The purpose is to show $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$ for any $x,y\in\Bbb{R}$.

Taking partial derivatives with respect to $x,y$ respectively $\frac{{\partial \frac{{\cos x - \cos y}}{{x - y}}}}{{\partial x}} = 0$, $\frac{{\partial \frac{{\cos x - \cos y}}{{x - y}}}}{{\partial y}} = 0$ gives $\sin x={ - \frac{{\cos x - \cos y}}{{x - y}}}$ and $\sin y={ - \frac{{\cos x - \cos y}}{{x - y}}}$. Plugging them back and we have the desired result $\left| {\frac{{\cos x - \cos y}}{{x - y}}} \right| \le 1$.

What I want to ask is, is there more decent way to show the inequity? Taking derivatives looks clumsy. Hope someone can help. Thank you!

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    $\begingroup$ The one variable mean value theorem solves this nicely. $\endgroup$ – zhw. Jul 10 '15 at 18:04

The mean value theorem can resolve this quickly, $$\cos(x)-\cos(y) = (-\sin(\eta))(x-y)$$ for some $\eta \in (x,y)$. Thus $$\left| \frac{\cos(x)-\cos(y)}{x-y} \right| = |\sin(\eta)| \le 1$$

  • $\begingroup$ Note I suggested this as a comment before your answer. $\endgroup$ – zhw. Jul 10 '15 at 18:16
  • $\begingroup$ Yes that is true. @zhw It is the first thing one should try. $\endgroup$ – Joel Jul 10 '15 at 18:22

Using the difference to product identity $\cos x - \cos y = -2\sin\dfrac{x-y}{2}\sin\dfrac{x+y}{2}$ along with the inequality $|\sin \theta| \le |\theta|$ gives:

$\left|\dfrac{\cos x - \cos y}{x-y}\right| = \left|\dfrac{-2\sin\frac{x-y}{2}\sin\frac{x+y}{2}}{x-y}\right| = \dfrac{2\left|\sin\frac{x-y}{2}\right|\left|\sin\frac{x+y}{2}\right|}{|x-y|} \le \dfrac{2\left|\frac{x-y}{2}\right| \cdot 1}{|x-y|} = 1$.

Note that both $\cos x - \cos y = -2\sin\dfrac{x-y}{2}\sin\dfrac{x+y}{2}$ and $|\sin \theta| \le |\theta|$ can be proven without calculus.

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    $\begingroup$ Your proof is nicer than mine. $\endgroup$ – marty cohen Jul 10 '15 at 19:21

This might not qualify as a proof, but you might appreciate this geometric reasoning:

Let $x, y \in \mathbb R$ be any two angles. The corresponding points on the unit circle are $(\cos x, \sin x)$ and $(\cos y, \sin y)$. Now observe that $|x-y|$ is the arc length from $x$ to $y$ along the circle (possibly wrapping around multiple times if $|x-y| > 2\pi$), while $|\cos x - \cos y|$ is just the horizontal distance between $x$ and $y$, which is clearly no larger than the straight-line distance, which is less than the arc length.


Hint: Apply $$|\cos (x)-\cos (y)|=|-2\sin \frac{x+y}{2}\sin \frac{x-y}{2}|$$

$$=|-2\sin \frac{x+y}{2}||\sin \frac{x-y}{2}|\le|x-y|$$


The only "analytic" inequality needed, aside from basic trigonometry, is $|\sin(x)| \le |x| $.

Let $x = y+h$, so

$\begin{array}\\ \cos(x)-\cos(y) &=\cos(y+h)-\cos(y)\\ &=\cos(y)\cos(h)-\sin(y)\sin(h)-\cos(y)\\ &=\cos(y)(\cos(h)-1)-\sin(y)\sin(h)\\ \end{array} $

We now use $|a\cos(y)+b\sin(y)| \le \sqrt{a^2+b^2} $. To show this, choose $v$ so that $\tan(v) = a/b$. Then

$\begin{array}\\ a\cos(y)+b\sin(y) &=b\tan(v)\cos(y)+b\sin(y)\\ &=b((\sin(v)/\cos(v))\cos(y)+\sin(y))\\ &=(b/\cos(v))(\sin(v)\cos(y)+\cos(v)\sin(y))\\ &=(b/\cos(v))\sin(v+y)\\ \end{array} $

But, since

$\begin{array}\\ \tan^2(v) &=\sin^2(v)/\cos^2(v)\\ &=(1-\cos^2(v))/\cos^2(v)\\ &=1/\cos^2(v)-1,\\ 1/\cos(v) &=\sqrt{1+\tan^2(v)}\\ &=\sqrt{1+a^2/b^2}\\ &=\sqrt{a^2+b^2}/b\\ \text{so}\\ b/\cos(v) &=\sqrt{a^2+b^2}\\ \end{array} $

Therefore $a\cos(y)+b\sin(y) =\sqrt{a^2+b^2}\sin(v+y) $. Since $|\sin(v+y)| \le 1$, $|a\cos(y)+b\sin(y)| \le \sqrt{a^2+b^2} $.

Finally, we get

$\begin{array}\\ |\cos(y+h)-\cos(y)| &=|\cos(y)(\cos(h)-1)-\sin(y)\sin(h)|\\ &\le \sqrt{(\cos(h)-1)^2+\sin^2{h}}\\ &= \sqrt{\cos^2(h)-2\cos(h)+1+\sin^2{h}}\\ &= \sqrt{2-2\cos(h)}\\ &= \sqrt{4\sin^2(h/2)} \quad\text{(since }1-\cos(h) = 2\sin^2(h/2))\\ &=2|\sin(h/2)|\\ &\le |h| \quad\text{(using } |\sin(h/2)| \le |h/2|)\\ \end{array} $

so that, as desired, $\left|\dfrac{\cos(x)-\cos(y)}{x-y}\right| \le 1 $.

Note that, just from basic trig, we have $|\cos(x)-\cos(y)| \le 2|\sin((x-y)/2)| $.


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