How do you explain the sine function of a basic triangle wave? I am working on an investigation focusing on mathematics in music. Modelling various different chords and their mathematical functions, I now need to understand (in relative detail) how the most basic triangle wave can be explained using maths.
Hence, can someone explain how exactly this formula works:
(2/pi) * arcsin[sin(Pi*x)]
I'm trying to understand how exactly the arcsin(sin(...)) part links to producing an actual triangle. (especially because sin(arcsin) cancels out).
Thank you very much
 A: If you're familiar with the definition of $\sin \theta$ using a unit circle,
the following visualization might be useful.
(And if you're not familiar with that definition, perhaps it would be useful
to become familiar with it.)
Let $\pi x$ be the angle of a radius of the unit circle,
measured counterclockwise from the positive horizontal axis.
(I do not say $x$-axis, because we are already using $x$ for something else.)
Then $\sin(\pi x)$ is the vertical coordinate of the point where the
ray intersects the circle.
As $x$ increases, the ray sweeps around the circle, so the vertical coordinate
of the intersecting point increases to $1$, then decreases to $-1$, then
increases to $0$ again as the ray returns to its starting position.
Then as $x$ continues to increase, the whole cycle repeats indefinitely.
Meanwhile, the $\arcsin$ function attempts to deduce the angle that could
have produced a given vertical coordinate on the circle.
But since there are many angles that reach the same point, and since
for most points (other than $(0,1)$ and $(0,-1)$) there are other points
with the same vertical coordinate, there are a lot of possible angles
to choose from. This is resolved by choosing only angles in the
range from $-\frac\pi2$ to $\frac\pi2$ inclusive.
Hence when $x = 0$, $\sin(\pi x) = 0$ and $\arcsin(\sin(\pi x)) = 0$
(since $0$ is the only angle in the range $-\frac\pi2$ to $\frac\pi2$
to produce a vertical coordinate of $0$).
Then as $x$ increases, as long as $\pi x < \frac\pi2$,
$\sin(\pi x)$ increases as the ray 
at angle $\pi x$ sweeps around the circle, and 
$\arcsin(\sin(\pi x))$ evaluates to $\pi x$ (the only angle in the range 
$-\frac\pi2$ to $\frac\pi2$ that produces the vertical coordinate
of the intersection point).
But when $\pi x$ passes $\frac\pi2$ and the ray starts two sweep through
decreasing vertical coordinates on the left side of the
vertical axis, $\arcsin(\sin(\pi x))$ is "trapped" on the right side of the
axis (it cannot produce angles greater than $\frac\pi2$), and therefore
$\arcsin(\sin(\pi x))$ starts to describe the angle of a mirror-image ray
that travels clockwise from angle $\frac\pi2$ to $-\frac\pi2$
at the same time the ray at angle $\pi x$ travels counterclockwise
from $\frac\pi2$ to $\frac{3\pi}2$.
But at angles $\frac{3\pi}2$ and $-\frac\pi2$, the rays intersect the same
point, and they then proceed to sweep counterclockwise together
back toward the top of the circle.
That is, as we increase $\pi x$ at a constant rate, 
$\arcsin(\sin(\pi x))$ initially increases at the same constant rate,
but it then decreases at the "mirror image" constant rate,
then increases again, and as the cycle repeats, it
decreases, increases, decreases, increases, and so forth.
That's the geometric explanation of how the triangular wave
gets to be triangular.
The factor of $2/\pi$ is a scaling factor to get the wave to have
the desired amplitude (peak value of $1$ rather than $\pi/2$).
A: It is easy to see that the expression is periodic with period $2$ due to $\sin \pi x$.
$\arcsin(\sin(\pi x))$ does not simply cancel itself out as the range of the $\arcsin$ function is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
We analyse the function in pieces.


*

*$0\leq \pi x\leq \frac{\pi}{2}$: $\arcsin(\sin(\pi x))=\pi x$

*$\frac{\pi}{2}\leq \pi x\leq \frac{3\pi}{2}$: $\arcsin(\sin(\pi x))=\frac{\pi}{2}-\pi x$

*$\frac{3\pi}{2}\leq \pi x\leq 2\pi$: $\arcsin(\sin(\pi x))=\pi x-2\pi$


These 3 linear piecewise functions create the triangle shape.
