knrt.net
当前位置:首页 >> 设F(x)=2AsinxCosx+2BCos2x%B,(A,B∈R+)若F(... >>

设F(x)=2AsinxCosx+2BCos2x%B,(A,B∈R+)若F(...

(1)∵f(x)=2asinxcosx+2bcos2x=asin2x+b(1+cos2x)=asin2x+bcos2x+b,∴f(0)=2b=8,f(π6)=32a+32b=12,解得a=43,b=4;(2)∵f(x)=43sin2x+4cos2x+4=8sin(2x+π6)+4,∴当x∈[0,π2]时,2x+π6∈[π6,7π6],∴-12≤sin(2x+π6)≤1,∴0≤8sin...

(1)f(x)=a*b=2cos??x+√3sin2x =cos2x+√3sin2x+1 =2sin(2x+π/6)+1 ∴f(x)最小正周期T=2π/2=π (2)f(A)=2sin(2A+π/6)+1=2 即sin(2A+π/6)=1/2 ∵0<A<π(三角形的内角) ∴π/6<2A+π/6<13π/6 因此 2A+π/6只能等于5π/6 即 A=(5π/6-π/6)/2=π/3 根...

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...

cos2x=2(cosx)^2-1是通过三角函数公式变换得来的。具体的换算过程如下。 解:根据三角函数两角和公式cos(A+B)=cosA*cosB-sinA*sinB可得, cos2x=cos(x+x)=cosx*cosx-sinx*sinx =(cosx)^2-(sinx)^2 =(cosx)^2-(1-(cosx)^2) =2(cosx)^2-1 所以cos2...

函数f(x)=2sinx1+cos2x=2sinx1+2cos2x?1=2sinx2cos2x=sinx|cosx|,当x=±π2时,函数f(x)无意义.当?π2<x<π2时,cosx>0,此时f(x)=sinxcosx=tanx,当?π≤x<?π2或π2<x≤π时,cosx<0,此时f(x)=?sinxcosx=?tanx.故对应的函数图象为B...

f(x)-g(x)=(1-a)cosx-(1+a)=0 (1-a)cosx-(1+a)=0 cosx=(1+a)/(1-a) 则 -10 a

1,f(x)=a·b=4sinx*2cosx-4√6(cosx+sinx)*cos(x+π/4) =4sin(2x)-4√6(cosx+sinx)*(√2/2cosx-√2/2sinx) =4sin(2x)-4√3[(cosx)^2-(sinx)^2] =8[1/2sin(2x)-√3/2cos(2x)] =8[sin(2x)cosπ/3-cos(2x)sinπ/3] =8sin(2x-π/3)。 当x∈[π/4,π/2]时,π/6

f(x)=a*b =√3cosxsinx-1/2cos2x =√3/2sin2x-1/2cos2x =sin(2x-π/6) 最小正周期为; T=2π/2=π 2.∵x∈[0,π/2] ∴2x-π/6∈[-π/6,5π/6] ∴当2x-π/6=-π/6时 f(x)取得最小值,f(x)=-1 当2x-π/6=π/2时 f(x)取得最大值,f(x)=1

解 f(x)=a*b =√3cosxsinx-1/2cos2x =√3/2(2sinxcosx)-1/2cos2x =√3/2sin2x-1/2cos2x =sin2xcosπ/6-sinπ/6cos2x =sin(2x-π/6) 最小正周期为; T=2π/2=π ∵x∈[0,π/2] ∴2x-π/6∈[-π/6,5π/6] ∴ 当2x-π/6=-π/6时 f(x)取得最小值,f(x)=-1 当2x-π/6=π/2...

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