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

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)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

(Ⅰ) f(x)=sin(2x+ π 6 )+cos2x =sin2xcos π 6 +cos2xsin π 6 +cos2x = 3 2 sin2x+ 3 2 cos2x= 3 ( 1 2 sin2x+ 3 2 cos2x)= 3 sin(2x+ π 3 ).令 2kπ- π 2 ≤2x+ π 3 ≤2kπ+ π 2 ,k∈z,求得 kπ- 5π 12 ≤x≤kπ+ π 12 ,函数f(x)的单调递增...

(1)由条件知12=34sinφ+14cosφ=12sin(φ+π6)∴φ+π6=π2?φ=π3(2)由(1)代入得f(x)=12sin2x32+cos2x12-12cosφ=12sin2x32+1+cos2x212-14=12sin(2x+π6)∴函数g(x)=12sin(4x+π6)∴函数y=g(x)的周期为T=π2递减区间为[π12+12kπ,π3+12kπ] &(k∈Z)

f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1 =2√3sinxcosx+2cos²x-1+2 =√3sin2x+cos2x+2 =2(sin2x*√3/2+cos2x*1/2)+2 =2(sin2xcosπ/6+cos2xsinπ/6)+2 =2sin(2x+π/6)+2 所以T=2π/2=π -π/6

(Ⅰ)∵函数f(x)=cos2(x-π6)-sin2x,∴f(π12)=cos2(-π12)-sin2π12=cosπ6=32. …(5分)(Ⅱ)∵f(x)=12[1+cos(2x-π3)]-12(1-cos2x)…(7分)=12[cos(2x-π3)+cos2x]=12(32sin2x+32cos2x) …(8分)=32sin(2x+π3). …(9分)因为 x∈[0,π2],所以 2x+π3∈...

多一个sin(2x+π/6)吧! f(x)=sin(2x+π/6)+2cos²x =sin2xcosπ/6+cos2xsinπ/6+1+cos2x =√3/2sin2x+3/2cos2x+1 =√3(1/2sin2x+√3/2cos2x)+1 =√3sin(2x+π/3)+1 函数最小正周期T=2π/2=π

f(x)=5(sinπ/6sin2x-cosπ/6cos2x) = 5sin(2x-π/6) sin(2x-π/6)=±1时,2x-π/6=kπ+π/2 ∴对称轴:x=kπ/2+π/3,其中k属于Z sin(2x-π/6)=0时,2x-π/6=kπ,x=kπ/2+π/12 ∴对称中心:(kπ/2+π/12,0),其中k属于Z

(1)f(x)=2sinφcos2x+cosφsin2x-sinφ=sinφ(1+cos2x)+cosφsin2x-sinφ=sin2xcosφ+cos2xsinφ=sin(2x+φ).∵x=π6时f(x)求得最大值,∴2×π6+φ=2kπ+π2,即φ=2kπ+π6.又因0<φ<π,所以=π6.于是函数f(x)的解析式为f(x)=sin(2x+π6),其最小...

f(x)= sin2x+cos(2x-π/6) = sin2x+ (√3/2)cos2x-(1/2)sin2x =(√3/2)cos2x+(1/2)sin2x =sin(2x+π/3) (1) f(x)最大值为1,当 x=kπ+π/12 时取得, k为整数。 (2) y=f(x) 的图像可由 y=sinx 的图像周期缩小一半,再向左平移 π/3 得到。

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