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
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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 $$ T_{12}(x)=2048x^{12} − 6144x^{10} + 6912x^8 − 3584x^6 + 840x^4 − 72x^2 + 1. $$ |
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You can always repeatedly use $$\begin{align*} \cos(a\pm b) &= \cos a\cos b \mp \sin a\sin b\\ \sin(a\pm b) &= \sin a\cos b \pm \cos a\sin b\\ \sin^2(r) &= 1-\cos^2(r). \end{align*}$$ For example, $$\begin{align*} \cos(4x) &= \cos(2x+2x)\\ &= \cos(2x)^2 - \sin^2(2x)\\ &= \cos(2x)^2 - (1-\cos^2(2x))\\ &= 2\cos(2x)^2 - 1\\ &= 2(\cos(x+x))^2 - 1\\ &= 2(\cos x\cos x - \sin x\sin x)^2 - 1\\ &= 2(\cos^2 x - \sin^2 x)^2 - 1\\ &= 2(\cos^2x - (1-\cos^2 x))^2 - 1\\ &= 2(2\cos^2x - 1)^2 - 1\\ &= 2(4\cos^4 x - 4\cos^2 x + 1) - 1\\ &= 8\cos^4 x - 8\cos^2 x + 1. \end{align*}$$ |
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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. |
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According to the Power-reduction formula, one can interchange between $\cos(x)^n$ and $\cos(nx)$ like the following: $$ \cos^n\theta = \frac{2}{2^n} \sum_{k=0}^{\frac{n-1}{2}} \binom{n}{k} \cos{((n-2k)\theta)} \tag{odd}\\ $$ $$ \cos^n\theta = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} \binom{n}{k} \cos{((n-2k)\theta)}. \tag{even} $$ Now combine $\sum_m c_m \cos^m\theta$, such that only $\cos(nx)$ is left. |
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There were two corrections in the posting ($1102$ should be $1120$ and $\cos x$ term should be $\cos^2 x$ term: $$\cos 10x = 512 (\cos x)^{10} -1280 (\cos x)^8 +1120 (\cos x)^6 -400(\cos x)^4+50(\cos x)^2-1$$ Follow Arturo's answer you should get the following $$ \begin{align*} \cos 2x &= 2 \cos^2 x -1 \\ \cos 3x &= 4 \cos^3 x - 3 \cos x\\ \cos 4x &= 8 \cos^4 x - 8 \cos^2 x +1 \\ \cos 5x &= 16 \cos^5 x - 20 \cos^3 x + 5 \cos x\\ \cos 6x &= 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x -1\\ \cos 8x &= 128 \cos^8 x - 256 \cos^6 x + 160 \cos^4 x -32 \cos^2 x +1\\ \cos 10x &= 512 \cos^{10} x -1280 \cos^8 x +1120 \cos^6 x -400 \cos^4 x +50 \cos^2 x-1\\ \end{align*} $$ (Corrected $-1120 $ to $1120$ ) |
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