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

f(x)=sin²x+√3sinxsin(x+½π) =(1-cos2x)/2+√3sinxcosx =-1/2cos2x+√3/2sin2x+1/2 =sin(2x-π/6)+1/2 最小正周期T=2π/2=π f(x)在区间【0,½π】上的取值范围 因为x∈[0,π/2] 则2x∈[0,π] 则2x-π/6∈[-π/6,5π/6] 所以根据三角函数的性...

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) = 3sin(2x-π/6) + 2 第一问: 2x-π/6属于(2kπ-π/2,2kπ+π/2)时单调增 故,增区间:(kπ-π/6,kπ+π/3),其中k属于Z 第二问: x 属于【-π/12,π/12】 2x 属于【-π/6,π/6】 2x-π/6 属于【-π/3,0】 2x-π/6=0时,最大值 = 3*sin0+2 = 0 + 2...

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)=√3/2sin2x-(cosx)²-½ =√3/2sin2x-1/2cos2x-1 =sin(2x-π/6)-1 T=2π/2=π 2x-π/6在[2kπ+π/2,2kπ+3π/2]是调递减 x在[kπ+π/3,kπ+5π/6]是调递减

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...

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

解: (1)根据题意得 T=2π/w=π得w=2 ∵ 对称轴x=π/3 ∴2*π/3+φ=π/2+kπ -π/2

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