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Trigonometric inequality

  1. Jul 19, 2013 #1
    1. The problem statement, all variables and given/known data
    If ##A+B+C=\pi##, prove that ##\cos A+\cos B+\cos C \leq 3/2##.


    2. Relevant equations



    3. The attempt at a solution
    I don't really know how to start. ##A+B=\pi-C##. Taking cos on both sides doesn't seem of much help. I need a few hints to start with.
     
  2. jcsd
  3. Jul 19, 2013 #2
    Taylor series expansion?
     
  4. Jul 19, 2013 #3
    What?

    This is basic trigonometry and I don't think I need to use calculus here.
     
  5. Jul 19, 2013 #4

    Zondrina

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    Hmm how about ##A = B = C = \frac{\pi}{3}##.

    So that ##cos(A) + cos(B) + cos(C) = \frac{3}{2}##
     
  6. Jul 19, 2013 #5
    I am not allowed to do that. I have to prove it in a proper way instead of substituting the values.
     
  7. Jul 19, 2013 #6

    Zondrina

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    That's a tough one then. Wolfram gives :

    cos(A+B+C)
    = cos(A)cos(B)cos(C) - sin(A)sin(B)cos(C) - sin(A)cos(B)sin(C) - cos(A)sin(B)sin(C)
    = cos(A)[cos(B)cos(C) - sin(B)sin(C)] - sin(A)[sin(B)cos(C) - sin(C)cos(B)]
    = cos(A)[cos(B+C)] - sin(A)[sin(B-C)]
     
  8. Jul 19, 2013 #7
    That's already in my notes. I am sure that I don't have to use this.
     
  9. Jul 19, 2013 #8

    verty

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    I've found a way to do this using calculus knowledge. I don't see an easier way.

    Start by assuming that we are talking about angles in triangle, so A,B,C are in the range [0,π]. Find the stationary points and show that all are ≤ 3/2. Include an argument that you have found all the stationary points.
     
  10. Jul 19, 2013 #9
    Sorry but I haven't done calculus problems involving three variables.

    Meanwhile, I was trying this. From the AM-GM inequality,
    [tex]\frac{\cos^2(A/2)+\cos^2(B/2)}{2} \geq \cos(A/2)\cos(B/2)[/tex]
    [tex]\Rightarrow \cos^2(A/2)+\cos^2(B/2) \geq \cos\left(\frac{A+B}{2}\right)+\cos\left(\frac{A-B}{2}\right)[/tex]

    Now I can covert A/2 and B/2 to A and B on LHS but at RHS, the term ##\cos\left(\frac{A-B}{2}\right)## troubles me. I can rewrite ##\cos\left(\frac{A+B}{2}\right)=\sin(C/2)##, add ##\cos C## on both the sides and make a perfect square at RHS plus some constant but the term ##\cos\left(\frac{A-B}{2}\right)## is a problem here.
     
  11. Jul 19, 2013 #10
    Here's a hint. On the interval 0 < theta < 90*, cos(x) is a concave function.

    That means that if I pick a cos(A) and a cos(B), the midpoint of these points has a y value of

    (cos(A)+cos(B))/2

    Then that is going to be less than cos(x) at that same point on the x axis.

    cos((A+B)/2)

    Unless of course, that midpoint is the point itself, or A=B

    I'll have to think of a way to do this outside of the 90* interval, but I bet some sort of comparison can be made considering that the sum of cosines will have one negative term.
     
  12. Jul 20, 2013 #11

    I like Serena

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    Hi Pranav! :wink:

    Since A, B, and C can be swapped without changing the equation, any extremum will have equal A, B, and C.
    Or rather, A, B, and C will be equal modulo ##2\pi##, since the cosine has a period of ##2\pi##.

    So we have:
    \begin{array}{lcl}
    A+B+C=\pi & \quad & (1)\\
    A \equiv B \equiv C \pmod{2\pi} && (2)
    \end{array}
    Substituing (2) in (1) gives:
    \begin{array}{lcl}
    3A &\equiv& \pi \pmod{2\pi} \\
    3A &=& \pi + 2\pi k \\
    A &=& \frac \pi 3 + \frac{2\pi}3 k \\
    A &\equiv& \pm \frac \pi 3, \pm \pi \pmod{2\pi}
    \end{array}

    When enumerating the possible solutions, we find 2 relevant and distinct solutions:
    \begin{array}{lcl}
    A=B=C=\frac \pi 3 & \quad & \cos(A)+\cos(B)+\cos(C) = \frac 3 2 \\
    A=B=\pi,\ C=-\pi && \cos(A)+\cos(B)+\cos(C) = -3
    \end{array}

    Since there are no boundary solutions we can conclude for this continuous function that:
    $$-3 \le \cos(A)+\cos(B)+\cos(C) \le \frac 3 2$$

    Alternatively, we can find the same result with the method of Lagrange multipliers.
    But then, you didn't want to use calculus. :smile:
     
  13. Jul 20, 2013 #12
    Sorry to hijack, but I didn't know we could conclude that. Could you explain this idea?
     
  14. Jul 20, 2013 #13
    You already know that I am not familiar with modular arithmetic. :rolleyes:

    Anyways, I asked my teacher about this. He told me to consider three points on the graph of y=cos x ##(A,\cos A), (B,\cos B)## and ##(C,\cos C)## and work on the centroid of the triangle formed.

    Centroid is ##\left(\frac{A+B+C}{3},\frac{\cos A+\cos B+\cos C}{3}\right)##. The ordinate on the graph of y=cos x corresponding to the abcissa of centroid is ##\cos\left(\frac{A+B+C}{3}\right)=3/2##. Since ordinate of centroid of is less than or equal to 3/2, the inequality is proved. I guess this is enough for the question.
     
  15. Jul 20, 2013 #14
    Yes, your teacher's method is exactly what I was getting at earlier. Since the midpoint of (A, cos A), (B, cos B) and the midpoint of (B, cos B), (C, cos B) and the midpoint of (A, cos B), (C, cos C) are always under the graph of cos(x) at those same points, then so is the center of the triangle. The only way it matches cos(x) is if the center of the triangle reduces to be cos(x), which happens when A=B=C.
     
  16. Jul 20, 2013 #15

    BruceW

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    I think you mean 1/2 here. But It looks like you understand the answer. It is quite a nice solution really.
     
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