<s>[INST] Maximizing $\sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta$ I need to maximize
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \tag{1}$$
where $\alpha, \beta \in [0, \frac{\pi}{2}]$.
With numerical methods I have found that
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \leq 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} - \sin^2 \frac{\alpha + \beta}{2}. \tag{2} $$
If $(2)$ is true then I can denote $x = \frac{\alpha + \beta}{2}$ and prove (using Cauchy inequality) that 
$$ 2 \sin x \cos x - \sin^2 x \leq \frac{\sqrt{5}-1}{2}. \tag{3}$$
But is $(2)$ true? How do I prove it? Maybe I need to use a different idea to maximize $(1)$?
 [/INST] Let $$f = \sin \alpha \cos \alpha + \sin \beta \cos \beta -\sin \alpha \sin \beta = \frac{1}{2}(\sin 2\alpha + \sin 2\beta) -\sin \alpha \sin \beta$$
$$\frac{\partial f}{\partial \alpha} = 0 \implies \cos 2\alpha - \sin \beta \cos \alpha = 0$$
$$\frac{\partial f}{\partial \beta} = 0 \implies \cos 2\beta - \sin \alpha \cos \beta = 0$$
From symmetry (also may not be difficult to show directly) $$\beta = \alpha$$
Finally, $$\alpha = \beta = \frac{1}{2} \tan^{-1} 2$$
Substituting in $f$, the maximum value is 0.61803398874989484820
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