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

trigonometry

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

Solution 1:

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