已知x屬于【-π/6,π/2）.求函數(shù)y=（sinx+1)*(cosx+1)的最值

數(shù)學(xué)人氣：731 ℃時(shí)間：2020-09-26 06:58:38

優(yōu)質(zhì)解答

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?yàn)閤∈［-π/6 ,π/2 ］,所以：
x+π/4∈[π/12,3π/4]
則當(dāng)x+π/4＝π/2即x＝π/4時(shí),函數(shù)y有最大值3/2 +√2
當(dāng)x+π/4＝π/12即x＝－π/6 時(shí),函數(shù)y有最小值1/2 +√3/4
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