已知函数F(x)=Cosxsin(x+π/6)%Cos2x%1/4,x∈R求F(x)的单调递增区间

f(x)=cosxsin(x+π=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]单调递增 所以有:x∈[kπ-π/12,kπ+5π/12]单调递增 因x∈[-π/6,π/4] 所以,f(x)=√3/2sin(2x-π/3) ,在2x-π/3∈[-2π/3,π/6]上 最大值=√3/4 最小值=-3/4

f(x)=4cosxsin(x+∏/6)-1=4cosx(√3/2sinx+1/2cosx)-1=2√3sinxcosx+2cos^2x-1=√3sin2x+cos2x=2sin(2x+π/6)此函数的最小正周期t=2π/2=π

解由fx=-f(x+π)知t=2π又由t=2π/w=2π即w=1故fx=sin(x+φ)又由f0=1/2则sin(0+φ)=1/2即φ=π/6故fx=sin(x+π/6)故gx=2cos(wx+φ)=2cos(x+π/6)由x属于(0,π/2)知2x属于(0,π)即2x+π/6属于(π/6,7π/6)故2x+π/6=π时,函数gx=2cos(wx+φ)=2cos(x+π/6)有最小值y=2cosπ=2*(-1)=-2.

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(Ⅰ)∵f(x)=4cosxsin(x+π6)-1=4cosx(32sinx+12cosx)-1=3sin2x+cos2x=2sin(2x+π6),∴f(x)的最小正周期T=2π2=π;(Ⅱ)∵x∈[-π6,π4],∴2x+π6∈[-π6,2π3],∴-12≤sin(2x+π6)≤1,-1≤2sin(2x+π6)≤2.∴f(x)max=2,f(x)min=-1.

解:(1)f(x)=4cosxsin(x+π/6)-1 =4cosx(√3/2sinx+1/2cosx)-1=2√3sinxcosx+2cosx-1=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6)所以f(x)的最小正周期t=2π/2=π;(2)x∈[-π/6,π/4]2x+π/6∈[-π/6,2π/3],当2x+π/6=π/2时,f(x)取到最大值2;当2x+π/6=-π/6时,f(x)取到最小值-1.

(1)f(x)=4cosx(sinxcosπ/6+sinπ/6cosx) -1 =2√3cosxsinx+2cosxcosx-1 =√3sin2x+cos2x =2(sin2xcosπ/6+sinπ/6cos2x) =2sin(2x+π/6)由此可得:∵ω=2∴ω/2π=π ∴最小正周期为π (2)由题意得:f(x)在[-π/6,π/6]单调递增,在[π/6,π/4]单调递减, ∴当x=π/6时,f(x)有最大值,最大值为2,当x= -π/6时,f(x)有最小值,最小值为-1

f(x)=4cosx*sin(x+π/6)-1 =4cosx(sinx*√3/2+cosx*1/2)-1 =2√3sinxcosx+2(cosx)^2-1 =√3sin2x+cos2x =2sin(2x+π/6) ①:w=2,T=2π/w=π ②:令t=2x+π/6,因为x∈[-π/6,π/4],推出t∈[-π/6,2π/3], 问题转化为求2sint在[-π/6,2π/3]的最大值和最小值,一画图,显然得到max=2,min=-1.这种问题,用这种方法是绝对不会错的.首先感谢楼上的人,我就不化简了,直接借用楼上化简的式子,但是关键部分的处理,楼上没有把这个思路写出来,等于就做了一个化简,就得到了答案.

1.)f(x)=4cosxsin(x+π/6)-1=2√3cosxsinx+2cosxcosx-1=√3sin2x+cos2x=2sin(2x+π/6)对称轴为2x+π/6=kπ+π/2对称轴为x=kπ/2+π/6.k为整数2)x在区间(-π/6,π/4),则2x+π/6在区间(-π/6,2π/3),2x+π/6=π/2时,f(x)= 2sin(2x+π/6) 取最大值2,2x+π/6=-π/6时,f(x)= 2sin(2x+π/6) 取最小值-1

1、 解:f(x)=4cosx(根号3/2*sinx+1/2*cosx)-1 =根号3*sin(2x)+2*(cosx)^2-1 =根号3*sin(2x)+cos(2x] =sin(2x+兀/3) 故最小正周期=2兀/2=兀 2、解:原式=2sin(x+兀/4)sin(兀/2一兀/4+x) =2sin(x+兀/4)sin(x+兀/4) =1-cos(2x+兀/2) =1+sin(2x) 故最小正周期=2兀/2=兀

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