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Trig. inequality

  1. Jan 17, 2007 #1
    Show that :

    (sinA/2 + sinB/2 + sinC/2)^2 >= (sinA)^2 + (sinB)^2 + (sinC)^2

    A,B and C are the angles of a triangle.
     
  2. jcsd
  3. Jan 18, 2007 #2

    Gib Z

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    Homework Helper

    Are you looking for an elementary proof using trig manipulations or identites?
    I can't see anything, Ill just right the question again that other people might find easier to help you with.

    Show that [tex]\frac{\sin A + \sin B + \sin C}{2} >= \sin^2 A + \sin^2 B + \sin^2 C[/tex]
    bounded by the condition [tex]A+B+C=\pi[/tex]
     
  4. Jan 18, 2007 #3
    Gib Z, i think you have wrote the question wrong. i think Jean-Louis wrote,

    [tex]\left(\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2}\right)^{2} \geq \sin^2 A + \sin^2 B + \sin^2 C[/tex] where [tex]A + B + C = \pi[/tex]
     
  5. Jan 18, 2007 #4

    cristo

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    Staff Emeritus
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    Have you tried doing anything? I'd multiply out the left hand side first, and see what happens.
     
  6. Jan 18, 2007 #5
    I think I got it. Here it goes :

    sin(x/2) = +/- sqrt((1-cos(x))/2)
    and sin^2(z) = 1 - cos^2(x)

    so that that (sinA)^2 + (sinB)^2 + (sinC)^2 = 3 - (cos^2(A) + cos^2(B) +
    cos^2(C))

    Now (sin A/2 + sin B/2 + sin C/2)^2 = (+/- sqrt((1-cos(A))/2) + +/-
    sqrt((1-cos(B))/2) + +/- sqrt((1-cos(C)/2) ) ^2

    If we let

    A = +/- sqrt((1-cos(A))/2)
    B = +/- sqrt((1-cos(B))/2)
    C = +/- sqrt((1-cos(C))/2)

    Then (sin A/2 + sin B/2 + sin C/2)^2 = A^2 + B^2 + C^2 + 2AB + 2AC + 2BC

    A^2 = (1 - cos(A))/2
    B^2 = (1 - cos(B))/2
    C^2 = (1 - cos(C))/2

    AB = 1/2 SQRT((1-cosA)(1-cosB))
    AC = 1/2 SQRT((1-cosA)(1-cosC))
    BC = 1/2 SQRT((1-cosB)(1-cosC))

    so you have

    (1 - cos(A))/2 + (1 - cos(B))/2 + (1 - cos(C))/2 +
    SQRT((1-cosA)(1-cosB)) + SQRT((1-cosA)(1-cosC)) + SQRT((1-cosB)(1-cosC))
    >= 3 - (cos^2(A) + cos^2(B) + cos^2(C))

    Now the Law of cosines tells us this....

    cosC = (a^2 + b^2 -c^2)/2ab
    cosA = (b^2 + c^2 - a^2)/2bc
    cosB = ((a^2 + c^2 - b^2)/2ac

    You have all the equations in terms of cosines - substitute simplify.....

    QED
     
  7. Jan 18, 2007 #6

    Gib Z

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    Homework Helper

    Kill me. Somebody kill me...
     
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