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已知函数F(x)=23sinxCosx%2sin2x+1(x∈R).(1...

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

f(x)=sin²x+sinxcosx+1 =(1-cos2x)/2+sin2x/2+1 =½(sin2x-cos2x)+3/2 =√2/2sin(2x-π/4)+3/2 ∴最小正周期是π 单调递增区间2x-π/4∈(2kπ-π/2,2kπ+π/2)→x∈(kπ-π/8,kπ+3π/8) 单调递减区间2x-π/4∈(2kπ+π/2,2kπ+3π/2)→x∈(kπ+3π/8,kπ+7π/8)

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

(sinx-cosx)sinx =(sinx)^2-sinxcosx =[-1+2(sinx)^2]/2+1/2-(sin2x)/2 =(-1/2)(cos2x+sin2x)+1/2 =(-√2/2)sin(2x+φ)+1/2 =(-√2/2)sin[2(x+φ/2)]+1/2 ∴最小正周期是(2π)/2=π

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

fx=4√3sinxcosx-4sin2x+1 =2√3sin2x-4sin2x+1 =(2√3-4)sin2x+1 f'x=(4√3-8)cos2x 单调增区间:f'(x)>0 ∴cos2x

f(x)=2√3sinxcosx-2sin²x+1 =√3sin2x+cos2x =2sin(2x+π/6) (2) 2kπ-π/2

f(x)=(1-cos2x)/2+√3/2sin2x+cos2x+1 =1/2cos2x+√3/2sin2x+3/2 =sin(2x+π/6)+3/2 1. 最小正周期T=2π/2=π 单调增区间 2kπ-π/2

(1)g(x)=2sin2x+sin2x-1=sin2x-cos2x=2sin(2x?π4)先将f(x)的图象向右平移π4个单位长度得到y=sin(x?π4)的图象;再将y=sin(x?π4)图象上各点的横坐标变为原来的12倍,得到函数y=sin(2x?π4)的图象;最后将曲线上各点的纵坐标变为原来的2倍...

(1)∵f(x)=2sin2x-23cos2x+1=4sin(2x-π3)+1.又∵π4≤x≤π2,∴π6≤2x-π3≤2π3,即3≤4sin(2x-π3)+1≤5∴f(x)max=5,f(x)min=3(2)∵|f(x)-m|<2,∴m-2<f(x)<m+2又p是q的充分不必要条件∴m?2<3m+2>5,∴3<m<5.∴m的取值范围为(3,5)

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