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已知函数F(x)=sin2x+2sinxsin(π2?x) +3sin2(3π2?...

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

亲,网友,您说的是不是下面的问题: 已知函数f x=sin(兀/2-x)sinx-根号3cos^2x,求周期、最值。 f(x)=1/2 sin2x-√3/2(cos2x+1) =sin(2x-π/3)-√3/2 T=2π/2=π。 f max=1-√3/2, f min=-1-√3/2. 送您 2015 夏祺 凉快

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

最小正周期2π/2=π 单调减区间 2kπ+π/2≤2x+π/6≤2kπ+3π/2 2kπ+π/3≤2x≤2kπ+4π/3 kπ+π/6≤x≤kπ+2π/3 y=sin(2x+π/6)+3/2 =sin[2(x+π/12)]+3/2 因此将sin2x向左移π/12得到sin[2(x+π/12)] 再向上移3/2个单位即可

1、 f(x)=sinx(sinxcosπ/6+cosxsinπ/6) =√3/2*sin²x+1/2*sinxcosx =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4+1/2*(1/2*sin2x-√3/2*cos2x) =√3/4+(sin2xcosπ/3-cos2xsinπ/3) =√3/4+sin(2x-π/3) 所以T=2π/2=π 2、 0≤x≤π/2 -π/3≤2x-π/3≤2π/3 -√3...

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。

对于函数f(x)=sin(2x+3π2)=-cos2x,它的周期等于2π2=π,故A正确.由于f(-x)=-cos(-2x)=-cos2x=f(x),故函数f(x)是偶函数,故B正确.令x=π4,则f(π4)=sin(2×π4+3π2)=0,故f(x)的一个对称中心,故C错误.由于0≤x≤π2,则0≤2x≤π,由于...

f(x)=sin(2x+π/3)+2sinxcosx+1 = sin2xcosπ/3+cos2xsinπ/3+sin2x+1 = 1/2sin2x+√3/2cos2x+sin2x+1 = 3/2sin2x+√3/2cos2x+1 = √3(sin2xcosπ/6+cos2xsinπ/6)+1 = √3sin(2x+π/6)+1 2x+π/6∈(2kπ+π/2,2kπ+3π/2)其中k∈Z时,单调减 即:x+π/12∈(kπ...

已知函数f(x)= 3cos ^ 2(WX)+(平方根3)sinwxcoswx +的(W> 0),和函数f(x)的相邻的图像之间的距离的两个坐标轴的对称π / 2 (1)的W (2)当x属于[π/ 6,5π/12] F(X)的最低值2,找到一个值 />(3)求函数在区间[0,π/ 2]间隔 特别要求...

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