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已知函数Fx=Cos2x%sin2x+4sinx·Cosx,求F(x)的最小...

解:f(x)=cos2x-sin2x+2sin2x. =cos2x+sin2x. ∴f(x)=√2sin(2x+π/4). 函数f(x)的最小值正周期T=2π/=π. 当sin(2x+π/4)=1时,函数f(x)取得最大值f(x)max=√2,此时x=2kπ+π/8, k∈Z.

∵f(x)=cos2x+23sinxcosx-sin2x=1+cos2x2+3sin2x?1?cos2x2=3sin2x+cos2x=2sin(2x+π6)∴函数f(x)的最小正周期T=2π2=π,值域为[-2,2].

(Ⅰ)f(x)=cos2x+3sin2x=2sin(2x+π6),(6分)∴f(π12)=2sin(π6+π6)=2sinπ3=3.(8分)(Ⅱ)由(Ⅰ)可知f(x)=2sin(2x+π6),∴函数f(x)的最小正周期T=2π2=π.(11分)函数f(x)的最大值为2.(13分)

(1)函数f(x)=4cos2x+43sinxcosx-1=2cos2x+23sin2x+1=4sin(2x+π6)+1,∴函数f(x) 的最小正周期T=2π2=π,∵sin(2x+π6)≤1,∴f(x)≤4+1=5,因此函数f(x)的最大值为5,对应的自变量x的取值集合为{x|x=kπ+π6,k∈Z}.(2)∵在△ABC中,a,b,c...

原式=f(x)=1+cos2x2+12sin2x=22(22sin2x+22cos2x)+12=22sin(2x+π4)+12,(1)f(3π4)=22sin(2×3π4+π4)+12=0;(2)T=2π2=π,∵y=sinx的递增区间为[2kπ?π2,2kπ+π2],∴2kπ?3π4≤2x≤2kπ+π4,即

(1)f(x)=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6) 所以T=2π/2=π 因为x∈[0,π/2] ==>2x+π/6∈[π/6,7π/6] 根据正弦曲线的单调性 f(x)max=2×1 当2x+π/6=π/2时取得 f(x)max=2×(-1/2)=-1 当2x+π/6=7π/6时取得 (2)由题意可得2sin(2x1+π/6)=...

(1)f(x)=12(1+cos2x)+12sin2x=12+12(sin2x+cos2x)=12+22sin(2x+π4),∵ω=2,∴T=π;∵-1≤sin(2x+π4)≤1,∴sin(2x+π4)的最小值为-1,则f(x)的最小值为1?22;(2)f(α+3π8)=12+22sin(2α+π)=12-22sin2α=2?64,∴sin2α=32,∵α∈(π4,...

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=2sin(2x-π4),∴T=2π2=π.(2)∵x∈[0,π2],∴2x-π4∈[-π4,3π4],∴sin(2x-π4)∈[-22,1],∴函数的值域为[-1,2].

(1)y=cos2x+sinxcosx=1+cos2x2+12sin2x=22 sin(2x+π4)+12∴T=2π2=π,由 2kπ?π2≤2x+π4≤π2+2kπ k∈Z,即 kπ?3π8≤x≤π8+kπ k∈Z,所以函数的单调增区间为:[?38π+kπ,π8+kπ] (k∈Z).(2)g(x)=2sin(x+π4)sin(x?π4)=-sin(2x+π2)=-cos2x,因为f(x...

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