Proving MVT: |sinx-siny| ≤ |x-y|

[Mstenbach]

Homework Statement

Prove for all real x and y that
|sinx - siny| <= |x-y|

Homework Equations

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

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!

 

[jgens]

Well, what does the mean value theorem say?

 

[Mstenbach]

Well, what does the mean value theorem say?

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?

 

[jgens]

Could the "real values x and y" have anything to do with the (a, b) interval?

Well, what do you think?

 

[Mstenbach]

Well, what do you think?

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?

 

[l'Hôpital]

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

 

[Mstenbach]

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

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.

 

[l'Hôpital]

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.

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?

 

[Mstenbach]

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?

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.

 

[l'Hôpital]

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.

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