Question

函数y=(1-sinx)(1-cosx)的最大值为

Answer 1

y=(1-sinx)(1-cosx)=1-sinx-cosx+sinxcosx
又2sinxcosx=（sinx+cosx）^2-1
所以设sinx+cosx=t
则sinxcosx=(t^2-1)/2
sinx+cosx=√2sin(x+π/4) 得到sinx+cosx属于[-√2,√2]
所以t属于[-√2,√2]
y=1-t+(t^2-1)/2=t*t/2-t+1/2 对称轴是x=1
所以当t=-√2时，y最大，最大是3/2+√2

Answer 2

y=(1-sinx)(1-cosx)

let y'=sinx-cosx+(cosx)^2-(sinx)^2=0

let tan(x/2)=u

2u/(1+u^2)-(1-u^2)/(1+u^2)+(2u)^2/(1+u^2)^2-(1-u^2)^2/(1+u^2)^2=0

(u+1)(u^2+2u-1)=0

u=-1,u=-1±√2

x=2arctan(-1)=3π/2+2nπ n∈Z

y=(1-sin(3π/2))(1-cos(3π/2))=2

x=2arctan(-1+√2)=π/4+2nπ 

y=(1-sin(π/4))(1-cos(π/4))≈0.09

x=2arctan(-1-√2)=-3π/4+2nπ

y=(1-sin(-3π/4))(1-cos(-3π/4))=3/2+√2

 

Ymax=y(3π/2)=3/2+√2