$\dfrac{\cos \alpha}{\cos (\beta-\gamma)}$

Question

I have a minimum and maximum problem connected to trigonometry. Please help, it is from an American olympiad from 2013.

Suppose that the angles of a triangle are $\alpha, \beta$ and $\gamma$. Let $H$ be the set $\left\{\dfrac{\cos \alpha}{\cos (\beta-\gamma)}, \dfrac{\cos \beta}{\cos (\gamma-\alpha)}, \dfrac{\cos \gamma}{\cos (\alpha-\beta)}\right\}$.

a) Find the minimal possible value of the largest item of $H$.

b) Find the maximal possible value of the smallest item of $H$.

Answer 1

WLOG, let $A=\alpha\ge B=\beta\ge C=\gamma$

$\cfrac {\cos A}{\cos (B-C)}=-\cfrac{\cos(B+C)}{\cos(B-C)}=1-\cfrac 2{1+\tan B\tan C}\tag 1$

1. If $A=\pi/2$, $\cfrac {\cos A}{\cos (B-C)}=0, \cfrac {\cos B}{\cos (C-A)}=\cfrac {\cos B}{\cos B}=1=\cfrac {\cos C}{\cos (A-B)}$. Hence minMax=$1$, maxMin=$0$;

2. If $A>\pi/2$, from $(1)$, it's easy to see the maximum in $H$ is $1-\cfrac 2{1+\tan A \tan B}$, and the minimum is $1-\cfrac 2{1+\tan B \tan C}$. When $A\to \pi/2$, minMax $\to 1$; when $B=C \to \pi/4$, maxMin $\to 0$;

3. If $A<\pi/2$, maximum in H is $1-\cfrac 2{1+\tan A \tan B}$, and the minimum is $1-\cfrac 2{1+\tan B \tan C}$. It can be easily shown mininum of maximum is $1/2$ when $A=B=\pi/3$ and maximum of minimum is $1/2$ when $B=C=\pi/3$.

Combine all cases, we have min max=max min = $1/2$ when $A=B=C=\pi/3$

PS: $f(x)=1-\cfrac 2{1+x}, x>0$, as $x$ increases, $\cfrac 2{1+x}$ decreases, and $f(x)$ increases. the minimum of $\tan A \tan B (\pi/2>A\ge B\ge C\ge \pi/3)$ occurs when $A=B=\pi/3$. etc, etc.