Calculus problem

[jessebu]

The question to to prove that abs(sin(x)-sin(y)) < (or equal to) abs(x-y).

Here's what I have so far:

abs(sin(x)-(sin(y)) = 2 sin((x-y)/2) cos ((x-y)/2)
= 2 sqrt((1-cos(x-y))/2) sqrt ((1+cos(x-y))/2)

I am unsure at how to go forward.

 

[galactus]

It would appear we can use the Mean Value Theorem on this one. Is that what you tried?.

Let \(\displaystyle f(x)=sin(x), \;\ x\neq y\)

The MVT sats there us a number, we can call c, between x and y such that:

\(\displaystyle \frac{sin(x)-sin(y)}{x-y}=cos(c)\)

\(\displaystyle \frac{|sin(x)-sin(y)|}{|x-y|}=|cos(c)|\leq 1\)

Now, can you finish?.

 

[jessebu]

Yeah, that did the trick. Thank you very much.