已知x∈［-π6 ，π2 ］，求y=（sinx+1）（cosx+1）的最值
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解：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
=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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