prove that maximum value of $\cos\alpha \cos\beta$ occurs when $\alpha=\beta=\frac{\sigma}{2}$

Question

If $\alpha,\beta \in(0,\frac{\pi}{2})$ and $\alpha+\beta=\sigma$(constant),then prove that maximum value of $\cos\alpha \cos\beta$ occurs when $\alpha=\beta=\frac{\sigma}{2}$

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$\cos\alpha \cos\beta=\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}$
As $\cos x$ is a concave function,so by Jensen inequality
$\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}\le\cos(\frac{\alpha+\beta+\alpha-\beta}{2})$
$\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}\le\cos(\alpha)$
I am stuck here

Answer 1

Hint:

As $\alpha+\beta=\sigma$(constant), $\cos(\alpha+\beta)=\cos(\sigma)=$(constant)

So, we need to maximize $\cos(\alpha-\beta)$

Answer 2

Write $$\cos(\alpha)\cos(\beta)$$ as a function of $$\alpha$$ $$f(\alpha)=\cos(\alpha)\cos(\sigma-\alpha)$$ then we get $$f'(\alpha)=-\sin(\alpha)\cos(\sigma-\alpha)+\cos(\alpha)(-\sin(\sigma-\alpha))\cdot(-1)$$ by the product and the chain rule.

Answer 3

$$\cos\alpha \cos \beta =\frac{\cos(\alpha + \beta )+\cos(\alpha -\beta)}{2}=\frac {\cos\sigma +\cos (\sigma-2\beta)}{2}$$ Hence, it is maximized when $\beta =\sigma/2$ so that $\cos(\sigma -2\beta )=1$

Answer 4

Total overkill: Lagrange multipliers. The local extrema of $\cos\alpha\cos\beta$ with the restriction $\alpha + \beta = \sigma$ are given by: $$-\sin\alpha\cos\beta = \lambda,$$ $$-\cos\alpha\sin\beta = \lambda.$$ Solution: $$\sin\alpha\cos\beta = \cos\alpha\sin\beta,$$ $$\tan\alpha = \tan\beta,$$ and as $\tan$ is injective in $(0,\pi/2)$: $$\alpha = \beta = \sigma/2.$$