# Proof Involving MVT?

1. Jan 6, 2010

### Mstenbach

1. The problem statement, all variables and given/known data
Prove for all real x and y that
|sinx - siny| <= |x-y|

2. Relevant equations
It's a question from the Mean Value Theorem/Rolle's Theorem section.

3. The attempt at a solution
Honestly, I've tried. It looks somewhat similar to the triangle inequality (I think?), but truth be told I can't get anywhere with this.

I'd appreciate if anyone could give me a hand and point me in the proper direction. Thanks!

2. Jan 6, 2010

### jgens

Well, what does the mean value theorem say?

3. Jan 7, 2010

### Mstenbach

Reading the hotlink I still fail to see the connection with MVT to the problem. Could the "real values x and y" have anything to do with the (a, b) interval?

4. Jan 7, 2010

### jgens

Well, what do you think?

5. Jan 7, 2010

### Mstenbach

Well, my guess is it does but I still fail to see any connection.

$| \sin \displaystyle x- \sin \displaystyle y| = 2 \left| \cos \left( \frac{\displaystyle x+ \displaystyle y}{2} \right) \sin \left( \frac{\displaystyle x- \displaystyle y}{2} \right) \right|$

Am I getting anywhere with this?

6. Jan 7, 2010

### l'Hôpital

How about you let $$x = b$$ and $$y = a$$?

7. Jan 7, 2010

### Mstenbach

Hmm, ok, thank you.

So I get now

|sinb - sina|
__________ <= 1
| b - a|

Which is similar to the theorem f(b)-f(a) / b-a ?

If it is, this basically says that the equation I came up with (which is the derivative) is equal to 1?

Can I work on this further by taking the derivative and getting cosb - cosa?

Sorry if I'm coming off as a bit.. stubborn. This question is really confusing me.

8. Jan 7, 2010

### l'Hôpital

You're close.
It's not exactly the derivative. I mean, it is the derivative at point c between a and b.

So, we have the equation in the form of
$$\frac{\sin b - \sin a}{b - a} = \cos c$$
As for your derivative idea, no. Derivative of sin a would be 0 because sin a is a constant. Instead, consider the following:
What do we know about $$\cos$$'s range?

9. Jan 7, 2010

### Mstenbach

Aha! I see it now. The range of cos is between 1 and -1, so it cannot exceed 1... I think I see that now.

Just one question, how did you find "cosc"?

Thank you very much.

10. Jan 7, 2010

### l'Hôpital

Once more, look up "Mean Value Theorem". It's all in there.

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