$解：過(guò)點(diǎn)?B?作?BE⊥AD，?交?AD?的延長(zhǎng)線于點(diǎn)?E?$
$設(shè)?DC=x，?則?BD=2x，??BC=BD+DC=3x?$
$∵?∠ADC=45°，??∠C=90°?$
$∴?△ACD?是等腰直角三角形$
$∴?AC=DC=x?$
$在?Rt△BCD?中，∵?BC=3x，??AC=x?$
$∴?AB=\sqrt {BC^2+AC^2}=\sqrt {10}x?$
$∴?cosB=\frac {BC}{AB}=\frac {3x}{\sqrt {10}x}=\frac {3\sqrt {10}}{10}?$
$∵?∠BDE=∠ADC=45°，??BE⊥AD?$
$∴?△BDE?是等腰直角三角形$
$∵?BD=2x?$
$∴?BE=DE=\frac {BD}{\sqrt 2}=\sqrt 2x?$
$∵?△ACD?是等腰直角三角形，?CD=x?$
$∴?AD=\sqrt 2CD=\sqrt 2x?$
$∴?AE=AD+DE=2\sqrt 2x?$
$在?Rt△ABE?中，∵?AE=2\sqrt 2x，??BE=\sqrt 2x?$
$∴?AB=\sqrt {AE^2+BE^2}=\sqrt {10}x?$
$∴?sin∠BAD=\frac {BE}{AB}=\frac {\sqrt 2x}{\sqrt {10}x}=\frac {\sqrt 5}5?$
$綜上所述，?cosB=\frac {3\sqrt {10}}{10}，??sin∠BAD=\frac {\sqrt 5}5?$
