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 $(1)$?

As Dan suggested, solve for the stationary points of the gradient. You'll end up solving two (identical) quartics in $\sin \alpha$, $\sin \beta$ (which are really just 2 quadratics due to the symmetry) and find that the only stationary point in the range is at $\alpha = \beta = \pi/6$. For this to not be the maximum on the domain, the max must occur on the boundary, which are much simpler sub-problems to solve (and by the symmetry you only have to solve 2), and in fact will prove that $\alpha = \beta = \pi/6$ gives the max over the domain.
