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Show that $$\left | \sin a - \sin b \right | \leq \left | a-b \right |$$

I saw a proof using The Mean Value Theorem, but I could not grasp it well. Are there any other proofs or can someone clarify it by using The Mean Value Theorem?

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    So by MVT, $\Bigl|{\sin a-\sin b\over a-b}\Bigr|=|\cos c|$ for some number $c$, right? – 2017-01-14
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    yes that was the way used – 2017-01-14
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    And $|\cos c|\le 1$, right? – 2017-01-14
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    that is right.. – 2017-01-14
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    Your desired inequality is one step away... – 2017-01-14
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    and what is that step? – 2017-01-14
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    Multiply both sides of $\Bigl|{\sin a-\sin b\over a-b}\Bigr|\le1$ by $|a-b|$. – 2017-01-14
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    Oh okay I see thanks – 2017-01-14
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    You're welcome. – 2017-01-14

2 Answers 2

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Using the Mean Value Theorem:

$$\frac{\sin{a}-\sin{b}}{a-b}=\cos{c}$$

$$\left|\frac{\sin{a}-\sin{b}}{a-b}\right|=|\cos{c}|$$

Note that $|\cos{c}|\leq 1$

Therefore:

$$\left|\frac{\sin{a}-\sin{b}}{a-b}\right|\leq 1$$

$$|\sin{a}-\sin{b}|\leq |a-b|$$

Which is what was to be obtained.

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    Thank you. That was explanatory enough to understand – 2017-01-14
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Without using the MVT:

$$\left|\sin a -\sin b\right|=\left|2\cos\frac{a+b}2\sin \frac{a-b}2\right|$$ $$\leq 2\left|\sin \frac{a-b}2\right|\leq 2\left|\frac{a-b}2\right|=\left|a-b\right|.$$

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    Seems pretty good. Thanks! – 2017-01-14