nynw.net
当前位置:首页 >> 已知函数F(x)=23sinxCosx+1%2sin2x,x∈R.(1)... >>

已知函数F(x)=23sinxCosx+1%2sin2x,x∈R.(1)...

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

(1)∵f(x)=sin2x+23sinxcosx+3cos2x,∴f(x)=1+3sin2x+2cos2x=3sin2x+cos2x+2=2sin(2x+π6)+2,故函数的最小正周期为T=2π2=π,由2kπ-π2≤2x+π6≤2kπ+π2,可得kπ-π3≤x≤kπ+π6,故函数单调递增区间为:[kπ-π3,kπ+π6](k∈Z)(2)∵x∈[-π6,π3],...

(1)f(x)=2sin2x+23sinxcosx+1=1-cos2x+3sin2x+1=2sin(2x-π6)+2∴f(x)的最小正周期T=2π2=π;(2)令π2+2kπ≤2x-π6≤3π2+2kπ(k∈Z)解得-π3+kπ≤x≤5π6+kπ(k∈Z),因此,f(x)的单调递减区间为[-π3+kπ,5π6+kπ],(k∈Z)(3)当x∈[0,π2]时,...

(1)f(x)=cos2x?sin2x+23sinxcosx+1=3sin2x+cos2x+1=2sin(2x+π6)+1.(4分)因此f(x)的最小正周期为π,最小值为-1.(6分)(2)由F(a)=2得2sin(2α+π6)+1=2,即2sin(2x+π6)=12,而由a∈[π4,π2],得2a+π6∈[23π,76π].(9分)故2a+π6=56...

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

求到2x+π/3∈[π/3,4π/3]这一步 2x+π/3=π时,sin最大=1 2x+π/3=4π/3时,sin最小=-√3/2

已知函数f(x)=-(√2)sin(2x+π/4)+6sinxcosx-2cos²x+1,x∈R; (1).求f(x)的最小正周期;(2).求f(x)在区间[0,π/2]上的最大值和最小值. 解:(1)。f(x)=-(sin2x+cos2x)+3sin2x+cos2x=2sin2x,故最小正周期T=π; (2)。在区间[0,π/2]上的最大...

f(x)=2√3sinxcosx-2sin²x+1 =√3sin2x+cos2x =2sin(2x+π/6) (2) 2kπ-π/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