# $\dfrac{\cos \alpha}{\cos (\beta-\gamma)}$ [closed]

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 $$H$$.

## closed as off-topic by Namaste, Eevee Trainer, Paul Frost, Carl Schildkraut, KReiserDec 26 '18 at 2:18

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• Without bothering to do any math, relying purely by instinct, my guess is that the equilateral triangle is relevant to both questions. Therefore, my first approach would be to attempt to prove this. – user2661923 Dec 25 '18 at 22:58
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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.