Mean Value Theorem

Kushballo7

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Oct 20, 2005
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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|)

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

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

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

sin(a)sin(b)ab=cos(c),sin(a)sin(b)ab=cos(c)1\displaystyle \frac{sin(a)-sin(b)}{a-b}=cos(c), \frac{|sin(a)-sin(b)|}{|a-b|}=|cos(c)|\leq1

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