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已知F(x)=12sin2x?Cosπ6+12Cos2xsinπ6(1)函数F...

f(x)=12(sin2xcosπ6+cos2xsinπ6)=12sin(2x+π6),(1)函数f(x)的最小正周期为T=2π2=π,最大值为12.(2)由?π2+2kπ≤2x+π6≤π2+2kπ,k∈z得,?π3+kπ≤x≤π6+kπ,k∈z,∴f(x)的单调递增区间为[-π3+kπ,π6+kπ],k∈Z,(3)由f(α2)=12得sin(α+π6...

(1)∵函数f(x)=12sin2xsinφ+cos2xcosφ-12sin(π2+φ),(0<φ<π),又因其图象过点(π6,12),∴12=12sin(2×π6)sinφ+cos2π6cosφ-12sin(π2+φ),(0<φ<π)解得Φ=π3,∴f(x)=12sin2xsinφ+cos2xcosφ-12sin(π2+φ)=12sin(2x+π6)由2x+π6=k...

(1)因为f(x)=12sin2xsinθ+1+cos2x2cosθ?12cosθ=12(sin2xsinθ+cos2xcosθ)=12cos(2x?θ)…(3分)由f(π6)=12得cos(π3?θ)=1∴π3?θ=2kπ又θ∈(0,π),∴θ=π3…(5分)∴f(θ)=12cosθ=14…(6分)(2)因为函数y=f(x)的图象上各点的横坐标缩短到...

f(x)=-12+sin(π6-2x)+cos(2x-π3)+cos2x=cos(2x+π3)+cos(2x-π3)+1+cos2x2-12=2cos2xcosπ3+cos2x2=32cos2x,故f(x)的周期是T=2π2=π.(2)∵x∈[-π8,3π8],2x∈[-π4,3π4],故当x=0时,f(x)的最大值是32

(1)f(x)=12sin2xsin?+1+cos2x2cos?? 12cos?=12cos(2x??)(0<?<π)∴2×π6??=kπ∴?=π3(2)f(A)=12cos(2A?π3)=34A∈(π6,π2)则2A?π3=π6所以A=π4由S△ABC=12bcsinA=2b4=12得b=2由余弦定理得a2=b2+c2-2bccosA=1所以a=1

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...

∵f(x)=sin(2x+π6)-2cos2x+1=sin(2x+π6)-cos2x=32sin2x-12cos2x=sin(2x-π6),∴f(x)的周期T=2π2=π,最大值为1.

(Ⅰ)因为f(x)=sin(2x?π6)+2cos2x?1=32sin2x?12cos2x+cos2x=32sin2x+12cos2x=sin(2x+π6)所以函数f(x)的单调递增区间是〔kπ?π3,kπ+π6〕(k∈Z)(Ⅱ)因为f(A)=12,所以sin(2A+π6)=12又0<A<π所以π6<2A+π6<13π6从而2A+π6=5π6故A=π3在△AB...

(1)f(x)=sin2x+cos(2x?π6)=sin2x+cos2xcosπ6+sin2xsinπ6=32sin2x+32cos2x=3(32sin2x+12cos2x)=3sin(2x+π6)…(4分)∴最小正周期T=2π2=π,…(5分)减区间:2kπ+π2≤2x+π6≤2kπ+3π2,k∈Z解得2kπ+π3≤2x≤2kπ+4π3,k∈Z所以单调减区间为[kπ+π6,kπ...

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