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

题目

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

答案

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)/2sinx+cosx = 2sin(x+π/4) 0≤x≤π/2 π/4 ≤x≤3π/4√ 2/2 ≤ sin(x+π/4) ≤1 1 ≤√ 2sin(...