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

Question

Let $x,y\in [-\pi,0]$. How to show that $$ x\mapsto \frac{\sin(x)-\sin(y)}{x-y} $$ is increasing? After differentiating it, I get another problem to prove: $$ \frac{\sin(x)-\sin(y)}{x-y}\leq \cos(x), $$ in which I do not know how to approach it. I think I should do something like that: $$ \frac{\sin(x)-\sin(y)}{x-y}=\frac{1}{x-y}\int_{y}^{x}\cos(m)\,\mathrm{d}m\leq \frac{\cos(x)}{x-y}\int_{y}^{x}\,\mathrm{d}m=\cos(x), $$ where $\leq$ follows from the fact that $\cos$ is increasing on $[-\pi,0]$. Is this correct?

Answer 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 $x$.

Answer 2

For each $y\neq x$, obviously $f$ is continuous on the interval $[-\pi,0]$. We will show that $f$ is increasing at any $\epsilon$-neighbourhood of $x$ when $x\in(-\pi,0)$. Let $\epsilon=|x-y|$. For every $x\in(-\pi,0)$ we choose $\epsilon$ to be sufficiently close to $0$, i.e., $y$ arbitary close to $x$, i.e., $y\rightarrow x$. Hence, at every $\epsilon$-neighbourhood of $x$, $f(x)=\lim_{y\rightarrow x}\frac{\sin(x)-\sin(y)}{x-y}=\cos(x)$ where $\cos(x)$ is increasing in $(-\pi,0)$ and because $f$ is continuous on the interval $[-\pi,0]$ it is increasing on $[-\pi,0]$.
Not really sure about this solution, please check this.

Answer 3

Another one solution that needs to be checked, please leave a comment.
Let $x_1<x_2$ and $x_1<y<x_2$. Also let $g(x)=\sin(x)$ and apply the mean value theorem for $g$ on the intervals $[x_1,y]$ and $[y,x_2]$. Then there exists $\xi_1\in(x_1,y)$ and $\xi_2\in(y,x_2)$ with $\xi_1<\xi_2$ such that $g'(\xi_1)=\frac{\sin(x_1)-\sin(y)}{x_1-y}$ and $g'(\xi_2)=\frac{\sin(x_2)-\sin(y)}{x_2-y}$ with $g'(\xi_1)<g'(\xi_2)$ because $\cos(x)$ is an increasing function on the interval $[-\pi,0]$.