y=√2sin(x+π/4)-1/tanx+cotx,求值域

1个回答

  • y=√2sin(x+π/4)-1/tanx+cotx=√2sin(x+π/4)-1/(sinx/cosx+cosx/sinx)

    =sinx+cosx-sinxcosx=sinx+cosx-[(1+2sinxcosx)-1]/2

    =sinx+cosx-[(sinx+cosx)^2-1]/2

    =-1/2*[(sinx+cosx)^2-2(sinx+cosx)-1]

    =-1/2*(sinx+cosx-1)^2+1

    =-1/2*[√2sin(x+π/4)-1]^2+1

    所以,函数的最大值是1

    最小值是-1/2*(-√2-1)^2+1=-1/2*(3+2√2)+1=-(3+2√2-2)/2=-(1+2√2)/2