∵tan(a+兀/4)=1/2
∴(2sin^2a+sin2a)/sin(a-兀/4)
=(2sin^2a+2sinacosa)/(sinacosπ/4-cosasinπ/4)
=2sina(sina+cosa)/.[√2/2(sina-cosa)]
=2√2(sina+cosa)/(sina-cosa)
=2√2（sina/cosa+1)/(sina/cosa-1)
=-2√2(1+tana)/(1-tana)
=-2√2(tanπ/4+tana)/(1-tanπ/4tana)
=-2√2tan(a+π/4)
=-2√2*1/2
=-√2問題的分母是cosa-打錯(cuò)了不好意思在幫我算一邊吧~∵tan(a+兀/4)=1/2∴(tana+1)/(1-tana)=1/2∴2tana+2=1-tana∴tana=-1/3 即sina/cosa=-1/3cosa=-3sina代入到sin^2a+cos^2a=1得10sin^a=1,sin^2a=1/10∵-兀/2＜a＜0∴sina>0∴sina=√10/10∴(2sin^2a+sin2a)/cos(a-兀/4)=(2sin^2a+2sinacosa)/(cosacosπ/4+sinasinπ/4)=2sina(sina+cosa)/.[√2/2(sina+cosa)]=2√2sina=2√2*√10/10=2√5/5嗯嗯，明白了，謝謝~OL