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已知函数 F(x)=Cos2x+2 3 sinxCosx(x∈...

(1)f(x)=cos2x+2 3 sinxcosx=cos2x+ 3 sin2x…(2分)=2( 1 2 cos2x+ 3 2 sin2x)=2sin(2x+ π 6 )…(4分)∴T=π,f(x) 最大值 =2…(6分)(2)先将y=2sinx(x∈R)的图象向左移 π 6 个单位,得到y=2sin(x+ π 6 )的图象;再将y=2sin(x+...

(1)∵2cos2x=1+cos2x,2sinxcosx=sin2x∴f(x)=1+cos2x+sin2x+3=sin2x+cos2x+4=2sin(2x+π4)+4…(5分)函数f(x)的表达式为:f(x)=2sin(2x+π4)+4;(2)函数的最小正周期为T=2π2=π由2x+π4=π2+kπ,解得x=π8+kπ2 (k∈Z)∴函数图象的对称轴方程...

(Ⅰ)f(x)=2cos2x+3sin2x=cos2x+3sin2x+1=2sin(2x+π6)+1令-π2+2kπ≤2x+π6≤π2+2kπ,k∈Z解得,kπ-π3≤x≤kπ+π6,k∈Z∴f(x)的递增区间为[kπ?π3,kπ+π6](k∈Z)(Ⅱ)由f(A)=2sin(2A+π6)+1=2,得A=π3S△ABC=12bcsinA=3∴c=4,由余弦定理得a2=1+4...

函数f(x)=2cos2x+23sinxcosx=1+cos2x+3sin2x=2sin(2x+π6)+1+1所以:函数的周期为:T=π(Ⅱ)由于x∈[?π6,π4]所以:2x+π6∈[?π6,2π3]sin(2x+π6)∈[?12,1]所以函数f(x)的值域为:[0,3]

(1)f(x)=5sinxcosx-53cos2x+532=52sin2x-532(1+cos2x)+532=5sin(2x-π3)…(4分)∴T=π.----(5分)(2)令2kπ-π2≤2x-π3≤2kπ+π2?在[kπ-π12,kπ+5π12](k∈Z),所以函数在[kπ-π12,kπ+5π12](k∈Z)上单调增函数,----(7分)在[kπ+5π12,kπ...

(Ⅰ)f(x)=3cos2x+sin2x=2(sinπ3cos2x+cosπ3sin2x)=2sin(2x+π3),…(5分)∴f(x)的表达式为y=2sin(2x+π3),周期为T=2π2=π…(7分)(Ⅱ)∵x∈[0, π2],∴2x+π3∈[π3, 4π3],…(9分)∴当2x+π3=4π3时,即x=π2时,f(x)的最小值为2sin4π3=-3...

(1)函数f(x)=2cos2x+23sinxcosx?1(x∈R)=cos2x+3sin2x=2(12cos2x+32sin2x)=2sin(π6+2x),∴周期T=2πω=2π2=π.(2)由 2kπ-π2≤π6+2x≤2kπ+π2,k∈z,可得 kπ-π3≤x≤kπ+π6,故函数f(x)单调增区间为[kπ-π3,kπ+π6].(3)∵0≤x≤π2,∴π6≤π6+2x≤7π...

(Ⅰ)∵f(x)=2cosxcos(π6-x)-3sin2x+sinxcosx=3(cos2x-sin2x)+2sinxcosx=3cos2x+sin2x=2sin(2x+π3).∴f(x)的最小正周期为π.(Ⅱ)∵x∈[?π3, π2],∴?π3≤2x+π3≤4π3,又f(x)=2sin(2x+π3),∴f(x)∈[?3, 2],f(x)的值域为[?3, 2].

f(x)=2cos²x+√3sin2x-1 =√3sin2x+cos2x =2sin(2x+π/6) 最小正周期是2π/2=π; 因为x∈[0,π/2],则:2x+π/6∈[π/6,7π/6],则: sin(2x+π/6)∈[-1/2,1],则:f(x)的最小值是f(0)=-1,f(x)的最大值是f(π/6)=2

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)...

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