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In a triangle, with $A, B, C$ are three angles, find the minimum and maximum of $$\cos(A-B)\cos(B-C)\cos(C-A)$$

lzutao
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    I'm guessing $A$, $B$, and $C$ are the three angles of the triangle. Right? – omegadot Jul 26 '15 at 04:21
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    multiplication of all cosinuses has absolute value less than one. So what you want to prove is completely wrong. – Arashium Jul 26 '15 at 04:22
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    Think about $A = \pi/6, B = \pi/3, C = \pi/2$, the result is strictly less than 1. – Zhanxiong Jul 26 '15 at 04:23
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    Sorry, I just want to prove it greater than $\frac18$ – lzutao Jul 26 '15 at 04:29
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    You can rewrite one of those angles as $\pi$ minus the other two, then it becomes a minimization problem of two variables. I would suggest Lagrange Multiplier as a "brute force" way to solve it. – Jared Jul 26 '15 at 04:31
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    If $A = \epsilon$, $B = \dfrac{\pi}{3}-\epsilon$, and $C = \dfrac{2\pi}{3}$ where $\epsilon > 0$ is tiny, then $\cos(A-B)\cos(B-C)\cos(C-A)$ is slightly larger than $-\dfrac{1}{8}$. So proving that $\cos(A-B)\cos(B-C)\cos(C-A)$ is greater than $\dfrac{1}{8}$ isn't possible. – JimmyK4542 Jul 26 '15 at 05:04

2 Answers2

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They are multiplication of cosinus of three angles which have sum of zero.

So:

$$\min(\cos(A-B)\cos(B-C)\cos(C-A))$$

Is the same as

$$\min(\cos(x)\cos(y)\cos(-x-y))$$

or in another form $$\min(\cos(x)\cos(y)\cos(x+y))$$

Due to symmetric position of $x$ and $y$, the candidates for minimum must be at the boundaries ($\cos . =\pm1$) or where $x=y$

We can check them.

For when $x=y$: $$A=\cos(x)\cos(y)\cos(x+y)=\cos^2(x)\cos(2x)$$

$$\frac{dA}{dx}=-\sin(2x)-\sin(4x)=0$$

$$\sin(2x)=-2\sin(2x)\cos(2x)$$

$$\sin(2x)(1+2\cos(2x))=0$$

if $\sin(2x)=0$ then $A=0$

if $\cos(2x)=\frac{-1}{2}$ then $A=\frac{-1}{8}$

So, the absolute mimimum is $\frac{-1}{8}$

Update

$\cos(2x)=\frac{-1}{2}$ means that $$x=k \pi \pm \frac{\pi}{3}$$

if we set $x$ as $(A-B)$ and $y$ as $(B-C)$ then since $x=y$ then:

$$A-B=B-C=k \pi \pm \frac{\pi}{3}$$

And there is always a degree of a freedom.

If you add any equal value to all A, B, and C the result of the multiplication remains the same. So A, B and C can be any number as far as they satisfy the equation above.

Just an example can be $B=0$, $A=\frac{\pi}{3}$ and $C=\frac{-\pi}{3}$

$$cos(\frac{\pi}{3})\times \cos(\frac{\pi}{3}) \times \cos(\frac{-2\pi}{3})=\frac{-1}{8}$$

Fix

For angles in a triangle, we need to add additional condition:

$$A+B+C=\pi$$

After, I did all my bests, I found these angles cannot be placed in any triangle. It means in a trangle, due to the following contrains:

$$0<A<\pi$$ $$0<B<\pi$$ $$0<C<\pi$$

we would have $A>\frac{-1}{8}$ instead of $A\ge\frac{-1}{8}$. So we can approach very close to $\frac{-1}{8}$ but we can never achieve it in a real triangle.

According to suggestion of JimmyK4542:

$$A=\varepsilon, B=\frac{\pi}{3}, C=\frac{2\pi}{3}-\varepsilon$$

Where $\varepsilon$ is very close to zero causes the multiplication reach very close to $\frac{-1}{8}$ but $\varepsilon$ never touches zero in a real triangle. It just approaches it.

Maximum

The maximum is when all $\cos(.)$ are at their maximum which means all angles are equivalent.

Symmetry

Also, this answer still relies on the fact that the minimum of cosxcosycos(x+y) "must be at the boundaries (cos.=±1) or where x=y", which needs to be justified. The fact that cosxcosycos(x+y) is symmetric with respect to x and y isn't enough to justify this

JimmyK4542

I agree. It looks like an open discussion: link

Community
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Arashium
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    For what angles $A,B,C$ is the absolute minimum $-\dfrac{1}{8}$ obtained? – JimmyK4542 Jul 26 '15 at 05:18
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    The problem specified that $A,B,C$ be angles in a triangle. This means that $A,B,C > 0$ (strictly greater than) and $A+B+C = \pi$. Also, just because the function $\cos x \cos y \cos(x+y)$ is symmetric with respect to $x$ and $y$ does not mean that the minimum must be at the boundaries of the region or where $x = y$. – JimmyK4542 Jul 26 '15 at 05:35
  • Fixed it. Thanks to Jimmy – Arashium Jul 26 '15 at 05:56
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    ^The maximum is trivially $1$, obtained when $A = B = C = \pi/3$ – JimmyK4542 Jul 26 '15 at 06:01
  • Also, this answer still relies on the fact that the minimum of $\cos x \cos y \cos(x+y)$ "must be at the boundaries ($cos.=±1$) or where $x=y$", which needs to be justified. The fact that $\cos x \cos y \cos(x+y)$ is symmetric with respect to $x$ and $y$ isn't enough to justify this. – JimmyK4542 Jul 26 '15 at 06:05
  • @JimmyK4542, I agree. – Arashium Jul 26 '15 at 06:06
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Let $A-B=x$ etc. $\implies x+y+z=0$

and $S=\cos x\cos y\cos z\iff2S=2\cos x\cos y\cos z=\cos x[\cos(y+z)+\cos(y-z)]$

$\implies2S=\cos x[\cos(-x)+\cos(y-z)]\iff \cos^2x+\cos x\cos(y-z)-2S=0$

which is a Quadratic Equation in $\cos x$

So, the discriminant $\cos^2(y-z)-4\cdot1(-2S)\ge0$

$\iff8S\ge-\cos^2(y-z)=\sin^2(y-z)-1\ge-1\implies S\ge?$

The equality occurs if $\sin^2(y-z)=0\iff\sin(y-z)=0$

lab bhattacharjee
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