In a proof I have encountered, this step is used:
|sinA-sinB| < |A-B|
But it was taken as a given and not proved itself. Why is this so? Am I missing something very obvious?
Yes, I posted in the trigonometry forum as I felt this was a trigonometry question :P
No, I do not know that theorem. I know derivatives and understand the basics of integrations (well, some of the concepts like the Riemann sum). But, is there no other way to prove that the relation I posted is always true?
I have attached a sketch which has the line y=x and the curve y=sin x.
Note that y=x is tangent to the sin x curve at origin. $\displaystyle l$ is a line parallel to y=x and passing through S=(A,sinA).
I have considered only first quadrant.
Note that PQRS is a parallelogram. ALSO:
B-A=$\displaystyle QB-PA=QM-PS=RM>LM=LB-SA=sinB-sinA. $
So we get B-A>sinB-sinA.
If $\displaystyle f$ is a differentiable function then
$\displaystyle A < B\, \Rightarrow \,\left( {\exists C \in \left( {A,B}\right)}\right)\left[{f'(C) = \frac{{f(B) - f(A)}}{{B - A}}} \right]$
That gives $\displaystyle \left| {\frac{{\sin (B) - \sin (A)}}{{B - A}}} \right| = \left| {\cos (C)} \right| \leqslant 1$.
That proves it.
Can you just subtract line segments like that? I mean, can you really write B-A = QB-PA?
Sorry, I am still in highschool, so I don't understand some of your notation. What does this mean:$\displaystyle \left( {\exists C \in \left( {A,B}\right)}\right)$ ?
Furthermore, doesn't this prove that |sin A - sin B| $\displaystyle \right| \leqslant$ |A-B| rather than |sinA - sinB| < |A-B| ?
I am sorry, I just don't follow it that well. Why does there HAVE to be a point C whose gradient is equal to the gradient of the secant AB? Also, is f'(C) the correct notation for the gradient of f(x) at point C? Furthermore, why does this simplify to |cos C|? Also, doesn't this prove that |sin A - sin B| $\displaystyle \right| \leqslant$ |A-B| rather than |sinA - sinB| < |A-B| ?
But I think, it has to be: $\displaystyle |\sin(A)-\sin(B)|\leqslant |A-B|$ in stead of $\displaystyle <|A-B|$, because if you take $\displaystyle A=B$, then you'll get: $\displaystyle |\sin(A)-\sin(A)|\leqslant |A-B| \Leftrightarrow 0\leqslant 0$ which is true, but not $\displaystyle 0<0$.
Beside the mean theorem, if $\displaystyle f$ is differentiable in the interval $\displaystyle ]A,B[$ and continuous in $\displaystyle [A,B]$ then there exists a number $\displaystyle C \in ]A,B[$ wherefore ... (see Plato's post) . To answer your questions, $\displaystyle f'(C)$ is indeed the correct notation and the reason you'll get $\displaystyle |\cos(C)|$ is because $\displaystyle \frac{d(\sin(x))}{dx}=\cos(x)$.
Important to understand the mean value theorem is to look at the graphic interpretation of it, that will help you more to understand this theorem.