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解:
f(x)=cos²x+sinx
=1-sin²x+sinx
=-(sinx-1/2)²+5/4
若sinx=1/2,即x=π/6+2kπ或者x=5π/6+2kπ,必定是最大值
∵[2π/5,3π/4]可分为[2π/5,π/2)∪[π/2,3π/4]
∵5π/6>3π/4,π/6<2π/5,且5π/6-3π/4<2π/5-π/6
故最大值在x=5π/6处
∴最大值为:
f(3π/4)=-[sin(3π/4)-1/2]²+5/4
=-[sin(π/4+π/2)-1/2]²+5/4
=-[cos(π/4)-1/2]²+5/4
=-[√2/2-1/2]²+5/4
=-(1/2+1/4-√2/2)+5/4
=-3/4+√2/2+5/4
=1/2+√2/2
f(x)=cos²x+sinx
=1-sin²x+sinx
=-(sinx-1/2)²+5/4
若sinx=1/2,即x=π/6+2kπ或者x=5π/6+2kπ,必定是最大值
∵[2π/5,3π/4]可分为[2π/5,π/2)∪[π/2,3π/4]
∵5π/6>3π/4,π/6<2π/5,且5π/6-3π/4<2π/5-π/6
故最大值在x=5π/6处
∴最大值为:
f(3π/4)=-[sin(3π/4)-1/2]²+5/4
=-[sin(π/4+π/2)-1/2]²+5/4
=-[cos(π/4)-1/2]²+5/4
=-[√2/2-1/2]²+5/4
=-(1/2+1/4-√2/2)+5/4
=-3/4+√2/2+5/4
=1/2+√2/2
更多追问追答
追问
能把sinX设为T吗 我这样看不懂 为什么要带3/4π
追答
解:
f(x)=cos²x+sinx
=1-sin²x+sinx
=-(sinx-1/2)²+5/4
设T=sinx
则f(x)=-(T-1/2)²+5/4
即f(x)转化为二次函数,对称轴为T=1/2
当T的值越接近1/2,f(x)的值越大。
∵[2π/5,3π/4]可分为[2π/5,π/2)∪[π/2,3π/4]
①在区间[2π/5,π/2)上,y=sinx是增函数
∵π/6sin(π/6)=1/2,即sin(2π/5)>1/2
②在区间[π/2,3π/4]上,y=sinx是减函数
∵3π/4sin(5π/6)=1/2,即sin(3π/4)>1/2
综合①②所述,T>1/2,故寻找T最接近1/2的值
∵5π/6-3π/4<2π/5-π/6
∴sin(3π/4)比sin(2π/5)更接近1/2
∴最大值为:
f(3π/4)=-[sin(3π/4)-1/2]²+5/4
=-[sin(π/4+π/2)-1/2]²+5/4
=-[cos(π/4)-1/2]²+5/4
=-[√2/2-1/2]²+5/4
=-(1/2+1/4-√2/2)+5/4
=-3/4+√2/2+5/4
=1/2+√2/2
如果哪里还不明白,可以继续问我,或者在hi里问我
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