Mean Value Theorem

[Kushballo7]

Use the Mean Value Theorem to prove the inequality |sin(a)-sin(b)| <= |a-b|

I just tried various things which seemed to get me no where.

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

??

 

[galactus]

You're on the right track, just keep going.

Let \(\displaystyle f(a)=sin(a)\) and \(\displaystyle a\not=b\).

By the MVT there is a number c between a and b such that:

\(\displaystyle \frac{sin(a)-sin(b)}{a-b}=cos(c), \frac{|sin(a)-sin(b)|}{|a-b|}=|cos(c)|\leq1\)

so \(\displaystyle |sin(a)-sin(b)|\leq|a-b|\), which also holds when \(\displaystyle a=b\)

 

[Kushballo7]

Thanks!