f(x)=2sin^2x-cos(2x+π/2)
=1-cos2x+sin2x
=√2sin(2x-π/4)+1
1)f(π/8)=√2sin(π/4-π/4)+1=0+1=1
2)最小正周期T=2π/2=π
单调递增区间:2x-π/4∈[2kπ-π/2,2kπ+π/2 ]
得:x∈[kπ-π/8,kπ+3π/8 ] 其中,k为整数
已知函数f(x)=2sin^2x-cos(2x+π/2)
已知函数f(x)=2sin^2x-cos(2x+π/2)
(1)求f(π/8)的值.(2)求函数f(x)的最小正周期及单调递增区间.
(1)求f(π/8)的值.(2)求函数f(x)的最小正周期及单调递增区间.
其他人气:401 ℃时间:2019-11-18 09:28:40
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