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I have to use the mean value theorem to prove this. I'll include my attempt below, for which I used thepost at the very bottom of this question as a guide. If my answer is correct could someone give me a hint about why we needed $f'(c)$ to prove this? Thank you.

${0 \le\lvert cos(a) - cos(b) \rvert \le 1}$

and $\lvert a - b \rvert \ge 0$ $$f(x) = cos(x)$$ and $$ f'(x) = \lvert-sin(x)\rvert$$ $0\le\lvert-sin(x)\rvert \le 1$

Thus, $f'(c) \le 1$

By the MVT, $${ {\lvert cos(b) - cos(a) \rvert} = f'(c){(b - a)}} $$ and $${\lvert cos(a) - cos(b) \rvert \le \lvert a - b \rvert} $$ for any real numbers $a,b$

Prove that for any real numbers $a,b$ we have $\lvert \arctan a−\arctan b\rvert\leq \lvert a−b\rvert$.

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    The mean value theorem gives an $f'(c)$ to the computation, all we do is bound it. Because you are using the MVT, you have to work with $f'(c)$. Of course, your answer is fine. – 2017-01-26
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    You have some errors in the use of the absolute value signs but your idea is correct. – 2017-01-26

1 Answers 1

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We have $(\cos )'=-\sin .$ By the MVT there exists $c$ in the closed interval whose endpoints are $a,b,$ such that $$\cos a -\cos b=(a-b)(\cos )'(c)$$ implying $\;|\cos a -\cos b|=|a-b|\cdot |(\cos)'(c)|=|a-b|\cdot |-\sin c|\leq |a-b|\cdot 1=|a-b|.$

Much easier than a trigonometric proof without calculus.

Footnote: When $a\ne b$ the MVT says there exists such a $c$ lying strictly between $a$ and $b$ but we did not need that.

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    this was great, I didn't really know what i was doing, but i get it now. thank you :) – 2017-01-26