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已知函数F(x)=23sinxCosx+2sin2x?1,x∈R.(I)求...

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...

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

(Ⅰ)∵f(x)=sinxcosx+sin2x,∴f(π4)=sinπ4cosπ4+sin2π4,…(1分)=(22)2+(22)2 …(4分)=1.…(6分)(Ⅱ)f(x)=sinxcosx+sin2x=12sin2x+1?cos2x2,…(8分)=12(sin2x?cos2x)+12=22sin(2x?π4)+12,…(9分)由x∈[0,π2]得 2x?π4∈[?π4,3π4],...

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)=√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

(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)=sinx(sinxcosπ/6+cosxsinπ/6) =√3/2*sin²x+1/2*sinxcosx =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4+1/2*(1/2*sin2x-√3/2*cos2x) =√3/4+(sin2xcosπ/3-cos2xsinπ/3) =√3/4+sin(2x-π/3) 所以T=2π/2=π 2、 0≤x≤π/2 -π/3≤2x-π/3≤2π/3 -√3...

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