Trigonometry Crackers

The equation can be rewritten as 2SinASinB+2SinBSinC+2SinCSinA +2CosACosB+2CosBCosC+2CosCCosA=-3. Now this can be rewritten as $Sin^2A+Sin^2B+Sin^2C+2SinASinB+2SinBSinC+2SinCSinA +Cos^2A+Cos^2B+Cos^2C+2CosACosB+2CosBCosC+2CosCCosA=0$ which implies that $(SinA+SinB+SinC)^2+(CosA+CosB+CosC)^2=0$ .Let SinA+SinB+SinC=a,CosA+CosB+CosC=b. It implies that $a^2+b^2=0$ which implies that a=b=0. Since a+b is what we need, a+b=0.

Consider the following sum:

$\small \begin{aligned} S & = (\cos A + \cos B + \cos C)^2 + (\sin A + \sin B + \sin C)^2 \\ & = \cos^2 A + \cos^2 B + \cos^2 C + 2(\cos A \cos B + \cos B\cos C + \cos C\cos A) \\ & \quad + \sin^2 A + \sin^2 B + \sin^2 C + 2(\sin A \sin B + \sin B\sin C + \sin C\cos A) & \small \blue{\text{As }\sin^2 \theta + \cos^2 \theta = 1} \\ & = 3 + 2(\cos A \cos B + \sin A \sin B + \cos B\cos C + \sin B \sin C + \cos C\cos A + \sin C \sin A) \\ & = 3 + 2(\blue{\cos (A-B) + \cos (B-C) + \cos (C-A)}) = 3 + 2\left(\blue{-\frac 32}\right) = 0 \end{aligned}$

Note that $(\cos A + \cos B + \cos C)^2 \ge 0$ and $(\sin A + \sin B + \sin C)^2 \ge 0$ . Therefore, $S=0$ only when $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$ . Hence $\sin A + \cos A + \sin B + \cos B + \sin C + \cos C = \boxed 0$ .

One quickie solution (just chalk it up to cursory observation) is $A = \frac{4\pi}{3}, B = \frac{2\pi}{3}, C = 0.$