Extreme Value with Triangle 3

Note first that $\sin(B)\sin(C) = \dfrac{1}{2}(\cos(B - C) - \cos(B + C)) = \dfrac{1}{2}(\cos(B - C) + \cos(A))$ , since $A = \pi - (B + C)$ . Thus

$\sin(A) + \sin(B)\sin(C) = \sin(A) + \dfrac{1}{2}\cos(A) + \dfrac{1}{2}\cos(B - C) = \dfrac{\sqrt{5}}{2}\left(\dfrac{2}{\sqrt{5}}\sin(A) + \dfrac{1}{\sqrt{5}}\cos(A)\right) + \dfrac{1}{2}\cos(B - C) = \dfrac{\sqrt{5}}{2}\sin(A + \theta) + \dfrac{1}{2}\cos(B - C)$ ,

where $\theta = \arcsin\left(\dfrac{1}{\sqrt{5}}\right)$ . This expression is maximized when $A + \theta = \dfrac{\pi}{2}$ and $B = C$ , resulting in a maximum of $\dfrac{\sqrt{5} + 1}{2} \approx \boxed{1.618}$ .