- 求函数y=(sinx*cosx)/(sinx
- 求y=(sinx*cosx)/(sinx-cosx+1) (0
- 令sinx-cosx=t,
两边平方得:1-2sinxcosx=t^2
sinxcosx=[1-(t^2)]/2
则y={[1-(t^2)]/2}/(t+1)
=(1-t)/2(t≠-1)
又t=sinx-cosx
=(√2)sin[x-(π/4)]
由0<x<π,得:-(π/4)<x-(π/4)<(3π/4)
-(√2)/2<sin[x-(π/4)]≤1
则 -1<t≤√2
由 y=(1-t)/2(t≠-1)
得 (1-√2)/2≤y<1
即值域为 [(1-√2)/2,1)