Aha! I see it now. The range of cos is between 1 and -1, so it cannot exceed 1... I think I see that now.
Just one question, how did you find "cosc"?
Thank you very much.
Hmm, ok, thank you.
So I get now
|sinb - sina|
__________ <= 1
| b - a|
Which is similar to the theorem f(b)-f(a) / b-a ?
If it is, this basically says that the equation I came up with (which is the derivative) is equal to 1?
Can I work on this further by taking the...
Well, my guess is it does but I still fail to see any connection.
| \sin \displaystyle x- \sin \displaystyle y| = 2 \left| \cos \left( \frac{\displaystyle x+ \displaystyle y}{2} \right) \sin \left( \frac{\displaystyle x- \displaystyle y}{2} \right) \right|
Am I getting anywhere with this?
Reading the hotlink I still fail to see the connection with MVT to the problem. Could the "real values x and y" have anything to do with the (a, b) interval?
Homework Statement
Prove for all real x and y that
|sinx - siny| <= |x-y|
Homework Equations
It's a question from the Mean Value Theorem/Rolle's Theorem section.
The Attempt at a Solution
Honestly, I've tried. It looks somewhat similar to the triangle inequality (I think?), but...