nynw.net
当前位置:首页 >> 设函数F(x)=Cos(2x+π6)+sin2x,(x∈R)(1)求... >>

设函数F(x)=Cos(2x+π6)+sin2x,(x∈R)(1)求...

(1)f(x)=sin2x+cos2x?32+sin2x?12=sin2x?32+cos2x?32=3sin(2x+ π 6). ∴f(x)的最小正周期T=2 π 2= π. (2)∵f(B)=32, ∴sin(2B+ π 6)=12. 又∵x∈(0, π 2), ∴2x+ π 6∈( π 6, 7π 6),∴2B+ π 6= 5π 6,故B= π 3. 在△ABC中,由余弦...

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)= 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。

(1)函数f(x)=sin(2x+π6)+sin(2x-π6)+2cos2x=sin2xcosπ6+cos2xsinπ6+sin2xcosπ6-cos2xsinπ6+1+cos2x=sin2x+cos2x+1=2sin(2x+π4)+1.由2kπ+π2≤2x+π4≤2kπ+3π2,解得kπ+π8≤x≤kπ+5π8(k∈Z).∴函数f(x)的单调递减区间[kπ+π8,kπ+5π8](k∈Z...

(I)∵函数f(x)=sin(7π6?2x)+2cos2x?1=sin7π6cos2x-cos7π6sin2x+cos2x=32sin2x+12cos2x=sin(2x+π6).故函数f(x)的周期为T=2π2=π.再令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,求得 kπ-π3≤x≤kπ+π6,k∈z,故单调递增区间为[kπ-π3,kπ+π6],k∈z.(II)...

见图 解:(I)f(x)==sin2x+cos2x=sin(2x+). 令 2kπ-≤(2x+)≤2kπ+,可得 kπ-≤x≤kπ+,k∈z. 即f(x)的单调递增区间为[kπ-,kπ+],k∈z. (II)在△ABC中,由,可得sin(2A+)=,∵<2A+<2π+, ∴<2A+= 或,∴A= (或A=0 舍去). ∵b,a,c成...

(1)∵f(x)=2cos2(π4?x)+sin(2x+π3)?1=12sin2x+32cos2x+cos(π2?2x)=32sin2x+32cos2x=3sin(2x+π6)∴函数f(x)的最小正周期是T=2π2=π;(2)当x∈[0,π2]时,2x+π6∈[π6,7π6],∴sin(2x+π6)∈[?12,1]∴3sin(2x+π6)∈[?

(1)f(x)=2cos2x-sin(2x-7π6)=1+cos2x+sin(2x-π6)=1+sin(2x+π6).令2kπ?π2≤2x+π6≤2kπ+π2,k∈Z可解得kπ?π3≤x≤kπ+π6,k∈Z故函数f(x)的单调递增区间为:[kπ?π3,kπ+π6],k∈Z.(2)∵x∈[-π6,π3]∴解得-π6≤2x+π6≤5π6∴由正弦函数的单调性知...

(Ⅰ)化简可得f (x)=12cos2x-32sin2x+3sin2x+2a=12cos2x+32sin2x+2a=sin(2x+π6)+2a.由2kπ-π2≤2x+π6≤2kπ+π2解得kπ-π3≤x≤kπ+π6(k∈Z).∴f (x)的单调递增区间为:[kπ-π3,kπ+π6](k∈Z).(Ⅱ)∵0≤x≤π4,∴π6≤2x+π6≤2π3,∴12≤sin(2x+π6)≤1...

∵f(x)=cos(2x-π3)+sin(2x+π6)=cos(π3-2x)+sin(2x+π6)=2sin(2x+π6).∴y=f(x)是以π为最小正周期的周期函数.命题①正确;当x=-π3时,f(?π3)=2sin[2×(?π3)+π6]=2sin(?π2)=?2.∴y=f(x)的一条对称轴为x=-π3.命题②正确;由π2+2kπ≤2...

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