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

(1)f(x)=2sinx?1+cosθ2+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=π2(2)由(1)知f(x)=cosx∵f(A)=cosA=32,且A为△ABC的内角∴A=π6由正弦定理得sinB=bsinAa=22知B=π4或B=3π4当B=π...

(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...

(Ⅰ)f(x)=2sinx1+cosθ2+cosxsinθ?sinx=sinx+sinxcosθ+cosxsinθ-sinx=sin(x+θ).因为f(x)在x=π时取最小值,所以sin(π+θ)=-1,故sinθ=1.又0<θ<π,所以θ=π2;(Ⅱ)由(Ⅰ)知f(x)=sin(x+π2)=cosx.因为f(A)=cosA=32,且A为△ABC的...

(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)=2sinxcos2φ2+cosxsinφ-sinx,∴f(x)=2sinx?1+cosφ2+cosxsinφ-sinx=sinx+sinxcosφ+cosxsinφ-sinx=sin(x+φ),∴f(x)=sin(x+φ),∵函数f(x)在x=π处取最小值.且0<φ<π,∴φ=π2.(2)根据(1)得f(x)=sin(x+π2)=cosx,∴f(...

(1)∵f(x)=sinxcos2φ2+cosxsinφ-sinxcosφ+cosxsinφ=sin(x+φ)∴sin(π+φ)=-1,又∵0<φ<π∴?=π2.(2)∵f(B)=?22,∴sin(B+π2)=cosB=?22∵0<B<π,∴B=3π4,∵asinA=bsinB?sinA=12,又∵A∈(0,π4),∴A=π6,C=π?A?B=π12,∴2sin(3C?θ)+si...

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。

化简 f(x)=sin(x+p) p=2kπ - 5π/2 k属于Z

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

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