Question

已知sinA-cosA=sinAcosA,求sin2A的值

Answer 1

令sinA-cosA=t

两边同时平方得

1-2sinAcosA=t^2

sinAcosA=(1-t^2)/2

故t=(1-t^2)/2

整理得 t^2+2t-1=0

解得t1=-1-√2【舍去】,t2=-1+√2

故sin2A=2sinAcosA=1-t^2=1-(-1+√2)^2=-2+2√2