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

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?

• It looks OK. (That $x\mapsto\frac{\sin(x)-\sin(y)}{x-y}$ is non-decreasing on $[-\pi,0]$ is equivalent to $\sin$ being convex there, which is also equivalent to the derivative $\sin'=\cos$ being non-decreasing.) Dec 10, 2021 at 16:41
• I’m pretty sure you can answer the original question without using derivatives, although I’m not sure this is what you want. Dec 10, 2021 at 16:48
• Are you sure about the domain. For $x$ starting at $-\pi$, the numerator is decreasing and the denominator is increasing, so the fraction is decreasing. Dec 10, 2021 at 17:56
• @AdamRubinson If I may ask, how would you answer the problem, without involving convexity? Dec 10, 2021 at 18:23
• Yeah actually you're right. It gets messy if we try to prove the original function is increasing from the definition of increasing alone. Dec 10, 2021 at 21:48

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 $$x$$.
• Please can you explain how you determined that $$\frac{\sin(x)-\sin(y)}{x-y} = \int_0^1 \cos(y + t(x - y))\,dt$$ is true? I don't get it at all... Dec 11, 2021 at 13:34
• It's like integrating $\cos(3 + 2t)$ and getting $(1/2) \sin(3 + 2t)$. Except instead of $3$ and $2$ you have $y$ and $x - y$. So the indefinite integral is $g(t) = {1 \over x - y} \sin(y + t(x - y))$. So the definite integral is $g(1) - g(0)$ which equals the left-hand side. Dec 11, 2021 at 14:35
Let $$x_1 and $$x_1. 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 $$g'(\xi_1) because $$\cos(x)$$ is an increasing function on the interval $$[-\pi,0]$$.
For each $$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]$$.