%5Ccos%20x%2BiC_n%5E1%5Ccos%5E%7Bn-1%7D%20x%5Csin%20x-C_n%5E2%5Ccos%5E%7Bn-2%7D%20x%5Csin%5E2%20x-%5Ccdots$

采用求和符号，我们得到如下等式

$%5Csum_%7Bk%3D0%7D%5Eni%5E%7Bk%7DC_n%5Ek%5Ccos%5E%7Bn-k%7Dx%20%5Csin%20%5Ekx%20%3D%5Ccos%7Bnx%7D%2Bi%5Csin%7Bnx%7D$

由于i是虚数单位，根据它的周期关系，有

$%5Csum_%7B%5Cfrac%7Bn%7D%7B2%7D%5Cge%20k%5Cge%200%7D(-1)%5E%7Bk%7DC_n%5E%7B2k%7D%5Ccos%5E%7Bn-2k%7Dx%20%5Csin%20%5E%7B2k%7Dx%20%2B$

$i%20%5Csum_%7B%5Cfrac%7Bn-1%7D%7B2%7D%5Cge%20k%5Cge%200%7D(-1)%5E%7Bk%7DC_n%5E%7B2k%2B1%7D%5Ccos%5E%7Bn-2k-1%7Dx%20%5Csin%20%5E%7B2k%2B1%7Dx$

$%3D%5Ccos%7Bnx%7D%2Bi%5Csin%7Bnx%7D$

比较实部能得到前面的棣莫弗公式，这里就不详述了。

下面推导一些高能公式

当n为奇数时

$%5Ccos%20%7Bnx%7D%3D%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5EkC_n%5E%7B2k%7D%5Ccos%20%5E%7Bn-2k%7Dx%5Csin%5E%7B2k%7Dx$

$%3D%5Ccos%20x%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5EkC_n%5E%7B2k%7D(1-%5Csin%5E2%20x)%20%5E%7B%5Cfrac%7Bn-1%7D%7B2%7D-k%7D%5Csin%5E%7B2k%7Dx$

除了前面有公因式cos x，后面可化成关于sin²x的式子，这样得出余弦n倍角的式子，

为了计算正弦系数，再引入系数数列R，设

$%5Ccos%20nx%3D%5Ccos%20x%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5EkR_k%5Csin%5E%7B2k%7Dx$

对比余弦n倍角的式子，可知

$R_0%3D1$

$R_1%3D%5Cfrac%7Bn%5E2-1%7D%7B2%7D$

两边求导，得

$-n%5Csin%20nx%3D-%5Csin%20x-%5Csum_%7Bk%3D1%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5EkR_k%5Csin%5E%7B2k%2B1%7Dx%2B$

$2(1-%5Csin%20%5E2x)%5Csum_%7Bk%3D1%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5EkR_kk%5Csin%5E%7B2k-1%7Dx$

$%3D-%5Csin%20x-%5Csum_%7Bk%3D1%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5EkR_k%5Csin%5E%7B2k%2B1%7Dx-$

$2%5Csum_%7Bk%3D0%7D%5E%7B%5Cfrac%7Bn-1%7D%7B2%7D-1%7D(-1)%5E%7Bk%7DR_%7Bk%2B1%7D(k%2B1)%5Csin%5E%7B2k%2B1%7Dx%20-2%5Csum_%7Bk%3D1%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5EkR_kk%5Csin%5E%7B2k%2B1%7Dx$

$%3D-%5Csin%20x-%5Csum_%7Bk%3D1%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5Ek%5Cleft%20%5B(2k%2B1)R_k%2B2R_%7Bk%2B1%7D(k%2B1)%5Cright%20%5D%5Csin%5E%7B2k%2B1%7Dx$

$%2B(-1)%5E%7B%5Cfrac%7Bn-1%7D%7B2%7D%7D(n%2B1)R_%7B%5Cfrac%7Bn%2B1%7D2%7D%5Csin%5E%7Bn%7Dx%20-(n%5E2-1)%5Csin%20x$

$%3D(-1)%5E%7B%5Cfrac%7Bn-1%7D%7B2%7D%7D(n%2B1)R_%7B%5Cfrac%7Bn%2B1%7D2%7D%5Csin%5E%7Bn%7Dx%20-$

$%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5Ek%5Cleft%20%5B(2k%2B1)R_k%2B2R_%7Bk%2B1%7D(k%2B1)%5Cright%20%5D%5Csin%5E%7B2k%2B1%7Dx$

为了得到递推关系，两边再求导得

$-n%5E2%5Ccos%20nx%3D(-1)%5E%7B%5Cfrac%7Bn-1%7D%7B2%7D%7Dn(n%2B1)R_%7B%5Cfrac%7Bn%2B1%7D2%7D%5Csin%5E%7Bn-1%7Dx%20%5Ccos%20x-$

$%5Csum_%7Bk%3D0%7D%5E%7B%5Cfrac%7Bn-1%7D%7B2%7D%7D(-1)%5Ek(2k%2B1)%5Cleft%20%5B(2k%2B1)R_k%2B2R_%7Bk%2B1%7D(k%2B1)%5Cright%20%5D%5Csin%5E%7B2k%7Dx%5Ccos%20x$

对比原式可知

$-n%5E2(-1)%5EkR_k%3D-(-1)%5Ek(2k%2B1)%5Cleft%20%5B(2k%2B1)R_k%2B2R_%7Bk%2B1%7D(k%2B1)%5Cright%20%5D$

$n%5E2R_k%3D(2k%2B1)%5E2R_k%2B(2k%2B1)(2k%2B2)R_%7Bk%2B1%7D$

$R_%7Bk%2B1%7D%3D%5Cfrac%7Bn%5E2-(2k%2B1)%5E2%7D%7B(2k%2B1)(2k%2B2)%7DR_%7Bk%7D$

$R_0%3D1$

$R_1%3D%5Cfrac%7Bn%5E2-1%7D%7B%202%7D$

$R_2%3D%5Cfrac%7B(n%5E2-1)(n%5E2-3%5E2)%7D%7B4!%7D$

$R_3%3D%5Cfrac%7B(n%5E2-1)(n%5E2-3%5E2)(n%5E2-5%5E2)%7D%7B6!%7D$

$%5Ccdots$

$R_k%3D%5Cfrac%7B(n%5E2-1)(n%5E2-3%5E2)%5Ccdots%20%5Bn%5E2-(2k-1)%5E2%5D%7D%7B(2k)!%7D$

经检验可知，当n为奇数时，有以下关系

$%5Ccos%20nx%3D%5Ccos%20x%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn-1%7D%7B2%7D(-1)%5EkR_k%5Csin%5E%7B2k%7D%20x$

其中

$R_k%3D%5Cfrac%7B(n%5E2-1)(n%5E2-3%5E2)%5Ccdots%20%5Bn%5E2-(2k-1)%5E2%5D%7D%7B(2k)!%7D$

$R_0%3D1$

同理，当n为偶数时，有

$%5Ccos%20nx%3D%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn%7D%7B2%7D(-1)%5EkC_n%5E%7B2k%7D%5Ccos%20%5E%7Bn-2k%7Dx%5Csin%20%5E%7B2k%7Dx$

$%3D%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn%7D%7B2%7D(-1)%5EkC_n%5E%7B2k%7D(1-%5Csin%5E2%20x)%5E%7B%5Cfrac%7Bn%7D%7B2%7D-k%7D%5Csin%20%5E%7B2k%7Dx$

引入数列R，使下列等式成立

$%5Ccos%20nx%3D%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn%7D%7B2%7D(-1)%5EkR_k%5Csin%5E%7B2k%7D%20x$

同理可知

$R_0%3D1$

$R_1%3D%5Cfrac%7Bn%5E2%7D%7B2%7D$

求二阶导得

$-n%5Ccos%5E2x$

$%3D%20%5Csum_%7Bk%3D1%7D%5E%5Cfrac%7Bn%7D%7B2%7D(-1)%5Ek2k(2k-1)R_k%5Csin%20%5E%7B2k-2%7Dx%5Ccos%5E2x$

$-%5Csum_%7Bk%3D1%7D%5E%7B%5Cfrac%7Bn%7D%7B2%7D%7D(-1)%5Ek2kR_k%5Csin%5E%7B2k%7Dx$

$%3D%5Csum_%7Bk%3D1%7D%5E%5Cfrac%7Bn%7D%7B2%7D(-1)%5Ek2k(2k-1)R_k%5Csin%20%5E%7B2k-2%7Dx$

$-%5Csum_%7Bk%3D1%7D%5E%7B%5Cfrac%7Bn%7D%7B2%7D%7D(-1)%5Ek4k%5E2R_k%5Csin%5E%7B2k%7Dx$

$%3D-%5Csum_%7Bk%3D0%7D%5E%7B%5Cfrac%7Bn%7D%7B2%7D-1%7D(-1)%5Ek(2k%2B2)(2k%2B1)R_%7Bk%2B1%7D%5Csin%20%5E%7B2k%7Dx$

$-%5Csum_%7Bk%3D1%7D%5E%7B%5Cfrac%7Bn%7D%7B2%7D%7D(-1)%5Ek4k%5E2R_k%5Csin%5E%7B2k%7Dx$

$%3D(-1)%5E%5Cfrac%7Bn%7D%7B2%7D(n%2B2)(n%2B1)R_%7B%5Cfrac%7Bn%7D%7B2%7D%2B1%7D%5Csin%5Enx-$

$%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn%7D%7B2%7D(-1)%5Ek%5Cleft%20%5B(2k%2B2)(2k%2B1)R_%7Bk%2B1%7D%2B4k%5E2R_k%5Cright%20%5D%5Csin%5E%7B2k%7Dx$

比较原式可知

$-n%5E2(-1)%5EkR_k%3D-(-1)%5Ek%5Cleft%20%5B(2k%2B2)(2k%2B1)R_%7Bk%2B1%7D%2B4k%5E2R_k%5Cright%20%5D$

$(n%5E2-4k%5E2)R_k%3D(2k%2B2)(2k%2B1)R_%7Bk%2B1%7D$

$R_%7Bk%2B1%7D%3D%5Cfrac%7Bn%5E2-4k%5E2%7D%7B(2k%2B2)(2k%2B1)%7DR_k$

$R_0%3D1$

$R_1%3D%5Cfrac%7Bn%5E2%7D%7B2%7D$

$R_2%3D%5Cfrac%7Bn%5E2(n%5E2-2%5E2)%7D%7B4!%7D$

$R_3%3D%5Cfrac%7Bn%5E2(n%5E2-2%5E2)(n%5E2-4%5E2)%7D%7B6!%7D$

$%5Ccdots$

$R_k%3D%5Cfrac%7Bn%5E2(n%5E2-2%5E2)%5Ccdots%20%5Bn%5E2-(2k-2)%5E2%5D%7D%7B(2k)!%7D$

经检验可知当n为偶数时，有以下关系

$%5Ccos%20nx%3D%5Csum_%7Bk%3D0%7D%5E%5Cfrac%7Bn%7D%7B2%7D(-1)%5EkR_k%5Csin%5E%7B2k%7D%20x$

其中

$R_k%3D%5Cfrac%7Bn%5E2(n%5E2-2%5E2)(n%5E2-4%5E2)%5Ccdots%20%5Bn%5E2-(2k-2)%5E2%5D%7D%7B(2k)!%7D$

$R_0%3D1$

这样就证明出余弦n倍角的两个n倍角公式，很冷门吧，只要有耐心、够执着，什么样的公式都能见得到。

根据这些思路，你能不能结合n为奇数和偶数的情况，写出任意整数n的余弦倍角通用公式呢？如果你是老朋友，会看过相关的视频，下面是证明过程，方法类似，就不赘述了，欢迎支持！

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