已知函数f(x)=2sin(x-π/3)cosx+sinxcosx+√3sin^2x(x属于R),求f(x)的单调递增区间
已知函数f(x)=2sin(x-π\/3)cosx+sinxcosx+√3sin^2x(x属于R),
f(x)=2sin(x-π\/3)cosx+sinxcosx+√3sin^2x =2(sinxcosπ\/3-cosxsinπ\/3)cosx+sinxcosx+√3sin^2x =sinxcosx-√3cos^2x+sinxcosx+√3sin^2x =2sinxcosx-√3(cos^2x-sin^2x)=sin2x-√3cos2x =2(1\/2sin2x-√3\/2cos2x)=2sin(2x-π\/3)由2kπ-π\/2≤2x-π\/3≤2kπ+...
已知函数f(x)=2sin(x-π\/3)cosx+sinxcosx+√3sin^2x
把括号展开,sin和cos的积换成sin的2倍角,在把sin 和cos的和合并为sin,括号里看成整体求区间;角B对应AC,角A对应BC,根据角B推出角A的范围,在用正弦定理解。。。
已知函数f(x)=2cosxsin(x+π\/3)-根号3sin^2x+sinxcosx+2(x∈R),
=2sin(2x+π\/3)所以最小正周期是π 单调增区间[-5π\/12+kπ,π\/12+kπ],k是整数。
已知函数fx=2sin(x-派\/3)cosx+sinxcosx+根号(3)sin^2 x
答:f(x)=2sin(x-π\/3)cosx+sinxcosx+√3(sinx)^2 =sin(x-π\/3+x)+sin(x-π\/3-x)+sinxcosx+(√3\/2)(1-cos2x)=sin(2x-π\/3)-√3\/2+(1\/2)sin2x+√3\/2-(√3\/2)cos2x =sin(2x-π\/3)+sin(2x-π\/3)=2sin(2x-π\/3)所以:单调递增区间满足:2kπ-π\/2<=2x-π...
已知函数f(x)=2cosxsin(x+π\/3)-√3sin²x+sinxcosx(x属于R)
=2sin[2x+(π\/3)]∴当2x+(π\/3)=π\/2时,函数f(x)有最大值且f(x)=2,此时,x=π\/12;当2x+(π\/3)=3π\/2时,函数f(x)有最小值且f(x)=-2,此时,x=7π\/12;又∵sinx的单调减区间为:2kπ+(π\/2)≤x≤2kπ+(3π\/2);∴2kπ+(π\/2)≤2x+(π\/3)≤2kπ+(3...
函数f(x)=[2sin(x+π\/3)+sinx]cosx-根3sin^2x,(x∈R)。1)求函数f(x...
f(x)=[2sin(x+π\/3)+sinx]cosx-√3sin^2x =[sinx+√3cosx+sinx]cosx-√3sin^2x =2sinxcosx+√3cos^2x-√3sin^2x =sin2x+√3cos2x =2sin(2x+π\/3)f(x)的最小正周期为π 根据题意,m只需大于f(x)在该区间的最小值就行 当x=5π\/12时,f(x)取最小值1 所以m>1 ...
已知f(x)=[2sin(x+3分之派)+sinx]cosx-根号3sin的平方x,x属于R
f(x)=[2sin(x+π\/3)+sinx]cosx-根号3sin^2x和差角公式展开f(x)=(sinx+根号3cosx+sinx)cosx-根号3sin2x=2sinx*cosx+根号3cosx*cosx-根号3sin2x降幂公式f(x)=sin2x+根号3*(1+cos2x)\/2-根号3sin2x=(1-根号3)sin2x+根号3\/2cos2x+根号3\/2 辅助角公式 不改变三角函数的周期 ...
已知函数f(x)=[2sin(x+π\/3)+sinx]cosx-根号3sin^2x,x属于R,求函数f...
f(x)=[2sin(x+π\/3)+sinx]cosx-根号3sin^2x 和差角公式展开 f(x)=(sinx+根号3cosx+sinx)cosx-根号3sin2x =2sinx*cosx+根号3cosx*cosx-根号3sin2x 降幂公式 f(x)=sin2x+根号3*(1+cos2x)\/2-根号3sin2x =(1-根号3)sin2x+根号3\/2cos2x+根号3\/2 辅助角公式 不改变三角函数...
已知函数f(x)=2cosxsin(x+π\/3)+sinxcosx-√3sin²x,x∈R
=2 sinxcosx+√3(cos²x-sin²x)=sin2x+√3cos2x =2sin(2x+π\/3)(1)最小正周期 =2π\/2=π (2)x∈[0,5π\/12]2x+π\/3∈[π\/3,7π\/6]sin(2x+π\/3)∈[-1\/2,1]2sin(2x+π\/3)∈[-1,2]f(x)>m成立 m<-1 如果您认可我的回答,请点击“采纳为满意...
函数f(x)=[2sin(x+π3)+sinx]cosx?3sin2x,(x∈R).(1)求函数f(x)的最...
=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f(x)<m成立,∴m>-1,故实数m的取值范围为(-1,+∞).