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