已知x属于【-π/6,π/2）.求函数y=（sinx+1)*(cosx+1)的最值
问题描述：
已知x属于【-π/6,π/2）.求函数y=（sinx+1)*(cosx+1)的最值
答
y=（sinx+1)(cosx+1）
=sinx+cosx+sinxcosx+1
=√2*sin(x+π/4)+1/2 *sin2x+1
=√2*sin(x+π/4) - 1/2 *cos(2x+π/2) +1
=√2*sin(x+π/4) - 1/2 *[1-2sin²(x+π/4)] +1
=sin²(x+π/4)+√2*sin(x+π/4)+1/2
=[sin(x+π/4)+√2/2]²
因为x∈［-π/6 ,π/2 ］,所以：
x+π/4∈[π/12,3π/4]
则当x+π/4＝π/2即x＝π/4时,函数y有最大值3/2 +√2
当x+π/4＝π/12即x＝－π/6 时,函数y有最小值1/2 +√3/4
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