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已知函数f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)
且函数y=f(x)的图像相邻两条对称轴之间的距离为π/2,(1)求f(π/6)的值 (2)若函数f(kx+π/12)(k>0)在[-π/6,π/3]上单调递增,求k的取值范围
人气:197 ℃ 时间:2019-08-20 20:06:52
解答
f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)=(1-cos2ωx)/2+(√3/2)sin2ωx=sin(2ωx-π/6)+1/2∵函数y=f(x)的图像相邻两条对称轴之间的距离为π/2即半周期=π/2∴T=2π/2ω=π,∴ω=1,∴f(x)=sin(2x-π/6)+1/2(1...
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