# Prove that $|\cos(\sin(x_1)) - \cos(\sin(x_2))| \leq |x_1 - x_2|, \forall x_1, x_2 \in \mathbb R$.

I asked this question without any limitation on methods that might be used. I believe it's turned out to be interesting to see a variety of different approaches. It turns out that the aim of the exercise that my students were given was to provide an example of the application of the Mean Value Theorem. I really like the MVT solution as it's a nice little application showing how powerful elementary calculus can be. None the less, I've enjoyed following other lines of thought.

I'm ticking Christian Blatter's solution as it's a particularly nice application of precalculus knowledge. It would be a good challenge question for brighter maths students in the final year of high school after introducing the various trigonometric formulae. It's a neat application of converting a difference to a product.

Finally, it's a good example of the contrast between pre and post calculus. Thanks to everyone who has contributed to this page.

• One has to show that $\cos(1)>0.5$. – Michael Hoppe Oct 8 '13 at 7:57
• @MichaelHoppe Would you mind expanding on your idea as a comment or even as an answer if you think it's an alternative to the three answers below. – Geoff Pointer Oct 9 '13 at 4:20
• I think this post has reached a reasonable conclusion now. The only reason I didn't finish this ages ago is because, as I had indicated, I didn't want any further solutions until today because of a conflict of interest with an assignment some of my students were doing. I've edited the main question to explain my choice of favourite answer. – Geoff Pointer Oct 21 '13 at 7:47

One has $$|\sin x|\leq|x|\qquad(x\in{\mathbb R})$$ and consequently $$|\cos\alpha-\cos\beta|=2\left|\sin{\alpha+\beta\over2}\right|\ \left|\sin{\alpha-\beta\over2}\right|\leq |\alpha-\beta|\ .$$ Similarly one proves $|\sin\alpha-\sin\beta|\leq|\alpha-\beta|$. Putting it together we obtain $$|\cos(\sin x_1)-\cos(\sin x_2)|\leq|\sin x_1-\sin x_2|\leq|x_1-x_2|\ .$$

It suffices to prove that $|\sin b - \sin a| \le |b - a|$, because then we will have: \begin{align*} |\cos \sin x_1 - \cos \sin x_2| &= |\sin (90 - \sin x_1) - \sin(90 - \sin x_2)| \\ &\le |(90 - \sin x_1) - (90 - \sin x_2)| \\ &= |\sin x_1 - \sin x_2| \\ &\le |x_1 - x_2| \end{align*}

To prove $|\sin b - \sin a| \le |b - a|$ (i.e. $\sin$ is Lipschitz with constant 1), either refer to Christian Blatter's explanation, or else assume WLOG $a \le b$ and use calculus:

$$|\sin b - \sin a| = \left| \int_a^b \cos x \; dx \right| \le \int_a^b |\cos x| \; dx \le \int_a^b 1 \; dx = |b - a|$$

• The mean value theorem suffices (we do not need its integral version): $|\cos x-\cos y|=|-\sin t||x-y|\le|x-y|$ for some $t$ in between $x$ and $y$. Hence, $|\cos\sin x_1-\cos\sin x_2|\le|\sin x_1-\sin x_2|=|\cos w||x_1-x_2|\le|x_1-x_2|$, for some $w$ in between $x_1$ and $x_2$. – Andrés E. Caicedo Oct 12 '13 at 3:28

Okay, the deadline has passed and the required solution for my students uses the Mean Value Theorem. I'll present that here, but there's more to it and I've edited the question as well to explain.

Calculus Solution Specifically Using the Mean Value Theorem

Let $f(x) = \cos(\sin(x))$ then $f'(x) = -\sin(\sin(x))\cos(x)$ which is defined for all $x \in \mathbb R$. So it is certainly continuous on any closed interval and differentiable on any open interval, so it satisfies the criteria for the Mean Value Theorem on $[x_1, x_2]$. So, there exists a $c \in [x_1, x_2]$ with

$$\frac{\cos(\sin(x_2)) - \cos(\sin(x_1))}{x_2 - x_1} = f'(c).$$

But, we can also see that $\left|f'(c)\right| \leq 1$ as $\left|\cos(c)\right| \leq 1, \forall\, c \in \mathbb R$ and $\left|\sin(\sin (c))\right| \leq 1, \forall\, c \in \mathbb R$, so together these imply

\begin{align*} &\left|\frac{\cos(\sin(x_2)) - \cos(\sin(x_1))}{x_2 - x_1}\right| = \left|f'(c)\right| \leq 1 \\ &\Rightarrow \left|\cos(\sin(x_2)) - \cos(\sin(x_1))\right| \leq \left|x_2 - x_1\right|\quad\blacksquare \end{align*}

• This seems to reproduce an already posted answer. – Did Oct 21 '13 at 7:31
• @Did Sorry, I didn't notice that as it's only a comment, not an actual answer. When I first posted this I wasn't looking for a calculus answer. I've presented the MVT solution as a contrast to Christian's pre calculus answer which I believe rounds this post off nicely. – Geoff Pointer Oct 21 '13 at 7:49
• Robert's answer looks like an actual answer to me. – Did Oct 21 '13 at 8:02
• @Did Robert doesn't mention the MVT. My second semester first year students have been taught the MVT in lectures and have been doing various related exercises for weeks. This was an assessed assignment question and the solution I've presented here is precisely how they're expected to answer it. (1) Verify that the function satisfies the prerequisites of the MVT. (2) Use it's promised result to come up with the inequality. Robert's solution is valid and uses elements of the MVT but not what my students were asked to demonstrate. I'm not competing with Robert – Geoff Pointer Oct 21 '13 at 9:39
• @Did Byzantine? Tell that to my head of first year. These first years have been introduced to the MVT for the first time. They've been set exercises which expect them to show that the prerequisites of the theorem either apply or don't as the case may be. The MVT proof is very succinct. I've mentioned the contrast between pre calculus and post calculus proofs for pedagogical interest. I've rewritten the OP to clarify this, maybe my previous attempt to edit it wasn't clear enough. I'm interested in all the answers that have been post here. It's not a competition. – Geoff Pointer Oct 21 '13 at 10:41

Set

$y(x) = \cos (\sin x); \tag{1}$

then

$y'(x) = -(\sin (\sin x)) \cos x. \tag{2}$

Note that, since $\vert \sin x \vert \le 1$, we have

$\vert \sin (\sin x) \vert <1; \tag{3}$

in fact, there exists a positive real $k$ with $\sin 1 < k < 1$ such that

$\vert \sin (\sin x) \vert < k; \tag{3}$

(3) follows from the facts that $1 < \frac{\pi}{2}$ and $\sin$ is monotonically increasing on $[0, \frac{\pi}{2}]$. From (3) we have, since $\cos x \le 1$ for all $x$,

$\vert y'(x) \vert = \vert (\sin (\sin x)) \cos x \vert = \vert (\sin (\sin x)) \vert \vert \cos x \vert < k \le 1, \tag{4}$

and now we simply integrate:

$\vert y(x_2) - y(x_1) \vert = \vert \int_{x_1}^{x_2} y'(s) ds \vert \le \int_{x_1}^{x_2} \vert y'(s) \vert ds \le k \vert x_2 - x_1 \vert < \vert x_2 -x_1 \vert, \tag{5}$

provided $x_1 \ne x_2$, so substituting (1) into (5) yields

$\vert \cos (\sin x_2) - \cos (\sin x_1) \vert < \vert x_2 - x_1 \vert; \tag{6}$

the inequality is apparently strict unless $x_2 = x_1$, in which case the two sides of (6) take the same value $0$. This proves the required inequality in all cases. QED.

Wow! Easier than I first thought it would be!

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

• I hope that other comments posted on this page haven't made you think I haven't valued your contribution. The up vote that your answer has, came from me in the first place. Cheers – Geoff Pointer Oct 22 '13 at 19:37
• @Geoff Pointer: not at all; glad to contribute. I think it an interesting problem and was impressed with the other answers. Thanks for the +1, and a "Cheerio" backatcha! – Robert Lewis Oct 22 '13 at 19:49