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已知函数F(x)=2sinx?Cos2θ2+Cosx?sinθ%sinx(0<...

(1) f(x)=2sinx? 1+cosθ 2 +cosx?sinθ-sinx=sin(x+θ) ∵当x=π时,f(x)取得最小值∴sin(π+θ)=-1即sinθ=1又∵0<θ<π,∴ θ= π 2 (2)由(1)知f(x)=cosx∵ f(A)=cosA= 3 2 ,且A为△ABC的内角∴ A= π 6 由正弦定理得 sinB= bsinA a = 2 2 知 B...

1)f(x)=2sinxcos^2φ/2+cosxsinφ-sinx, f(x)=sinx(2cos^2φ/2-1)+cosxsinφ =sinxcosφ+cosxsinφ =sin(x+φ) 在x=π处取得最小值(0

f(x)=2sinxcosx-sin²x+cos²x =sin2x+cos2x =√2sin(2x+π/4) 在[0, π]上,2x+π/4∈[π/4, 2π+π/4] 其中单调增区间为: 2x+π/4 ∈[π/4, π/2]U[3π/2,9π/4] 单调减区间: 2x+π/4 ∈[π/2, 3π/2] 即f(x)的单调增区间为:[0, π/8]U[5π/8, π] f(x)...

f(x)=√5[sinx*(1/√5)-cosx*(2/√5)]=√5sin[x-arccos(1/√5)] sin[x-arccos(1/√5)]=1时,x-arccos(1/√5)=2kπ+π/2, k∈N cosθ=cos(2kπ+π/2+arccos(1/√5)=-sin[arccos(1/√5)]=-(2√5)/5

f(X)=cos2X+sin2X =√2(sinπ/4cos2X+cosπ/4sin2X) =√2sin(2X+π/4) ∵X属于[0,π/2], ∴-√2/2≤sin(2X+π/4)≤1, ∴-1≤f(X)≤√2。 即当X=π/2时,f(X)最小=-1, 当X=π/4时,f(X)最大=√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)=sin²x+2sinxcosx+3cos²x =(1-cos2x)/2+sin2x+3(1+cos2x)/2 =√2sin(2x+π/4)+2 (I) 函数f(x) 的最小正周期T=2π/2=π ∵x∈(0,π) ∴2x+π/4∈(π/4,9π/4) 当2x+π/4=π/2.即x=π/x时,f(x)=2+√2为最大值, 当2x+π/4=3π/2.即x=5π/8时,f(x)...

1)f(x)=2sinxcos^2φ/2+cosxsinφ-sinx=f(x)=sinx(2cos^2φ/2-1)+cosxsinφ=sinxcosφ+cosxsinφ=sin(x+φ) 在x=π处取得最小值(0

(1)f(x)=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6) 所以T=2π/2=π 因为x∈[0,π/2] ==>2x+π/6∈[π/6,7π/6] 根据正弦曲线的单调性 f(x)max=2×1 当2x+π/6=π/2时取得 f(x)max=2×(-1/2)=-1 当2x+π/6=7π/6时取得 (2)由题意可得2sin(2x1+π/6)=...

亲,你传个手写的吧,这个看不太明白

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