当x属于[0,π⼀2]则函数=[sin(x+π⼀2)+1]*[cos(π⼀2-x)+1]的最小值为？

函数=[sin(x+π/2)+1]*[cos(π/2-x)+1]=(cosx+1)(sinx+1)=sinxcosx+sinx+cosx+1
令sinx+cosx=t sinxcosx=( t²-1)/2
sinx+cosx = 2sin(x+π/4) 0≤x≤π/2 π/4 ≤x≤3π/4
√ 2/2 ≤ sin(x+π/4) ≤1 1 ≤√ 2sin(x+π/4)≤√2 1 ≤t≤√2
函数=[sin(x+π/2)+1]*[cos(π/2-x)+1]=(cosx+1)(sinx+1)=sinxcosx+sinx+cosx+1
=( t²-1)/2+t+1=1/2(t+1)^2
f(t)=1/2(t+1)^2 在（-1.+ ∞ ）上为增函数 1 ≤t≤√2
所以 函数=[sin(x+π/2)+1]*[cos(π/2-x)+1]的最小值为f(1)=1/2x4=2

y=(cosx+1)(sinx+1)=1+sinx+cosx+sinxcosx
令t=sinx+cosx=√2*sin(x+π/4)
当x∈[0,π/2]时，t∈[1,√2]
则：t²=1+2sinxcosx，即：sinxcosx=t²/2-1/2
所以：y=1+t+t²/2-1/2=1/2(t+1)²
由t∈[1,√2] 得：y∈[2,1.5+√2]
故ymin=2