Results 1 to 13 of 13

Math Help - |sinA - sinB| < |A-B|

  1. #1
    Newbie
    Joined
    Jul 2011
    Posts
    16

    |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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by Supernova008 View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2011
    Posts
    16

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by Plato View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by Supernova008 View Post
    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.
    Attached Thumbnails Attached Thumbnails |sinA - sinB| &lt; |A-B|-neu.draw-1.jpg  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by Supernova008 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2011
    Posts
    16

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by abhishekkgp View Post
    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?

    Quote Originally Posted by Plato View Post
    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| ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by Supernova008 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by Supernova008 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Jul 2011
    Posts
    16

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by abhishekkgp View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jul 2011
    Posts
    16

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by Plato View Post
    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| ?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by Supernova008 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Jul 2011
    Posts
    16

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by abhishekkgp View Post
    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!
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: |sinA - sinB| < |A-B|

    Quote Originally Posted by Supernova008 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find Exact Value of SinB
    Posted in the Trigonometry Forum
    Replies: 17
    Last Post: October 8th 2011, 01:32 PM
  2. Prove sinA+sinB>sinC>|sinA-sinB|
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: August 21st 2010, 05:49 PM
  3. Sina alpha, Cosine beta...
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 22nd 2010, 05:26 PM
  4. sina*cosa=4/7 , a=?
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 2nd 2010, 10:50 PM
  5. tan(A-B)/tan(A) + (sinC)^2/(sinA)^2=1
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 28th 2009, 12:29 AM

Search Tags


/mathhelpforum @mathhelpforum