解：?$(1)$?由題意?$AB=\sqrt {AC^2+BC^2}=\sqrt {13}$?
∴?$sin A=\frac {BC}{AB}=\frac {3\sqrt {13}}{13}，$??$cos B=\frac {BC}{AB}=\frac {3\sqrt {13}}{13}$?
?$(2)cos A=\frac {AC}{AB}=\frac {2\sqrt {13}}{13}，$??$sin B=\frac {AC}{AB}=\frac {2\sqrt {13}}{13}$?
?$(3)$?發(fā)現(xiàn)：?$sin A=cos (90°-∠A)，$??$cos A=sin (90°-∠A)$?
理由：∵?$sin A=\frac {∠A的對(duì)邊}{斜邊}=\frac ac，$??$cos B=\frac {∠B的鄰邊}{斜邊}=\frac ac$?
∴?$sin A=cos B=cos (90°-∠A)$?
