Results 1 to 6 of 6

Math Help - Trigonometric Inequality

  1. #1
    Member great_math's Avatar
    Joined
    Oct 2008
    Posts
    132

    Trigonometric Inequality

    If \alpha,\beta and \gamma be the angles in any acute triangle. Prove that

    2<\sin \alpha + \sin \beta + \sin \gamma \le \dfrac{3\sqrt{3}}{2}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Using Jensen's inequality,

    \frac{sin \alpha+sin \beta+sin \gamma}{3} \leq sin (\frac{\alpha + \beta + \gamma}{3})
    \frac{sin \alpha+sin \beta+sin \gamma}{3} \leq sin (\frac{\pi}{3})
    \frac{sin \alpha+sin \beta+sin \gamma}{3} \leq \frac{\sqrt3}{2}
    \sin \alpha + \sin \beta + \sin \gamma \le \dfrac{3\sqrt{3}}{2}
    Last edited by alexmahone; June 7th 2009 at 09:12 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Rachel.F's Avatar
    Joined
    Jun 2009
    Posts
    26
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by Rachel.F View Post
    Thats wrong!

    \sin \beta < 1, \sin \gamma < 1 does not imply the last step in your proof. This part of the question was posted by fardeen_gen recently.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Rachel.F's Avatar
    Joined
    Jun 2009
    Posts
    26
    Quote Originally Posted by Isomorphism View Post
    Thats wrong!

    \sin \beta < 1, \sin \gamma < 1 does not imply the last step in your proof. This part of the question was posted by fardeen_gen recently.
    why? i didn't understand
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by Isomorphism View Post
    Thats wrong!

    \sin \beta < 1, \sin \gamma < 1 does not imply the last step in your proof. This part of the question was posted by fardeen_gen recently.
    Quote Originally Posted by Rachel.F View Post
    why? i didn't understand
    \sin \beta < 1, \sin \gamma < 1 \implies \sin \beta + \sin \gamma < 2

    Now \sin \alpha + \sin \beta + \sin \gamma > \sin \beta + \sin \gamma , however the right hand side of the inequality is less than 2. This says nothing about the sum of the sines.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inequality trigonometric
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 3rd 2011, 11:42 AM
  2. Trigonometric Inequality
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: August 28th 2010, 03:56 AM
  3. Trigonometric inequality.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 10th 2010, 01:06 AM
  4. another trigonometric inequality
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 2nd 2009, 10:03 PM
  5. Trigonometric inequality
    Posted in the Algebra Forum
    Replies: 11
    Last Post: December 21st 2008, 12:23 AM

Search Tags


/mathhelpforum @mathhelpforum