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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