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设函数F(x)=23sinxCosx?2sin2x+1(x∈R),则F(x)...

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

依题意得:f(x)=cos2x-sin2x+23sinxcosx=cos2x+3sin2x=2sin(2x+π6),(1)∵x∈R,∴f(x)max=M=2,最小正周期T=2π2=π;(2)由f(xi)=M=2得:2xi+π6=2kπ+π2,k∈Z,解得:xi=kπ+π6,k∈Z,又0<xi<10π,∴k=0,1,2,…,9,∴x1+x2+…+x10=(1+2...

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(...

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

(1)∵f(x)=32sin2x-1+cos2x2+12=32sin2x-12cos2x=sin(2x-π6),∴f(x)的最小正周期为T=2π2=π.(2)由f(A2+π3)=13,得sin[2(A2+π3)-π6]=sin(A+π2)=cosA=13,∴在△ABC中,sinA=223,又因为a=2,b=1,∴由正弦定理asinA

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

(1)代入x=π/4,y=2 m(√2/2+√2/2)^2+1-2(√2/2)^2=2 m=1 (2)f(x)=(cosx+sinx)^2+1-2(sinx)^2 =1+sin2x+cos2x =1+√2sin(2x+π/4) 所以f(x)的最小值为1-√2 此时2x+π/4=2kπ+3π/2 x=kπ+5π/8 f(x)取最小值时x的集合为{x|x=kπ+5π/8,k∈Z}

解:f(x)=½sin(2x)+(√3/2)(cosx-sinx)=½sin(2x)+(√6/2)cos(x+π/4)cos(x+π/4)的最小正周期T₁=2π/1=2πf(x+2π)=½sin[2(x+2π)]+(√6/2)cos(x+2π+π/4)=½sin(2x+4π)+(√6/2)cos(x+2π+π/4)=½sin(2x)+(√6/2)cos(x+π/4)=f(x...

f(x)=2cosx(sinx-cosx)+1 =2sinxcosx-2(cosx)^2+1 =sin2x-cos2x =根号2sin(2x-Pai/4) 故最小正周期T=2Pai/2=Pai 最大值=根号2,最小值=-根号2

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