$解：?(1)①?由勾股定理可得?AB=\sqrt {AC^2+BC^2}=10?$
$∴?sin A=\frac {BC}{AB}=cosB=\frac {6}{10}=\frac {3}{5}，??cos A=\frac {AC}{AB}=sinB=\frac {8}{10}=\frac {4}{5}?$
$②由勾股定理可得?BC=\sqrt {AB^2-AC^2}=24?$
$∴?sin A=cos B=\frac {BC}{AB}=\frac {24}{25}，??cos A=sin B=\frac {AC}{AB}=\frac {7}{25}?$
$③由勾股定理可得?AB=\sqrt {AC^2+BC^2}=\sqrt {29}?$
$∴?sin A=cos B=\frac {BC}{AB}=\frac 5{\sqrt {29}}=\frac {5\sqrt {29}}{29}，??cos A=sin B=\frac {AC}{AB}=\frac 2{\sqrt {29}}=\frac {2\sqrt {29}}{29}?$
$?(2)?當(dāng)?∠A+∠B=90°?時(shí)，?sin A=cos B，??cosA=sinB$
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