∵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
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024