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It's not hard to find multiple trigonometric functions of period $2\pi$ that added to self shifted by some constant offset result in a constant.

In classic pythagorean identity, you have

$F(x)+F\left(x+\frac{\pi}{2} \right) = 1 $

where $F(x)=\sin^2 x$

Or you can use symmetry of the sine wave and create

$F(x)+F(x+\pi ) = 0$

where $F(x) = \sin x$

Now what I'm looking for is a transformation of the sine function that while retaining the $2\pi$ period, gives you identity if repeated three times, with $\frac 2 3 \pi$ shift in each appearance:

$F(x) + F\left(x + \frac 2 3 \pi \right) + F\left(x+\frac 4 3 \pi\right) = \mathrm{const}$

Can you find such a function?

Reason and purpose:

I've been trying to develop a better RGB$<=>$HSV color space conversion - all the common ones use sawtooth style variant functions with variant equations, and I think using trigonometric functions could result in more smooth color passages, never mind much simpler algorithm.

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    @HenningMakholm: Yes, it either won't cover the whole RGB cube or will exceed its borders, but that's a problem with the RGB cube and not with HSV model or my conversion; In RGB the most saturaded C, M and Y are less saturated than most saturated R, G and B; RGB doesn't really have a color wheel; it has a color hexagon, and trying to fit HSV (which is naturally inclined towards color wheel) into RGB's hexagonal nature is "clipping its wings". I choose to have a smooth transition over all hues at <85% saturation than suffer sharp transitions over all saturations.2012-09-02

2 Answers 2

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$ e^{i\theta} + e^{i(\theta+2\pi/3)} + e^{i(\theta+4\pi/3)} = 0. $ (Draw the picture and this is obvious.)

Therefore, the sum of the three real parts is $0$.

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Let $F(x)=\sin(ax+b)$ where a,b are indeterminate constants.

$F(x) + F\left(x + \frac 2 3 \pi \right) + F\left(x+\frac 4 3 \pi\right) $

$=\sin(ax+b)+\sin(a(x+\frac{2\pi}{3})+b)+\sin(a(x+\frac{4\pi}{3})+b)$

$=\sin(ax+b)+\sin(a(x+\frac{4\pi}{3})+b)+\sin(a(x+\frac{2\pi}{3})+b)$

$=(1+2\cos\frac{2\pi a}{3})\sin(a(x+\frac{2\pi}{3})+b)$

This expression can not be constant for all $x$ unless $1+2\cos\frac{2\pi a}{3}=0$

Or, $\cos\frac{2\pi a}{3}=-\frac{1}{2}=\cos\frac{2\pi}{3}$

$\implies \frac{2\pi a}{3}=2m\pi±\frac{2\pi}{3}$ where $m$ is an integer,

$\implies a=3m±1$

$F(x)=\sin(ax+b)$ where $a$ is an integer with $(a,3)=1$.

As $b$ is an indeterminate constant, we can have some other indeterminate constant $c=b-\frac{\pi}{2}$

$F(x)=\sin(ax+\frac{\pi}{2}+c)=\cos(ax+c)$