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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?

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    $\begingroup$ 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.) $\endgroup$
    – nejimban
    Dec 10, 2021 at 16:41
  • $\begingroup$ I’m pretty sure you can answer the original question without using derivatives, although I’m not sure this is what you want. $\endgroup$ Dec 10, 2021 at 16:48
  • $\begingroup$ 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. $\endgroup$ Dec 10, 2021 at 17:56
  • $\begingroup$ @AdamRubinson If I may ask, how would you answer the problem, without involving convexity? $\endgroup$
    – UnknownW
    Dec 10, 2021 at 18:23
  • $\begingroup$ Yeah actually you're right. It gets messy if we try to prove the original function is increasing from the definition of increasing alone. $\endgroup$ Dec 10, 2021 at 21:48

3 Answers 3

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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 $x$.

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  • $\begingroup$ 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... $\endgroup$ Dec 11, 2021 at 13:34
  • $\begingroup$ 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. $\endgroup$
    – Zarrax
    Dec 11, 2021 at 14:35
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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 $[-\pi,0]$.

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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.

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