21、(6分) 如图在Rt△ABC中,∠C=90°, AC=BC=4,点D是BC边上-动点,DE⊥AD交AB于点E
(1)当CD=DB时,求BE的值;
(2)求AE的最小值.
21ã解ï¼è§ä¸å¾ï¼1ï¼AB=AEï¼BE=AB-AE=0ï¼
(2)AE=AD/cosa=[AC/cos(Ï/4-a)]/cosa=AC/cosacos(Ï/4-a)=4â2/[cos^2a+(1/2)sin2a];
AE'=-4â2[2cosa(-sina)+cos2a]/[(cos^2a+(1/2)sin(2a)]^2
=4â2[sin(2a)-cos(2a)]/[cos^2a+sin(2a)]^2=0; å³ï¼
sin(2a)-cos(2a)=â{[1-cos(4a)]/2}-â{[1+cos(4a)/2}
={â{[1-cos(4a)]/2}-â{[1+cos(4a)/2}}{â{[1-cos(4a)]/2}+â{[1+cos(4a)/2}}/{â{[1-cos(4a)]/2}+â{[1+cos(4a)/2}}
=(â2/2){[1-cos(4a)]-[1+cos(4a)}/{â[1-cos(4a)]+â[1+cos(4a)]}
=-â2cos(4a)/{â[1-cos(4a)]+â[1+cos(4a)]}=0; 4a=Ï/2. a=Ï/8
AEmin=4â2/[cos^2(Ï/8)+(1/2)sin(Ï/4)]=4â2/{[1+cos(Ï/4)]/2+(1/2)sin(Ï/4)]}
=4â2/[(1+â2/2)/2+(1/2)(â2/2)]=8â2/[(1+â2]=8(2-â2)ã