# Thread: |sinA - sinB| < |A-B|

1. ## |sinA - sinB| < |A-B|

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?

2. ## Re: |sinA - sinB| < |A-B|

Originally Posted by Supernova008
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?
You posted in the trigonometry forum.
The usual proof is done using the mean value theorem. But that is calculus topic. You know that theorem?

3. ## Re: |sinA - sinB| < |A-B|

Originally Posted by Plato
You posted in the trigonometry forum.
The usual proof is done using the mean value theorem. But that is calculus topic. You know that theorem?
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?

4. ## Re: |sinA - sinB| < |A-B|

Originally Posted by Supernova008
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?
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. $l$ 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= $QB-PA=QM-PS=RM>LM=LB-SA=sinB-sinA.$
So we get B-A>sinB-sinA.

5. ## Re: |sinA - sinB| < |A-B|

Originally Posted by Supernova008
Yes, I posted in the trigonometry forum as I felt this was a trigonometry question. 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?
If $f$ is a differentiable function then
$A < B\, \Rightarrow \,\left( {\exists C \in \left( {A,B}\right)}\right)\left[{f'(C) = \frac{{f(B) - f(A)}}{{B - A}}} \right]$

That gives $\left| {\frac{{\sin (B) - \sin (A)}}{{B - A}}} \right| = \left| {\cos (C)} \right| \leqslant 1$.

That proves it.

6. ## Re: |sinA - sinB| < |A-B|

Originally Posted by abhishekkgp
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. $l$ 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= $QB-PA=QM-PS=RM>LM=LB-SA=sinB-sinA.$
So we get B-A>sinB-sinA.
Can you just subtract line segments like that? I mean, can you really write B-A = QB-PA?

Originally Posted by Plato
If $f$ is a differentiable function then
$A < B\, \Rightarrow \,\left( {\exists C \in \left( {A,B}\right)}\right)\left[{f'(C) = \frac{{f(B) - f(A)}}{{B - A}}} \right]$

That gives $\left| {\frac{{\sin (B) - \sin (A)}}{{B - A}}} \right| = \left| {\cos (C)} \right| \leqslant 1$.

That proves it.
Sorry, I am still in highschool, so I don't understand some of your notation. What does this mean: $\left( {\exists C \in \left( {A,B}\right)}\right)$ ?

Furthermore, doesn't this prove that |sin A - sin B| $\right| \leqslant$ |A-B| rather than |sinA - sinB| < |A-B| ?

7. ## Re: |sinA - sinB| < |A-B|

Originally Posted by Supernova008
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: $\left( {\exists C \in \left( {A,B}\right)}\right)$ ?

Furthermore, doesn't this prove that |sin A - sin B| $\right| \leqslant$ |A-B| rather than |sinA - sinB| < |A-B| ?
QB= angle B and PA= angle A.

8. ## Re: |sinA - sinB| < |A-B|

Originally Posted by Supernova008
Sorry, I am still in highschool, so I don't understand some of your notation. What does this mean: $\left( {\exists C \in \left( {A,B}\right)}\right)\$ ?
That just means the is a point between A & B where the slope is the same as the slope of the secant.

9. ## Re: |sinA - sinB| < |A-B|

Originally Posted by abhishekkgp
QB= angle B and PA= angle A.
I must admit that I am a little confused. How does B-A = QB-PA? You seem to subtract these segments horizontally, but then you subtract PS from QM vertically.

10. ## Re: |sinA - sinB| < |A-B|

Originally Posted by Plato
If $f$ is a differentiable function then
$A < B\, \Rightarrow \,\left( {\exists C \in \left( {A,B}\right)}\right)\left[{f'(C) = \frac{{f(B) - f(A)}}{{B - A}}} \right]$

That gives $\left| {\frac{{\sin (B) - \sin (A)}}{{B - A}}} \right| = \left| {\cos (C)} \right| \leqslant 1$.

That proves it.
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| $\right| \leqslant$ |A-B| rather than |sinA - sinB| < |A-B| ?

11. ## Re: |sinA - sinB| < |A-B|

Originally Posted by Supernova008
I must admit that I am a little confused. How does B-A = QB-PA? You seem to subtract these segments horizontally, but then you subtract PS from QM vertically.
Point Q lies on the curve y=x. the perpendicular dropped from Q on the x-axis meets the x-axis at B. hence OB=QB. Now OB=measure of angle B. hence QB=B.

12. ## Re: |sinA - sinB| < |A-B|

Originally Posted by abhishekkgp
Point Q lies on the curve y=x. the perpendicular dropped from Q on the x-axis meets the x-axis at B. hence OB=QB. Now OB=measure of angle B. hence QB=B.
D'oh. Sorry, I had confused the notations... Thanks!

13. ## Re: |sinA - sinB| < |A-B|

Originally Posted by Supernova008
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| $\right| \leqslant$ |A-B| rather than |sinA - sinB| < |A-B| ?
But I think, it has to be: $|\sin(A)-\sin(B)|\leqslant |A-B|$ in stead of $<|A-B|$, because if you take $A=B$, then you'll get: $|\sin(A)-\sin(A)|\leqslant |A-B| \Leftrightarrow 0\leqslant 0$ which is true, but not $0<0$.

Beside the mean theorem, if $f$ is differentiable in the interval $]A,B[$ and continuous in $[A,B]$ then there exists a number $C \in ]A,B[$ wherefore ... (see Plato's post) . To answer your questions, $f'(C)$ is indeed the correct notation and the reason you'll get $|\cos(C)|$ is because $\frac{d(\sin(x))}{dx}=\cos(x)$.

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.

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