Linearizing the trigonometric functions or: Squaring the circle by Fourier transformation It's an easy exercise to approximate the cosine and the sine function by a piecewise linear function on the unit interval $[0,1]$. Let $\tau = 2\pi$.
Let
$$\boxed{\cos_\bigcirc(x) = \cos(\tau x)\\\sin_\bigcirc(x) = \sin(\tau x)}$$
and compare this to
$$\boxed{\cos_\square(x) = \begin{cases}
+1 & \text{ for } \frac{0}{8} \leq x \leq \frac{1}{8} \\
+2 - 8x & \text{ for } \frac{1}{8} \leq x \leq \frac{3}{8} \\
-1  & \text{ for } \frac{3}{8} \leq x \leq \frac{5}{8} \\
-6 + 8x & \text{ for } \frac{5}{8} \leq x \leq \frac{7}{8} \\
+1 & \text{ for } \frac{7}{8} \leq x \leq \frac{8}{8} \\
\end{cases}
\\ \\\sin_\square(x) = \begin{cases}
+0 + 8x & \text{ for } \frac{0}{8} \leq x \leq \frac{1}{8} \\
+1 & \text{ for } \frac{1}{8} \leq x \leq \frac{3}{8} \\
+4 - 8x & \text{ for } \frac{3}{8} \leq x \leq \frac{5}{8} \\
-1 & \text{ for } \frac{5}{8} \leq x \leq \frac{7}{8} \\
-8 + 8x & \text{ for } \frac{7}{8} \leq x \leq \frac{8}{8} \\
\end{cases}}$$
These are the plots:

Observations


*

*It may come as a surprise or not that while $\cos_\bigcirc(x)$ and $\sin_\bigcirc(x)$ yield the unit circle by $x_\bigcirc(x) = \cos_\bigcirc(x)$ and $y_\bigcirc(x) = \sin_\bigcirc(x)$, the functions  $\cos_\square(x)$ and $\sin_\square(x)$ yield the unit square by $x_\square(x) = \cos_\square(x)$ and 
$y_\square(x) = \sin_\square(x)$. By "unit square" I mean the square with "radius" $1$, not with side length $1$. The unit circle in the incircle of this square:





*

*While the circumference of the unit circle is just the "number of the circle" $\tau$, the circumference of the unit square is $8$ (the "number of the square"). Note how $8$ is used in the definition of $\cos_\square(x)$ and $\sin_\square(x)$, compared to $\tau$ in the definition of $\cos_\bigcirc(x)$ and $\sin_\bigcirc(x)$. Note further that not by accident $8 \approx \tau$ and that not by accident $8 = 3^2 - 1^2$:





*

*There are natural generalizations of piecewise linear approximations of the cosine and the sine for arbitrary regular $n$-polygons which will approximate the true functions better and better as the polygons will approximate the circle better and better.

*The tangens $\tan_\bigcirc(x) = \tan(\tau x)$ is very well approximated already by 
$\tan_\square(x) = \frac{\sin_\square(x)}{\cos_\square(x)}$:



*

*The functions $\cos_\square(x)$ and $\sin_\square(x)$ can be used to parametrize the square spiral analoguous to how $\cos_\bigcirc(x)$ and $\sin_\bigcirc(x)$ can be used to parametrize the Archimedean spiral.


Questions

  
*
  
*Under which name and in which contexts have the functions
  $\cos_\square(x)$ and $\sin_\square(x)$ been studied before?
  
*Is there an elegant and/or more compact way to write the equations for
  $\cos_\square(x)$ and $\sin_\square(x)$ in one closed expression, e.g.
  by using the Heaviside function?
  
*Is there a closed formula for the Fourier transform of $\cos_\square(x)$ and $\sin_\square(x)$ (the "Fourier transform of the quadrature of the circle")? 
  

This is how the Fourier transforms of $\cos_\square(x)$ and $\sin_\square(x)$ look like:

Summary
Thanks to user J.M. we now know the Fourier coefficients $\widehat{\cos}_\square(k)$
$$\boxed{\widehat{\cos}_\square(k) = 8 \cdot \begin{cases}
\ \ \ \ \ 0 & \text{ for } k \equiv 0 \mod 2 \\
+(\pi k)^{-2} & \text{ for } k \equiv 1 \mod 8 \text{ or } k \equiv 7 \mod 8\\
-(\pi k)^{-2} & \text{ for } k \equiv 3 \mod 8 \text{ or } k \equiv 5 \mod 8\\
\end{cases}}$$
and $\widehat{\sin}_\square(k)$ accordingly. 
Now can perform the squaring of the circle by these steps:


*

*Consider $\cos_\bigcirc(x)$ and $\sin_\bigcirc(x)$ which "draw" the unit circle (with diameter $2$).

*Consider the functions $\cos_\square(x)$ and $\sin_\square(x)$ defined by
$$\cos_\square(x) := \sum_{k=0}^{\infty} \widehat{\cos}_\square(k)\cos_\bigcirc(kx)\\
\sin_\square(x) := \sum_{k=0}^{\infty} \widehat{\sin}_\square(k)\sin_\bigcirc(kx)$$


*The functions $\cos_\square(x)$ and $\sin_\square(x)$ "draw" the unit square (with diameter $2$).

 A: For #2, we can do this via floor, absolute value, min, and max.
First, we need a triangle wave.  $\left|x - \left\lfloor x \right\rfloor - 1/2\right|$ gives a triangle wave of period $1$ and range $[0,1/2]$.  We need range $[-2,2]$ because that's where the diagonals will meet, so multiply by $8$ and subtract $2$:  $8\left|x - \left\lfloor x \right\rfloor - 1/2\right|-2$.  Then we cut off the top and bottom with min and max:
$$\cos_\square(x) = \min\left(\max\left(-1,8\left|x - \left\lfloor x \right\rfloor - 1/2\right|-2\right),1\right)$$
THen since $\sin_\square(x)=\cos_\square\left(x-\frac{1}{4}\right)$ we can just ... do that and get 
$$\sin_\square(x) = \min\left(\max\left(-1,8\left|x - \left\lfloor x-1/4 \right\rfloor - 3/4\right|-2\right),1\right)$$
A: The work has been done by user J.M. (who seems to be a mathematician of sorts), so let me just put it in order.
First let me rewrite the definition of $\cos_\square(x)$ a bit:
$$\cos_\square(x) = \begin{cases}
+1 & \text{ for } -1 \leq 8x \leq 1 \\
+2 - 8x & \text{ for }\ \ \ \ \ 1 \leq 8x \leq 3 \\
-1  & \text{ for }\ \ \ \ \ 3 \leq 8x \leq 5 \\
-6 + 8x & \text{ for }\ \ \ \ \ 5 \leq 8x \leq7 \\
\end{cases}$$
The unit square has diagonal (= "diameter") $2\sqrt{2}$, so let's normalize $\cos_\square(x)$ by a factor $\frac{1}{\sqrt{2}}$ so the unit square has diameter $2$ (just like the unit circle has diameter $2$):
$$\boxed{\cos_\square(x) = \frac{1}{\sqrt{2}}\cdot\begin{cases}
+1 & \text{ for } -1 \leq 8x \leq 1 \\
+2 - 8x & \text{ for }\ \ \ \ \ 1 \leq 8x \leq 3 \\
-1  & \text{ for }\ \ \ \ \ 3 \leq 8x \leq 5 \\
-6 + 8x & \text{ for }\ \ \ \ \ 5 \leq 8x \leq7 \\
\end{cases}}$$
User J.M.'s expression for the Fourier coefficients $\widehat{\cos}_\square(k)$ can then be written in this form:
$$\boxed{\widehat{\cos}_\square(k) = 8 \cdot \begin{cases}
\ \ \ \ \ 0 & \text{ for } k \equiv 0 \mod 2 \\
+(\pi k)^{-2} & \text{ for } k \equiv 1 \mod 8 \text{ or } k \equiv 7 \mod 8\\
-(\pi k)^{-2} & \text{ for } k \equiv 3 \mod 8 \text{ or } k \equiv 5 \mod 8\\
\end{cases}}$$
A: To those who don't have the function $f(x) = \arcsin(\sin(2\pi x))$ before their inner eye, this is how it looks like:

It's essentially this function that user J.M. cuts off at $\min = -1$ and $\max = +1$ to yield $\sin_\square(x)$.
