[help]How to proof this-------

[handsomeboy]

Hello all
How to proof Abs[cos[a]-sin]<=Abs[a-b]
thanks a lot

 

[zorro]

Hi handsomeboy,

The above inequality doesn’t hold for all pairs a,b .
A simple counterexample is a=b=0.
Regards

Z

 

[knmaheshy2k]

hi handsomeboy,
i think there is something wrong in the equation or else there is something missing in the equation...
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happy learnig.........

 

[handsomeboy]

OH sorry
my question is
How to proof Abs[cos[a]-cos]<=Abs[a-b]

 

[steve10]

You need the following inequality, for any x>=0,
x>=sinx.

There are many different ways to prove it, and one of them I like very much is through geometry. But it needs some sketches and, therefore, I'll forget it.

Here is another way by analysis. Set f(x)=x-sinx. Then f'(x)=1-cosx>=0 which means that it is a non decreasing function for x>=0. Since f(0)=0, then f(x)>=0, which means x>=sinx for x>=0.
Now,
|cos[a]-cos|
=|cos[(a+b)/2+(a-b)/2]-cos[(a+b)/2-(a-b)/2]|
=|2sin[(a+b)/2]sin[(a-b)/2]|
=|sin[(a+b)/2]||2sin[(a-b)/2]|

Obviously, |sin[(a+b)/2]|<=1. Besides, according to the inequality we proved a while ago, we have
|2sin[(a-b)/2]|=|2sin|(a-b)/2||<=2*|(a-b)/2|=|a-b|
Therefore,
|cos[a]-cos|<=|a-b|.