(MVT) f(x)=sinx. Show that, for any given a and b, |sina-sinb|<=|b-a|

[NWeid1]

I'm so bad with the Mean Value Theorem. Can someone help me prove that, if f(x)=sinx, that, for any given a and b, |sinb-sina|<=|b-a|. Explain if you could too, please. Thanks a lot.

 

[gb7nash]

Start with what you know. What does the mean value theorem say?

 

[NWeid1]

First of all, I just realized I should've posted this in the homework section. And I know the function is cont. on [a,b] and f'(x) exists on (a,b) so f'(c)=(f(b)-f(a))/(b-a). So f'(c)=(sinb-sina)/(b-a). But I'm not sure how f'(c) comes into play. And I don't know how to interperate the absolute value either.

 

[gb7nash]

Ok, so we have:

(sin(b)-sin(a)) = f'(c)(b-a)

It's a well-known fact that x = y → |x| = |y|.

How can we apply that?

 

[NWeid1]

I honestly don't know...lol :\

 

[gb7nash]

x = y → |x| = |y|

This says that:

left-hand side = right-hand side

can be rewritten as:

|left-hand side| = |right-hand side|

For example:

a+b = c+d → |a+b| = |c+d|

Now, let's look at the problem at hand. We have:

(sin(b)-sin(a)) = f'(c)(b-a)

Applying the absolute value on both sides, how can we rewrite this?

 

[NWeid1]

|sin(b)-sin(a)| = |f'(c)(b-a)|. I understand this, I just read it wrong, lol sorry

 

[gb7nash]

|sin(b)-sin(a)| = |f'(c)(b-a)|

Ok, good. Now, let's calculate what f'(c) is. What's f'(x)? After you have that, plug c in.

 

[NWeid1]

f'(x) = cosx so f'(c) = cos(c).

 

[gb7nash]

f'(x) = cosx so f'(c) = cos(c).

Correct. So now, we have:

|sin(b)-sin(a)| = |cos(c)(b-a)|

It's also a well-known fact that |xy| = |x||y|. How could we rewrite the right-hand side?

 

[NWeid1]

|sin(b)-sin(a)| = |cos(c)||b-a|. And -1<=cosx<=1 so would it be |sin(b)-sin(a)| <= 1|b-a| ?

 

[gb7nash]

|sin(b)-sin(a)| = |cos(c)||b-a|. And -1<=cosx<=1 so would it be |sin(b)-sin(a)| <= 1|b-a| ?

Correct.

 

[NWeid1]

Thank you so much for working with me and not just posting an unexplained proof. Props!