nynw.net
当前位置:首页 >> 已知F(x)=12sin2x?Cosπ6+12Cos2xsinπ6(1)函数F... >>

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)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...

(1)依题意,f(x)=2(32sinx?12cosx)(12sinx+32cosx)=sinxcosx-32(cos2x-sin2x) …(3分)=12sin2x?32cos2x=sin(2x-π3),…(5分)因为π6≤x≤5π12,所以0≤2x?π3≤π2,从而0≤sin(2x?π3)≤1,所以函数f(x)的值域为[0,1];…(7分)(2)依题意...

(Ⅰ)f(x)=sin(2x+π6)-1+cos2x2-12cos2x+12=32sin2x+12cos2x-cos2x=32sin2x-12cos2x=sin(2x-π6),∵ω=2,∴T=π,∵x∈[0,π2],∴2x-π6∈[-12,1],则f(x)在区间[0,π2]上的取值范围是[-12,1];(Ⅱ)f(B)=sin(2B-π6)=1,由0<B<π,得-π6...

∵y=sin2xsinπ6-cos2xcos5π6=12sin2x-(-32)cos2x=sin(2x+π3),∴最小正周期T=2π2=π.故答案为:π.

(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)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...

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

网站首页 | 网站地图
All rights reserved Powered by www.nynw.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com