Prove that |sin(a)-sin(b)|<=|a-b| for all a and b.

[MarkSA]

Prove that |sin(a)-sin(b)|<=|a-b| for all a and b.

I'm really stumped on this one. Not sure to begin, and i'm not entirely sure it's a calculus problem (?)

I'm guessing I need to make use of some theorem to prove this but I can't think what I might use. Any suggestions?,

Thanks

 

[galactus]

Let \(\displaystyle \L\\f(a)=sin(a) \;\ and \;\ a\neq{b}\)

By the Mean Value Theorem there is a number c between a and b such that

\(\displaystyle \L\\\frac{sin(a)-sin(b)}{a-b}=cos(c), \;\ \frac{|sin(a)-sin(b)|}{|a-b|}=|cos(c)|\leq{1}\)

So, \(\displaystyle \L\\|sin(a)-sin(b)|\leq{|a-b|}\), which also holds when a=b.

 

[MarkSA]

thanks for the reply.

I'm a little confused how you arrived at |sin(a)-sin(b)|<=|a-b|

I understand the procedure up until the:

|sin(a)-sin(b)|/(a-b) = |cos(c)| <= 1

I see cos(c) must be <= 1, but how did you take the above equation and derive |sin(a)-sin(b)|<=|a-b| from it?

If |(a-b)| is multiplied on both sides of the equation, the first part becomes:
|sin(a)- sin(b)|=|cos(c)||(a-b)|?

I think i'm not familiar working with equations with more than one equal sign or greater/less than.

 

[o_O]

Well if:
\(\displaystyle \frac{|sina - sinb|}{|a-b|} = |cos c|\) and \(\displaystyle |cos c| \leq 1\)

then that would mean:

\(\displaystyle \frac{|sina - sinb|}{|a-b|} \leq 1\) as well. Then you multiply both sides by |a - b| to arrive at your conclusion.