How to expand $\cos nx$ with $\cos x$?

Multiple Angle Identities:

How to expand $\cos nx$ with $\cos x$, such as $$\cos10x=512(\cos x)^{10}-1280(\cos x)^8+1120(\cos x)^6-400(\cos x)^4+50(\cos x)^2-1$$ See a list of trigonometric identities in english/ chinese

Solution 1:

These are usually denoted $T_n$ and called Chebyshev polynomials of the first kind. For every $n\geqslant0$, $\cos(nu)=T_n(\cos(u))$ with $$ T_n(x)= \sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} (x^2-1)^k x^{n-2k}. $$ For example, $$ T_8(x) = 128x^8 - 256x^6 + 160x^4 - 32x^2 + 1, $$ and $$ T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x. $$ Likewise, $$ T_{10}(x)=512x^{10} − 1280x^8 + 1120x^6 − 400x^4 + 50x^2 − 1, $$ $$ T_{11}(x)=1024x^{11} − 2816x^9 + 2816x^7 − 1232x^5 + 220x^3 − 11x, $$ and $$ T_{12}(x)=2048x^{12} − 6144x^{10} + 6912x^8 − 3584x^6 + 840x^4 − 72x^2 + 1. $$

Solution 2:

Here is a neat way to derive what the answer will be, using Euler's formula $e^{ix}=\cos x + i \sin x$.

We have that $\cos nx$ is the real part of $e^{i(nx)}=\left(e^{ix}\right)^n=(\cos x + i \sin x)^n$

By the binomial formula, $(\cos x + i \sin x)^n=\displaystyle\sum_{k=0}^n i^k \binom{n}{k} \sin^k(x) \cos^{n-k}(x)$. Since $i^k$ is real if and only if $k$ is even, we therefore have (replacing $k$ with a new indexing variable $2\ell$)

$$\cos nx= \sum_{\ell=0}^{\lfloor n/2\rfloor} (-1)^{\ell} \binom{n}{2\ell} (\sin^2(x))^{\ell} \cos^{n-2\ell}(x).$$

Finally, we use the Pythagorean identity $\sin^2=1-\cos^2$ to rewrite

$$\cos nx= \sum_{\ell=0}^{\lfloor n/2\rfloor} (-1)^{\ell} \binom{n}{2\ell} (1-\cos^2(x))^{\ell} \cos^{n-2\ell}(x)=\sum_{\ell=0}^{\lfloor n/2\rfloor} \binom{n}{2\ell} (\cos^2(x)-1)^{\ell} \cos^{n-2\ell}(x).$$

This agrees with Didier's answer.