Prove Inequality with Mean Value Theorem: |\sin a - \sin b| \leq |a - b|

[endeavor]

Use the Mean Value Theorem to prove the inequality
$$|\sin a - \sin b| \leq |a - b|$$ for all a and b.

I know by the Mean Value Theorem, I can say:
$$\sin a - \sin b = \cos c(a - b)$$

I've been trying to figure it out for awhile, but could not, so I peeked at my solution's manual. They assumed b < a, and said
$$|\sin a - \sin b| \leq |\cos c||b - a| \leq |a - b|$$

how did they arrive at this?

 

[Galileo]

What can you say about the bounds of $|\cos c|$?

 

[HallsofIvy]

Are you aware that $-1\le cos(x)\le 1$?

 

[endeavor]

Yes, but wouldn't that make it:
$$|\sin a - \sin b| \geq |\cos c||b - a|$$ ?
and
$$|\cos c||b - a| \leq |a - b|$$

And doesn't
$$|b - a| = |a - b|$$ ?

 

[devious_]

I think you're getting mixed up. Maybe it will help if you write your equation as:

$$|\sin a - \sin b| = |\cos c||a - b|$$

 

[Benny]

If two numbers are equal then their absolute values are certainly also equal. So

$$\sin \left( a \right) - \sin \left( b \right) = \cos \left( c \right)\left( {a - b} \right)$$

$$\Rightarrow \left| {\sin \left( a \right) - \sin \left( b \right)} \right| = \left| {\cos \left( c \right)} \right|\left| {a - b} \right|$$ (1)

I would say that assuming a < b or vice versa is just to allow you to use the MVT. It makes no difference whether a or b is greater since you are dealing with absolute values. After all, |a| = |-a|.

Anyway once you get to equation (1), all you need has been given to you. If you combine HallsofIvy and Galileo's hints then you should be able to draw the required conclusion.

If you still can't see what they are suggesting then consider the following

2 = 2.
2 < 2(2) = 4
2 < 3(2) = 6.