Question

f(x)=1-sinx-cosx+2sinx cosx最大值和最小值

f(x)=1-sinx-cosx+2sinx cosx最大值和最小值

Answer 1

令sinx+cosx=t (-√2<=t<=√2)
sin²x+cos²x+2sinxcosx=t²
2sinxcosx=t²-1
f(t)=1-t+t²-1
=t²-t
=(t-1/2)²-1/4
当t=1/2时f(1/2)min=-1/4
当t=-√2时f(-√2)max=2+√2

Answer 2

解：由sin^2a+cos^2a=1,有
f(x)=(sin^2a+2sinx cosx+cos^2a)-(sinx+cosx)
=(sinx+cosx)^2+(sinx+cosx)
又根据三角函数y=sinx，y=cosx的性质，有
f(x)max=2+√2(当x=2kπ+π/4,k∈Z)；f(x)min=2-√2（当x=2kπ-3π/4,k∈Z）