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2sin2x 3sinx 0

2sinx 与 sin2x 同阶, 同阶的两个函数相减,再用同阶的函数代替,不就减完了么?? 只有在它们相除时,可以用同阶的代替 。

f(x)=2sin²x+2√3sinx×sin(x+π/2) =1-cos2x+2√3sinxcosx =1-cos2x+√3sin2x =2(√3/2*sin2x-1/2*cos2x)+1 =2sin(2x-π/6)+1 最小正周期:T=2π/2=π 行家正解,不明白可以追问!祝您学习进步 满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢

f(x)=2cosx(3sinx?cosx)+1=2sin(2x-π6)∴要得到函数f(x)=2sin(2x-π6),需将函数y=2sin2x的图象向右平移π12k(k∈N)个单位,∴φ的最小值为π12故答案为:π12

f(x)= - √3sin^2(x)+sinxcosx = -√3[(1-cos2x)/2]+(1/2)sinxcosx =(1/2)sin2x+(√3/2)cos2x-(√3/2) =sin2xcos60º+cos2xsin60º - (√3/2) f(x)=sin(2x+60º)-(√3/2) f(x/2)=sin[2(x/2)+60º] - (√3/2) =sin(x+60º) - (√3/2)...

因为f(x)=3sinxsin(x+π2)+sin2x=3sinxcosx+1?cos2x2=32sin2x?12cos2x+12=sin(2x?π6)+12,x∈[0,2π3]?2x?π6∈[?π6,7π6]?sin(2x?π6)∈[?12,1],故f(x)∈[0,32]故答案为:[0,32]

f(x)=√3sin2x+2sin²x =√3sin2x-(1-2sin²x)+1 =√3sin2x-cos2x+1 =2[sin2xcos(π/6)-cos2xsin(π/6)]+1 =2sin(2x-π/6)+1. ∵-1≤sin(2x-π/6)≤1, ∴sin(2x-π/6)=-1时,f(x)|min=-1. sin(2x-π/6)=1时,f(x)|max=3。

cos²x-sin²x-√3sin2x =cos2x-√3sin2x =2×[(1/2)×cos2x-(√3/2)×sin2x] =2×(cos60°×cos2x-sin60°×sin2x) =2cos(2x+60°)>0, 2x+60°∈(-180°+360°k,180°+360°k),k∈Z, x∈(-120°+180°k,60°+180°k),k∈Z

∵f(x)=sin2x-2sinxcosx+3cos2x=1-2sinxcosx+2cos2x=1+1+cos2x-sin2x=-(sin2x-cos2x)+2=-2sin(2x-π4)+2.∴把函数f(x)的图象沿x轴向左平移m(m>0)个单位,得到函数g(x)的图象的解析式为:g(x)=-2sin(2x+2m-π4)+2.∵函数g(x)的图象关...

f(x)=2sin^2x+2√3sinxcosx-1 =1-cos2x+√3sin2x-1 =√3sin2x-cos2x =2sin(2x-π/6) f(x)是正弦函数,其中心对称点位于函数与x轴的交点上 根据题意,f(x)的图像关于点(x,0)对称 所以2sin(2x-π/6)=0 2x-π/6=kπ x=kπ/2+π/12,其中k是任意整数

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