$$sinB+sinC=sinA \iff 2sin\dfrac{B+C}{2}cos\dfrac{B-C}{2}=2sin\dfrac{A}{2}cos\dfrac{A}{2}$$
$$\iff cos\dfrac{A}{2}cos\dfrac{B-C}{2}=sin\dfrac{A}{2}cos\dfrac{A}{2}$$
$$ \iff cos\dfrac{B-C}{2}=sin\dfrac{A}{2}=cos\dfrac{B+C}{2}$$
$$\iff cos\dfrac{B+C}{2}-cos\dfrac{B-C}{2}=0 $$
$$\iff -2sinBsinC=0$$
$$\iff \begin{bmatrix} & sinB=0 & \\ & sinC=0 & \end{bmatrix}$$
$$ \iff \begin{bmatrix} & \hat{B}=0 & \\ & \hat{C}=0 & \end{bmatrix}$$
Điều này không xảy ra do $B,C$ là góc của 1 tam giác.