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?
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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?
![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?](/uploads/image/z/19291050-18-0.jpg?t=%E5%BD%93x%E5%B1%9E%E4%BA%8E%5B0%2C%CF%80%2F2%5D%E5%88%99%E5%87%BD%E6%95%B0%3D%5Bsin%28x%2B%CF%80%2F2%29%2B1%5D%2A%5Bcos%28%CF%80%2F2-x%29%2B1%5D%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA%3F)
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
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
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