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已知函数f(x)=sin(x-π/6)cosx,x属于(0,π/2) (1)求函数f(x)的值域 (2)若曲线f(x)在x0处的切线的倾斜角
已知函数f(x)=sin(x-π/6)cosx,x属于(0,π/2)
(1)求函数f(x)的值域
(2)若曲线f(x)在x0处的切线的倾斜角α属于[arctan1/2,π/4],求x0的取值范围
人气:232 ℃ 时间:2020-09-29 08:18:40
解答
(1)f(x)=(sinxcosπ/6+cosxsinπ/6)cosx=sinxcosxcosπ/6+cos²xsinπ/6=1/2*sin2xcosπ/6+1/2(cos2x+1)sinπ/6=1/2(sin2xcosπ/6+cos2xsinπ/6)+1/2sinπ/6=1/2sin(2x+π/6)+1/4∵-1第二问有点问题什么问题?请在追问中说明,或hi我提出问题第一问答案是[-1/2,1/4] 我算的是(-1/2,1/4]第二问答案是[π/12,π/4]我算的是(0,π/4]求正解(1)不好意思,x-π/6 我当x+π/6计算了,还漏看:x属于(0,π/2)这个了。正解如下:f(x)=(sinxcosπ/6-cosxsinπ/6)cosx=sinxcosxcosπ/6-cos2xsinπ/6=1/2*sin2xcosπ/6-1/2(cos2x+1)sinπ/6=1/2(sin2xcosπ/6-cos2xsinπ/6)-1/2sinπ/6=1/2sin(2x-π/6)-1/4x∈(0,π/2)∴-π/6<2x-π/6<5π/6∴-1/2
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