Show that $x\mapsto \frac{\sin(x)-\sin(y)}{x-y}$ is increasing

Solution 1:

This is just a fancy way of saying $\sin x$ is convex on $[-\pi,0]$. Note that ${\frac{\sin(x)-\sin(y)}{x-y}}$ is the slope of the chord of the graph of $\sin x$ between $(x,\sin x)$ and $(y,\sin y)$, which will increase in $x$ for a convex function. An explicit way to show this is to note that $$\frac{\sin(x)-\sin(y)}{x-y} = \int_0^1 \cos(y + t(x - y))\,dt$$ Differentiating this under the integral sign in $x$ results in $$-\int_0^1 t\sin(y + t(x - y))\,dt$$ Since $\sin$ is negative inside the interval of integration, the above quantity is positive. Thus the quotient $\frac{\sin(x)-\sin(y)}{x-y} $ is increasing in $x$.