History of the general formula for linearising $\cos^n(x)$

Question

I was wondering where the formula: $$\cos^n(x)=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\cos(x(2k-n))$$ Was first originally published. I accidentally derived it a few days ago and was wondering where it was first established for historical context. I derived it in the following way:

Using Euler's formulas, we can write $\cos^n(x)$ out as: $$\cos^n(x)=\frac{1}{2^n}(e^{ix}+e^{-ix})^n$$ Which, when expanded using Newton's binomial formula yields: $$\cos^n(x)=\frac{1}{2^n}\sum_{k=0}^n\frac{n!}{k!(n-k)!}e^{ikx}\cdot e^{-ix(n-k)}$$ Which simplifies down to: $$\cos^n(x)=\frac{1}{2^n}\left(\sum_{k=0}^n(\frac{n!}{k!(n-k)!}\cos(x(2k-n)))+i\sum_{k=0}^n(\frac{n!}{k!(n-k)!}\sin(x(2k-n)))\right)$$ $$=\frac{1}{2^n}C_n+\frac{i}{2^n}S_n$$ Because $S_n\in\mathbb{R}$ we can write that $\Im(\cos^n(x))=S_n=0$ And consequently we yield: $$\cos^n(x)=\frac{1}{2^n}\sum_{k=0}^n\frac{n!}{k!(n-k)!}\cos(x(2k-n))$$ Where does this originally come from? I'm sure it's been done before but I've looked everywhere and can't find it at all. This is as usual purely recreational.

Answer 1

Note: This does not address the question of origin of these formulas. But, in the spirit of recreation, here are some other ways to rearrange the sum, which are easier to parse if we treat odd and even $n$ separately.

Because the cosine is an even function and the binomial coefficients are symmetric, we can get away with half as many terms and some other forms.

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When $n$ is odd, we fold the sum in half: $$ \cos^n(x) = \frac{1}{2^{n-1}} \sum_{k=0}^{(n-1)/2} \binom{n}{k} \cos\bigl( (n-2k)x \bigr). $$ Equivalently, with $n = 2m+1$, $$ \cos^{2m+1}(x) = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m+1}{k} \cos\bigl( (2m-2k+1)x \bigr), $$ or by reindexing with $j = m - k$, $$ \cos^{2m+1}(x) = \frac{1}{2^{2m}} \sum_{j=0}^{m} \binom{2m+1}{m-j} \cos\bigl( (2j+1)x \bigr). $$

For example, with $n = 7$, equivalently $m = 3$, $$ \cos^7(x) = \frac{1}{2^6} \biggl[ \binom{7}{3} \cos(1x) + \binom{7}{2} \cos(3x) + \binom{7}{1} \cos(5x) + \binom{7}{0} \cos(7x) \biggr]. $$

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When $n$ is even, we peel off the middle term before folding in half: $$ \cos^n(x) = \frac{1}{2^n} \binom{n}{n/2} + \frac{1}{2^{n-1}} \sum_{k=0}^{n/2-1} \binom{n}{k} \cos\bigl( (n-2k)x \bigr). $$ Equivalently, with $n = 2m$, $$ \cos^{2m}(x) = \frac{1}{2^{2m}} \binom{2m}{m} + \frac{1}{2^{2m-1}} \sum_{k=0}^{m-1} \binom{2m}{k} \cos\bigl( (2m-2k)x \bigr), $$ or by reindexing with $j = m - k$, $$ \cos^{2m}(x) = \frac{1}{2^{2m}} \binom{2m}{m} + \frac{1}{2^{2m-1}} \sum_{j=1}^{m} \binom{2m}{m-j} \cos( 2jx ). $$

For example, with $n = 6$, equivalently $m = 3$, $$ \cos^6(x) = \frac{1}{2^6} \binom{6}{3} + \frac{1}{2^5} \biggl[ \binom{6}{2} \cos(2x) + \binom{6}{1} \cos(4x) + \binom{6}{0} \cos(6x) \biggr]. $$

Answer 2

Not specificaly an answer about historical roots, but nevertheless using a technique developed circa 200 years ago, Fourier series.

Indeed, you can consider your expansion of the even function $\cos^n(x)$ as a Fourier expansion on the basis of functions $\cos(kx), \ k=0,1,2...$ (no need for the $\sin(kx)$) with the particularity that it is a finite expansion.

Using the classical formulas for the coefficients $a_k$ of a Fourier series, we have the double indexing :

$$a_{n,k}=\frac{1}{2\pi}\int_{-\pi}^{ \pi}\cos^n(x) \cos(kx)dx=\begin{cases}\frac{1}{2^n}\binom{n}{\tfrac12(n-k)}&\text{if } n-k \text{ is even}\\0& \text{otherwise.}\end{cases}\tag{1}$$

(see the answer here).

i.e., one out of two of these coefficients is $O$ (as in the expressions given by Sammy Black).

Making a distinction between the even and the odd cases for the exponent, in the same way Sammy Black has done in his excellent answer :

• $n=2m$ where there is constant $a_0\ne 0$ {the "stripped off" term in the answer by S. B.) for the Fourier series because $a_0$ is the average value of a positive function.

• $n=2m+1$ where there is in particular no $a_0$ term.

It is always interesting to have a graphical idea of the functions we are working on ; here are the two kinds (even exponent in green, odd exponent in red) :

Answer 3

Chebyshev Polynomials page is the most visited Wikiedia page by me. $$t^n=2^{1-n}\sum_{k=0\\j\equiv n\pmod 2}^n'\,\,\large{n\choose \frac{n-j}{2}}T_j(t)$$ where the prime at the summation symbol indicates that the contribution of $j = 0$ needs to be halved if it appears. Let $t=\cos x$. Then, $$\cos^nx=2^{1-n}\sum_{k=0\\j\equiv n\pmod 2}^n'\,\,\large{n\choose \frac{n-j}{2}}\cos(jx).$$ Example: $\cos^3x=2^{1-3}({3 \choose 1}\cos x+{3 \choose 0}\cos 3x)=\frac{3\cos x+\cos3x}{4}$.