1. ## Mean Value Theorum

Hi
I'm studying for my midterm and am really confused.
One ofthe practice questions is that I am supposed to show that
lsinx-sinyl 'is less than or equal to' lx-yl
using the Mean Value Theorum.
I thought the mean value theorum is pretty much this.
so lsinx-sinyl/lx-yl is less than or equal to zero
meaning the derivative of f(t)=sint is less than zero
so cos(t) is less that zero. Okay, so cos(t)=lsinx-sinyl/lx-yl
and now I'm stuck. How do I prove this statement. Isn't it a proof of itself?
Also, do the absolute value signs matter or can I basically just ignore them.
Thanks!

2. Originally Posted by sberxa
Hi
I'm studying for my midterm and am really confused.
One ofthe practice questions is that I am supposed to show that
lsinx-sinyl 'is less than or equal to' lx-yl
using the Mean Value Theorum.
I thought the mean value theorum is pretty much this.
so lsinx-sinyl/lx-yl is less than or equal to zero
meaning the derivative of f(t)=sint is less than zero
so cos(t) is less that zero. Okay, so cos(t)=lsinx-sinyl/lx-yl
and now I'm stuck. How do I prove this statement. Isn't it a proof of itself?
Also, do the absolute value signs matter or can I basically just ignore them.
Thanks!
If $\displaystyle x<y$ then MVT on $\displaystyle [x,y]$ gives $\displaystyle \sin y - \sin x = \cos ( z) (y-x) \implies |\sin x - \sin y | = |x-y| |\cos z|$ where $\displaystyle x<z<y$. But $\displaystyle |\cos z|\leq 1$. Thus, $\displaystyle |\sin x - \sin y| \leq |x-y|$.