函数y=1/2sinx+根号3/2cox的单调性x∈[0,π/2]
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解决时间 2021-02-23 14:46
- 提问者网友：缘字诀
- 2021-02-22 15:37
函数y=1/2sinx+根号3/2cox的单调性x∈[0,π/2]
最佳答案
- 五星知识达人网友：酒醒三更
- 2021-02-22 17:13
y=1/2sinx+√3/2cosx
=sinxcosπ/3+cosxsinπ/3
=sin(x+π/3)
x∈[0,π/2]
x+π/3∈[π/3，5π/6]
x+π/3∈（π/3，2π/2）时单调增；x+π/3∈（π/2，5π/6）时单调减
∴单调增区间（0，π/6）；单调减区间（π/6，π/2)
=sinxcosπ/3+cosxsinπ/3
=sin(x+π/3)
x∈[0,π/2]
x+π/3∈[π/3，5π/6]
x+π/3∈（π/3，2π/2）时单调增；x+π/3∈（π/2，5π/6）时单调减
∴单调增区间（0，π/6）；单调减区间（π/6，π/2)
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