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How to prove that the minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$

user127.0.0.1
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maths lover
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    Can $A,B,C$ be anything, or just angles of a triangle? (Or neither?) What are the restrictions (if any) on $A,B,C$? – John Jan 16 '14 at 00:48
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    any kind of angles – maths lover Jan 16 '14 at 00:59
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    Let $f(x,y):=\cos(x)+\cos(y)+\cos(x+y)$ (use substitutions $x=A-B$ and $y=B-C$ and $\cos(z)=\cos(-z)$). We are looking for its minimum value, so look for zeroes of its differential. – Berci Jan 16 '14 at 01:10
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    If $A$, $B$ and $C$ are separated by $120^\circ$ (modulo $360^\circ$), then the cosine of the difference of any two of them is $-1/2$, so the minimum is $\leq -3/2$. – Stefan Smith Jan 16 '14 at 02:30
  • @John : No matter what angles $A$, $B$, and $C$ are, the three angles $A-B$, $B-C$, and $C-A$ satisfy the constraint that their sum is $0$. So we have two degrees of freedom. – Michael Hardy Jan 16 '14 at 02:32

4 Answers4

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Starting with $F=\cos(a-b)+\cos(b-c)+\cos(c-a)$ expand the cosines via the addition formula for cosine, and get $$\cos a \cos b + \cos b \cos c + \cos c \cos a \\ +\sin a \sin b + \sin b \sin c + \sin c \sin a.$$ Then after applying $(u+v+w)^2-u^2-v^2-w^2=2uv+2vw+2wu,$ the double $2F$ of our objective function may be seen to be $$2F=(\cos a+\cos b +\cos c)^2+(\sin a +\sin b + \sin c)^2-3.$$ Note we have combined the terms e.g. $-\cos^2 a -\sin^2 a=-1$ to obtain the final $-3.$ Thus $2F \ge -3$ i.e. $F \ge -3/2.$ Since there are values of $a,b,c$ which achieve $F=-3/2$ this finishes a proof.

coffeemath
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  • @mathslover Is this last comment really just a separate question, or are you saying it would be another way to show the claim of the question in your posted question above? [I think the only thing I left out of the above answer was the values of $a,b,c$ which would give $-3/2$, for example $0,2\pi/3,4\pi/3$] – coffeemath Jan 21 '14 at 18:05
  • will it only take values 0 and 2pi/3 .4pi/3,,of these types..?? the above comment is the extension of this question.. – maths lover Jan 22 '14 at 03:23
  • @mathslover One can add any constant $c$ to all three angles and get other examples, i.e. $(0+c,2\pi/3+c,4\pi/3+c)$ are points for which the objective function $F$ is $-3/2$. I think these are the only ones. – coffeemath Jan 22 '14 at 09:50
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This result can be used (substituting $(A-B)$ as $X$, $(B-C)$ as $Y$)

napstablook
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As Berci suggested ,$x=A-B,y=B-C \to f=\cos {x} +\cos{y}+cos{(x+y)}$

$f=2\cos{\dfrac{x+y}{2}} \cos{\dfrac{x-y}{2}}+2\left(\cos{\dfrac{x+y}{2}}\right)^2-1 \ge 2\cos{\dfrac{x+y}{2}}+2\left(\cos{\dfrac{x+y}{2}}\right)^2-1 =2\left(\cos{\dfrac{x+y}{2}}+\dfrac{1}{2}\right)^2-\dfrac{3}{2} \ge -\dfrac{3}{2}$

first " $\ge$ " : $\cos{\dfrac{x+y}{2}}<0,x=y$

second " $\ge$ " : $\cos{\dfrac{x+y}{2}}=-\dfrac{1}{2}$

so the "=" will hold when $x=y= \dfrac{2\pi}{3}(or \dfrac{4\pi}{3})+2k\pi $

chenbai
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If $\displaystyle2x+2y+2z=n\pi,$

$$F=\cos2x+\cos2y+\cos2z=2\cos(x+y)\cos(x-y)+2\cos^2z-1$$

As $\displaystyle x+y=n\pi-y, \cos(x+y)=(-1)^n\cos z$ $$\implies2\cos^2z\pm2\cos z\cos(x-y)-(1+F)=0$$ which is a Quadratic Equation in $\cos z$

So, the discriminant must be $\ge0$

$$\implies (2\cos(x-y))^2\ge 4.\cdot2(-1-F)\iff 2F\ge-2-2\cos^2(x-y)\ge-3 $$

The equality occurs if $\cos^2(x-y)=1\implies \cos2(x-y)=2\cos^2(x-y)-1=1\iff 2(x-y)=2m\pi$ where $m$ is an integer

lab bhattacharjee
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