已知函数y=2sin(2x+π/3) 求:1.振幅,周期,初相 2.求他的对称轴方程及单调递增区间

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已知函数y=2sin(2x+π/3) 求:
1.振幅,周期,初相 2.求他的对称轴方程及单调递增区间

1 A=2 T=2π/2=π 初相=π/3 2 2x+π/3=π/2+kπ 所以对称轴为x=π/12+kπ/2 -π/2+2kπ≦2x+π/3≦π/2+2kπ 所以-5π/12+kπ≦x≦π/12+kπ 所以递增区间为[-5π/12+kπ,π/12+kπ ]

1) A=2 T=2π/2=π 初相=π/3 2)由题意得2x+π/3=π/2+kπ ∴对称轴为x=π/12+kπ/2 ∵ -π/2+2kπ≦2x+π/3≦π/2+2kπ ∴-5π/12+kπ≦x≦π/12+kπ ∴递增区间为[-5π/12+kπ,π/12+kπ ]