Proving triangle inequality of $\sin|x-y|$ on $[0,\pi/2).$

Question

I've been struggling with this inequality for a while. I'm doing this to show that $d(x,y) = \sin|x-y| $ is a metric on $[0,\pi/2).$

I came across this solution: https://math.stackexchange.com/a/2315556/616064, however I can't understand from where this came from: $$|\sin(x-y)|=|\sin(x-z)\cos(z-y)+\sin(z-y)\cos(x-z)|$$

I'd really appreciate some clarification on this.

Answer 1

You can check it out on Wikipedia: https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities (Angle-sum and difference identities)

Here, of course, $x-y=(x-z)+(z-y)$.

Answer 2

we used that $z-z = 0$ and $\sin(a+b)=\sin a\cos b+\sin b\cos a$ $$\sin(x-y)=\sin(x-z+z-y)\\ =\sin((x-z)+(z-y))\\ =\sin(x-z)\cos(z-y)+\sin(z-y)\cos(x-z)$$

Answer 3

$x-y = (x-z)+(z-y)$ $$\therefore \sin(x-y) = \sin(x - z)\cos(z-y) + \sin(z-y)\cos(x-z)$$ $$\therefore |\sin(x-y)| = |\sin(x - z)\cos(z-y) + \sin(z-y)\cos(x-z)|$$