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已知函数F(x)=sin2x+3sinxsin(x+π2),x∈R.(1...

(1)f(x)=sin2x+3sinxsin(x+π2)=1?cos2x2+32sin2x=sin(2x-π6)+12∴当π2+2kπ≤2x-π6≤3π2+2kπ(k∈Z),时函数单调减,即2π3+kπ≤x≤5π3+kπ(k∈Z),函数单调减,∴函数的单调递减区间为[2π3+kπ,5π3+kπ](k∈Z),(2)∵f(x)=sin(2x-π6)+12∴...

第二问在下图写~

(1)f(x)=cosπ3?cos(2x+π3)=12?cos(2x+π3),所以,f(x)的最大值M=12+1=32.(2)f(A)=1,即12?cos(2A+π3)=1,cos(2A+π3)=?12,因为△ABC是锐角三角形,0<A<π2,π3<2A+π3<4π3,所以2A+π3=2π3,A=π6,所以△ABC的面积S=12×AB×AC×si...

(1)∵f(x)=sin2x+3sinxcosx-1=1?cos2x2+32sin2x-1=32sin2x-12cos2x-12=sin(2x-π6)-12,∴函数f(x)=sin2x+3sinxcosx-1的最小正周期T=2π2=π;(2)∵x∈[-π12,π2],∴2x-π6[-π3,5π6],∴sin(2x-π6)∈[-

f(x)=sin2x-2√3sin^2x+√3+1 =sin2x+√3(-2sin^2x+1)+1 =(sin2x+√3cos2x)+1 =(sin2xcos(π/3)+cos2xsin(π/3))*2+1 =2sin(2x+π/3)+1 最小正周期=π -π/2+2kπ

(1)f(x)=3(1?cos2x)2+12sin2x?32=12sin2x?32cos2x=sin(2x-π3)∵x∈(0,π2)∴?π3<2x?π3<2π3.∴当?2x?π3=π2时,即x=5π12时,f(x)的最大值为1.(2)由f(x)=sin(2x-π3),若x是三角形的内角,则0<x<π,∴?π3<2x?π3<5π3.令f(x)=12...

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)∵f(x)=[2sin(x+π3)+sinx]cosx?3sin2x =2sinxcosx+3cos2x?3sin2x =sin2x+3cos2x=2sin(2x+π3).∴最小正周期T=2π2=π.(2)∵x 0∈[0,5π12],∴2x 0+π3∈[23,7π6],∴sin(2x 0+π3)∈[?12,1],∴f(x0)的值域为[-1,2].∵存在x 0∈[0,5π12],使...

(1)解法一:∵f(x)=1?cos2x2+sin2x+3(1+cos2x)2=2+sin2x+cos2x=2+2sin(2x+π4)(4分)∴当2x+π4=2kπ+π2,即x=kπ+π8(k∈Z)时,f(x)取得最大值2+2.因此,f(x)取得最大值的自变量x的集合是{x|x=kπ+π8,k∈Z}. (8分)解法二:∵f(x)=(sin2...

(1)∵f(x)=sin(π4+x)sin(π4-x)+3sinxcosx=12cos2x+32sin2x…(2分)=sin(2x+π6),…(4分)∴f(π6)=1.…(6分)(2)由f(A2)=sin(A+π6)=1,而0<A<π可得:A+π6=π2,即A=π3.(8分)∴sinB+sinC=sinB+sin(2π3-B)=32sinB+32cosB=3si...

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