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

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)=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

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(...

(1) f(x)=2 3 sinxcosx+2co s 2 x+a= 3 sin2x+(2co s 2 x-1)+a+1 …(2分)= 3 sin2x+cos2x+a+1=2sin(2x+ π 6 )+a+1 …(5分)所以f(x) max =a+3=1,得a=-2.…(7分)(2)由(1)得 f(x)=2sin(2x+ π 6 )-1 ,因为f(x)≥0,所以, sin(2x+ π...

f(x)=√3sinxcosx+cos²x =(√3/2)·2sinxcosx+½(2cos²x-1)+½ =(√3/2)sin2x+½cos2x+½ =sin(2x+π/6) +½ 最小正周期T=2π/2=π

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

f(x)=2cosxcos(x?π6)?3sin2x+sinxcosx=2cosx(32cosx+12sinx)-3sin2x+sinxcosx=3(cos2x-sin2x)+2sinxcosx=3cos2x+sin2x=2(32cos2x+12sin2x)=2sin(2x+π3),(1)∵ω=2,∴T=2π2=π;(2)∵f(x)=1,即2sin(2x+π3)=1,∴sin(2x+π3)=12,...

f(x)=sin^2x+√3sinxcosx+2cos^2x =cos^2x+√3/2*2sinxcosx+1 =1/2cos2x+√3/2sin2x+3/2 =sinx(2x+π/6)+3/2 T=2π/2=π 2x+π/6在[2kπ-π/2,2kπ+π/2]上单调递增 x在[kπ-5π/12,kπ+π/6]上单调递增 2)y=sin2x的图像经X轴向左平移π/12个单位得到y=sinx(2x+...

(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)

求到2x+π/3∈[π/3,4π/3]这一步 2x+π/3=π时,sin最大=1 2x+π/3=4π/3时,sin最小=-√3/2

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