1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sum of cosines

  1. Sep 29, 2014 #1
  2. jcsd
  3. Sep 29, 2014 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Are you interested in absolute value (easy = 0) or most negative (harder)?
     
  4. Sep 30, 2014 #3
    Thank You Mathman! Actually I am interested to find minimum value not absolute . What I am thinking Geometrically is that the possibility seems by imagining unit vectors inclined at 120 degrees to each other in 3D space.Am I correct?It gives thevalue -3/2.
     
  5. Sep 30, 2014 #4

    mathman

    User Avatar
    Science Advisor
    Gold Member

    You can put 3 vectors at 120 deg. apart in a plane. I haven't looked at the problem in any detail, but I would guess you could do better in 3d.
     
  6. Sep 30, 2014 #5
    In the general case, it's easiest to do the problem in vector fashion. The angle cosines are easy: ## \cos \theta_{12} = n_1 \cdot n_2 ##. For unit vector n relative to unit vectors n1, n2, n3, we get
    $$ S = \cos \theta_1 + \cos \theta_2 + \cos \theta_3 = n \cdot n_1 + n \cdot n_2 + n \cdot n_3 = n \cdot (n_1 + n_2 + n_3) $$
    We now want to maximize ## S = n \cdot n_{total} ## and we wish to do so with the constraint that ## |n| = 1 ##. This is easiest to do with Lagrange multipliers:
    $$ S' = S - \frac12 \lambda (n^2 - 1) $$
    Taking derivatives with respect to all components of n, we get ## \lambda n = n_1 + n_2 + n_3 = n_{total} ##, meaning that n is proportional to ntotal.
     
  7. Oct 2, 2014 #6
    It seems to me that -3/2 is indeed the minimum. Denote three unit vectors in 3-space by a,b,c with components a(i), b(i), c(i) where i=1,2,3. Then we wish to minimize a.b + b.c + c.a subject to a^2=b^2=c^2=1. Using Lagrange multipliers L,M,N we wish to minimize:

    a(i)b(i) + b(i)c(i) + c(i)a(i) - L(a^2-1) - M(b^2-1) - N(c^2-1) where I have used summation convention for repeated indices.

    Differentiating by a(i), b(i), c(i) respectively we get

    b(i) + c(i) =2La(i)
    a(i) + c(i) =2Mb(i)
    b(i) + a(i) =2Nc(i)

    it is immediately apparent from this that all the vectors are coplanar. furthermore, subtracting the first two of these gives
    b(i) (1+2M) = a(i) (1+2L)

    But by inspection the minimum is not achieved by having a(i) and b(i) be the same vector, since the cosine is then maximized, so we must have L=M=N= -1/2

    Therefore b(i) + c(i) = -a(i), or a + b + c = 0. Hence at the extremum, a,b,c are at the vertices of an equilateral triangle, as conjectured.
     
  8. Oct 5, 2014 #7
    Thank You Friends!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook