functionstrigonometry
I came up with this function:
$$2\left(\frac{1}{1+e^{\textstyle\frac{-6\sin^{-1}(\cos(x))}{\pi/2}}}-\frac12\right)$$
to mimic a 'cosine'-esque function with flat peaks and valleys. Here it is as plotted by Wolfram Alpha:
What I was wondering is, is there a more elegant way to achieve this effect? (The values the function outputs need not be the same as those of this function – it only needs to look cosine-esque and have flat peaks and valleys).
You can also use $y = \frac{1}{1+ax^n}$ to give yourself another parameter to play with. It will still go from 0 to 1 as x goes from -infinity to infinity, but you can change how quickly it gets close.
Start with a "triangle" function of period $2 \pi/n$, minimum value $0$ and maximum value $\pi/n$: a suitable choice is $T_n(t) = \frac{1}{n} \arccos(\cos(n t))$. A regular $n$-gon of inradius $r_0$ is given, in polar coordinates, by $r = r_0/\cos(T_n(\theta - \theta_0))$. Its $y$ coordinate is then $$y = \dfrac{r_0 \sin(\theta)}{\cos(T_n(\theta - \theta_0))}$$
Here is a similar animation using a pentagon ($n=5$):
Best Answer
How about
$$\sqrt{\frac{1+b^2}{1+b^2 \cos^2 v}}\cos\,v$$
where $b$ is an adjustable parameter?