# Sum of cosines

1. Sep 29, 2014

### gianeshwar

Dear Friends! How can I find the minimum value of

2. Sep 29, 2014

### mathman

Are you interested in absolute value (easy = 0) or most negative (harder)?

3. Sep 30, 2014

### gianeshwar

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.

4. Sep 30, 2014

### mathman

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.

5. Sep 30, 2014

### lpetrich

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.

6. Oct 2, 2014

### davidmoore63@y

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.

7. Oct 5, 2014

### gianeshwar

Thank You Friends!