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I have no idea about it.Your help will be appreciated!

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  • $\begingroup$ Do you know that $| \sin x | \leq | x |, \: \forall x \in \mathbf{R}$? $\endgroup$ – user305860 Jun 7 '16 at 15:14
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    $\begingroup$ You can use the mean value inequality. $\endgroup$ – BrL Jun 7 '16 at 15:17
  • $\begingroup$ @ErikJoensson Yes $\endgroup$ – Man Big Jun 7 '16 at 15:20
  • $\begingroup$ @BrL Can you show me specifically? $\endgroup$ – Man Big Jun 7 '16 at 15:25
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I am not particularly fond of this proof, but it does work. I will just give instructions.

Prove that $$\sin \phi - \sin \psi = 2 \cos \frac{\phi + \psi}{2} \sin \frac{\phi - \psi}{2}$$ by using $\sin (\phi + \psi) - \sin (\phi - \psi) = 2 \cos \phi \sin \psi$. Then with $|\sin \phi | \leq |\phi|$ and the formula above, you obtain the result. Note that this is the general formula, where both $\phi$ and $\psi$ are variables.

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  • $\begingroup$ The $+$ in the last $\sin$ should be a $-$. $\endgroup$ – BrL Jun 7 '16 at 15:33
  • $\begingroup$ Fixed it, thanks. $\endgroup$ – user305860 Jun 7 '16 at 15:35
  • $\begingroup$ FYI, the method you give was the common method for establishing this in 1800s texts, but back then students had these kinds of trig. formulas at their fingertips more than today (not counting Wikipedia and internet searches). $\endgroup$ – Dave L. Renfro Jun 7 '16 at 15:36
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@Erik Joensson's proof is original and only uses basic trigonometry. You can also use the mean value theorem : there exists some $t$ such that $\sin(x)-\sin(c) = (x-c)\sin'(t) = (x-c)\cos(t)$. But $\cos$ is bounded by $1$.

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Let us first consider when $x\ge c$, we see that we can drop the absolute value bars on the right hand side of the inequality leaving us with $$\pm(\sin x - \sin c)\le x-c$$ Notice that when $x=c$ they are both $0$ and equal. Taking the directional derivative in the direction of $<1,0>$ (just a regular derivative) of both sides we have $$\frac{d}{dx}\pm(\sin x - \sin c)=\pm\cos x \le 1=\frac{d}{dx}(x-c)$$ thus, since they are both equal at $x=c$, and the value of $x-c$ is increasing at a greater rate than $\sin x - \sin c$, we have shown that $$|\sin x-\sin c|\le|x-c|\qquad\text{for}\qquad x\ge c$$ Now when $x\le c$, we can drop the absolute value bars, but we must multiply both sides by $-1$ giving $$\sin c -\sin x\le c-x$$ Again at $x=c$ they are both $0$ and equal. This time taking the directional derivative in the direction of $<-1,0>$ of both sides we have $$\frac{d}{dx}\pm(\sin c -\sin x)\cdot<-1,0>=\pm\cos x\le 1=\frac{d}{dx}(c-x)\cdot<-1,0>$$ Thus we have shown the equality also holds for $x\le c$ and we are done.

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