已知函数f(x)=1 2sin(2x π 6)(1)求f(x)的振幅
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①原式=f(x)=2cos2x+sinx^2=2cos2x+1-cos2x/2=3/2cos2x+1/2故f(π/3)=3/2*cos2π/3+1/2=-3/4+1/2=-1/4②依f(x)=3/2c
f(x)=sin²x+sinxcosx=[1-cos(2x)]/2+sin(2x)/2=sin(2x)/2-cos(2x)/2+1/2=(√2/2)sin(2x-π/4)+1/2最小正周期T
其图像经过点M(π/3,1/2)代入f(x)=sin(x+φ)1/2=sin(π/3+φ)∵0<φ<π∴π/3<π/3+φ<4π/3∵1/2=sin(π/3+φ)∴π/3+φ=
1.sin(x-π/6)2.可知半周期为2π/3,又在区间(0,π/3)上是增函数,故ω>0,2π/ω=4π/3从而ω=1.5
1、由于函数g(x)=sin(2(x-a)+π/3)为偶函数,所以g(x)的图像关于y轴对称,即函数g(x)当x=0时取得最值,所以g(0)=±1,解得sin(π/3-2a)=±1,sin(2a-π/
∵f(x)=2sin(π-x)cosx=2sinxcosx=sin2x1、最小正周期T=2π/2=π.2、∵-π/6≤x≤π/2∴-π/3≤2x≤π,∴-√3/2≤f(x)≤1,∴最大值1,最小值-√
f(x)=sin2x-2sin^2x=sin2x+cos2x-1=√2sin(2x+π/4)-1.(1)T=2π/2=π.(2).当2x+π/4=2kπ+π/2,k∈Z,即x=kπ+π/8,k∈Z时,
f(x)=cosx+sinxf(x)=√2sin(x+π/4)(1)递增区间:2kπ-π/2≤x+π/4≤2kπ+π/2得:2kπ-3/4π≤x≤2kπ+π/4递增区间是:[2kπ-3π/4,2kπ+
函数g(x)=sin(x+π6),f(x)=2cosx•g(x)−12=32sin2x+12cos2x=sin(2x+π6).(1)函数f(x)的最小正周期T=π,因为2x+π6=kπ,所以对称中心坐
f(-11π/12)=4cos^4(11π/12)-2cos(11π/6)+1/sin(7π/6)sin(2π/3)sin2π/3sin(7π/6)=-sinπ/6sinπ/3=-√3/42cos11
函数f(x)=sin(ωx+π4)的图象向左平移π6个单位后得到函数f(x)=sin(ωx+π4+ωπ6)的图象,由已知可知,它的与函数g(x)=sin(ωx+π6)的图象重合,所以π4+ωπ6=2k
f(x)=sin2x+cos2x-1=√2sin(2x+π/4)-1.1、最小正周期是π,最大值时2x+π/4=2kπ+π/2,即x=kπ+π/4,k是整数.再问:已知函数f(x)=2sin(∏-X)
f(-1)+f(1)=sin(-1)cos(-1)+sin1cos1=-sin1cos1+sin1cos1=0
因为f(x)=sinx+cosx=√2sin(x+π/4)第一题T=2π/1=2π第二题当sin(x+π/4)=1时,为最大值,即f(x)=√2sin(x+π/4)=-1时,为最小值,即f(x)=-√
由题意得f(x)=2sinωxcosωx+23sin2ωx−3=sin2ωx−3cos2ωx=2sin(2ωx−π3)…(2分)由周期为π,得ω=1.得f(x)=2sin(2x−π3)…(4分)由正弦
∵x∈[0,π3],∴π3≤x+π3≤2π3,根据正弦函数的性质得,32≤sin(x+π3)≤1,则3≤2sin(x+π3)≤2,∴f(x)的值域是[3,2].故答案为:[3,2].
f(x)=√3sin(2x-π/6)+2sin^2(x-π/12)=√3sin(2x-π/6)-cos(2x-π/6)+1=2sin[(2x-π/6)-π/6]+1=2sin(2x-π/3)+1(1)
(1)∵f(x)=3sinπx+cosπx=2(32sinπx+12cosπx)=2sin(πx+π6),∴函数f(x)的最小正周期T=2ππ=2,又∵x∈R,∴−1≤sin(πx+π6)≤1,∴−2
1:(sinwx)^2+√3sinwxsin(wx+π\2)=(sinwx)^2+√3sinwxcoswx=2[(sinwx)^2+(√3\2)sin2wx]\2=[2(sinwx)^2+√3sin2
(Ⅰ)f(x)=2cosx•sin(x−π6)−12=3sinxcosx-cos2x-12=32sin2x−12cos2x-1=sin(2x−π6)-1∴f(x)的最小值是-2,最小正周期为T=2π2