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已知函数F(x)=sin2x?Cos(2x?π6),其中x∈R.(1)...

(1)f(x)=sin2x?cos2xcosπ6?sin2xsinπ6=12sin2x?32cos2x=sin2xcosπ3?cos2xsinπ3=sin(2x?π3)∴最小正周期T=2π2=π(2)由题意,解不等式?π2+2kπ≤2x?π3≤π2+2kπ得 ?π12+kπ≤x≤5π12+kπ,(k∈Z)∴f(x)的递增区间是[?π12+kπ,5π12+kπ](k∈Z)

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

(1)f(x)=sin2x+cos2x?32+sin2x?12=sin2x?32+cos2x?32=3sin(2x+ π 6). ∴f(x)的最小正周期T=2 π 2= π. (2)∵f(B)=32, ∴sin(2B+ π 6)=12. 又∵x∈(0, π 2), ∴2x+ π 6∈( π 6, 7π 6),∴2B+ π 6= 5π 6,故B= π 3. 在△ABC中,由余弦...

由题意得,f(x)=32cos2x-12sin2x+sin2x=32cos2x+12sin2x=sin(2x+π3),(1)f(x)的最小正周期T=2π2=π;(2)由x∈[0,π2]得,2x+π3∈[π3,4π3],当2x+π3=π2时,此时x=π12,函数f(x)取到最大值1,当2x+π3=4π3时,此时x=π2,函数f(x)取到最...

(1)函数f(x)=sin(2x+π6)+sin(2x-π6)+2cos2x=sin2xcosπ6+cos2xsinπ6+sin2xcosπ6-cos2xsinπ6+1+cos2x=sin2x+cos2x+1=2sin(2x+π4)+1.由2kπ+π2≤2x+π4≤2kπ+3π2,解得kπ+π8≤x≤kπ+5π8(k∈Z).∴函数f(x)的单调递减区间[kπ+π8,kπ+5π8](k∈Z...

(I)∵函数f(x)=sin(7π6?2x)+2cos2x?1=sin7π6cos2x-cos7π6sin2x+cos2x=32sin2x+12cos2x=sin(2x+π6).故函数f(x)的周期为T=2π2=π.再令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,求得 kπ-π3≤x≤kπ+π6,k∈z,故单调递增区间为[kπ-π3,kπ+π6],k∈z.(II)...

见图 解:(I)f(x)==sin2x+cos2x=sin(2x+). 令 2kπ-≤(2x+)≤2kπ+,可得 kπ-≤x≤kπ+,k∈z. 即f(x)的单调递增区间为[kπ-,kπ+],k∈z. (II)在△ABC中,由,可得sin(2A+)=,∵<2A+<2π+, ∴<2A+= 或,∴A= (或A=0 舍去). ∵b,a,c成...

函数f(x)=sin(2x-π6)-2cos(x-π4)cos(x+π4)+1=32sin2x?12cos2x?2(22cosx+22sinx)(22cosx?22sinx)+1=32sin2x?12cos2x?cos2x +1=

(1)函数f(x)=sin(2x-π3)+cos(2x-π6)+2cos2x-1=sin2xcosπ3-cos2xsinπ3+cos2xcosπ6+sin2xsinπ6+cos2x (3分)=sin2x+cos2x (4分)=2sin(2x+π4) (5分)所以函数f(x)的最小正周期T=2π2=π.(6分)(2)∵f(x)在区间[-π4,π4]上是增...

(1)f(x)=sin(2x+π6)+sin(2x?π6)?cos2x+a=3sin2x-cos2x+a=2sin(2x?π6)+a.故函数f(x)的最小正周期为T=π,由 2x-π6=kπ+π2,k∈z,求得 对称轴方程为 x=kπ2+π3(k∈Z).(2)当x∈[0,π2]时,-π6≤2x-π6≤5π6,f(x)min=2sin(-π6)+a=-1+a=-2,所...

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