<s>[INST] solving for sin of the sum of two angles of a triangle In triangle $ABC$, 
$3\sin B+4\cos C=6$ and $4\sin C+3\cos B=1$. 
Show that $\sin(B+C)=0.5$.
Can we assume $\angle A = 180 - ( B + C)$ and use sum formula.
 [/INST] Hint: By squaring both sides of the equation:
$$9\sin^2 B + 24 \sin B\cos C + 16 \cos^2 C = 36$$
$$9\cos^2 B + 24 \cos B\sin C + 16 \sin^2 C = 1$$
By adding the equations:
$$9(\sin^2B+\cos^2B)+24(\sin B\cos C+\cos B\sin C)+16(\cos^2 C+\sin^2C)=37$$
Using $\sin^2x+\cos^2x=1$
$$9(1)+24(\sin B\cos C+\cos B\sin C)+16(1)=37\implies$$
$$\sin B\cos C+\cos B\sin C=\frac 12$$
Using $\sin B\cos C+\cos B\sin C=\sin(B+C)$
$${\sin(B+C)=\frac 12}$$
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