TITLE: Picture/intuitive proof of $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$?
QUESTION [9 upvotes]: Is there a nice geometric, intuitive or picture proof as to why the easily algebraically provable identity $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$ is true?
Note I'm not looking for a computational proof like the one linked to, more a proof without words or intuitive style proof, thanks.

REPLY [2 votes]: This is not a direct proof but a link to the Fourier series: $4\cos^3\theta=3\cos\theta+\cos3\theta$.

Similarly, for the fifth power: $16\cos^5\theta=10\cos\theta+5\cos3\theta+\cos5\theta$.

The main lesson is that to represent the $k^{th}$ power, the odd harmonics from $1$ to $k$ suffice.

The explanation is not so difficult: the Fourier coefficients are computed from the integrals
$$\int_0^{2\pi}\cos^k\theta\ e^{in\theta}d\theta=\int_0^{2\pi}\left(\frac{e^{i\theta}-e^{-i\theta}}2\right)^ke^{in\theta}d\theta.$$
When developing, the lowest exponent of $e^{i\theta}$ is $n-k$. If $n>k$, the integrand is oscillatory and the integral vanishes.

If you admit that, then $\cos^3\theta=a\cos\theta+b\cos3\theta$.
From the figure, at $\theta=0$ the values add up to $1$ and at $\theta=\pi/2$ the slopes cancel out, so that
$$a+b=1\\a-3b=0,$$
$$\cos^3\theta=\frac34\cos\theta+\frac14\cos3\theta.$$
Similarly, $\cos^5\theta=a\cos\theta+b\cos3\theta+c\cos5\theta$, at $\theta=0$ the values add up to $1$ and at $\theta=\pi/2$ the first and third derivatives cancel out, so that
$$a+b+c=1\\
a-3b+5c=0\\
a-27b+125c=0,$$
$$\cos^5\theta=\frac{10}{16}\cos\theta+\frac{5}{16}\cos3\theta+\frac{1}{16}\cos5\theta.$$