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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) f(x)=cos2x+ 3 sin2x =2(sin π 6 cos2x+cos π 6 sin2x)= 2sin(2x+ π 6 ) ∴M=2,T= 2π 2 =π(2)当 2x+ π 6 ∈[ π 2 +2kπ, 3π 2 +2kπ] ,即 x∈[ π 6 +kπ, 2π 3 +kπ](k∈R) 时,f(x)单调递减.∴f(x)的单调递减区间为 [ π 6 +kπ, 2π ...

(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...

(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...

(1)f(x)=3cos2x+2sinxcosx=3cos2x+sin2x=2sin(2x+π3),T=2π2=π,(2)∵x∈[0,π4],∴2x+π3∈[π3,5π6],∴12≤sin(2x+π3)≤1∴1≤f(x)≤2,即函数的值域为[1,2]

(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...

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]

(Ⅰ)f(x)=cos2x+3sin2x=2sin(2x+π6),令2kπ-π2≤2x+π6≤2kπ+π2,k∈Z,得到kπ-π3≤x≤kπ+π6,k∈Z,则f(x)的单调递增区间为[kπ-π3,kπ+π6],k∈Z;(Ⅱ)由f(A)=2sin(2A+π6)=1,即sin(2A+π6)=12,得:2A+π6=5π6,即A=π3,由余弦定理a2=b2+c...

(法一)f(x)=3×1+c0s2x2+12sin2x?32=12sin2x+32cos2x=sin(2x+π3)(I )f(π4)=sin(2×π4+π3 )=cosπ3=12(II)函数的基本性质如下①奇偶性:函数f(x)既不是奇函数,也不是偶函数②单调性:函数f(x)的单调增区间为[kπ?5π12,kπ+π12] k∈Z,单...

(Ⅰ)∵函数f(x)=2(cos2x+3sinxcosx)+1=cos2x+3sin2x+2=2sin(π6+2x)+2,故它的最小正周期等于 2π2=π.令 2kπ-π2≤π6+2x≤2kπ+π2,k∈z,可得kπ-π3≤x≤kπ+π6,k∈z,故函数的单调增区间[kπ?π3,kπ+π6](k∈Z).(Ⅱ)当x∈[0,π2]时,π6+2x∈[π6,7π6],s...

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