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

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解 (1)f(x)=sin2x+2sinx?cosx+3cos2x=sin2x+2sinxcosx+3cos2xsin2x+cos2x=tan2x+2tanx+3tan2x+1=175;(2)f(x)=sin2x+2sinx?cosx+3cos2x=sin2x+cos2x+2=2sin(2x+π4)+2,∵ω=2,∴f(x)的最小正周期为T=2π2=π;由π2+2kπ≤2x+π4≤3π2+2kπ,k...

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

(Ⅰ)化简可得f(x)=sin2x+3sinxsin(x+π2)=1?cos2x2+3sinxcosx=32sin2x-12cos2x+12=sin(2x-π6)+12,可得周期T=2π2=π;(Ⅱ)由?π2+2kπ≤2x?π6≤π2+2kπ得?π6+kπ≤x≤π3+kπ,k∈z∴函数f(x)的单调递增区间是[?π6+kπ,π3+kπ],k∈z;(Ⅲ)由x∈[0,2π...

f(x)=2sinxcos(π/2-x)-√3sin(π+x)cosx+sin(π/2+x)cosx =2sinxsinx+√3sinxcosx+cosxcosx =2sin^2x+√3/2*sin2x+cos^2x =sin^2x+√3/2*sin2x+1 =(1-cos2x)/2+√3/2*sin2x+1 =√3/2*sin2x-1/2cos2x+1/2+1 =√3/2*sin2x-1/2cos2x+3/2 =sin2xcosπ/6-cos2xs...

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

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

(1)f(x)=sin2x+3sinxsin(x+π2)=1?cos2x2+32sin2x=sin(2x-π6)+12∴当π2+2kπ≤2x-π6≤3π2+2kπ(k∈Z),时函数单调减,即2π3+kπ≤x≤5π3+kπ(k∈Z),函数单调减,∴函数的单调递减区间为[2π3+kπ,5π3+kπ](k∈Z),(2)∵f(x)=sin(2x-π6)+12∴...

f(x) = sin(π/2-x)sinx - √3cos²x = cosxsinx - √3cos²x = 1/2sin2x - √3/2cos2x - √3/2 = sin2xcosπ/3-cos2xsinπ/3 - √3/2 = sin(2x-π/3) - √3/2 最小正周期:2π/2 = π 最大值:1 - √3/2 = (2-√3)/2

(Ⅰ)∵f(x)=3cos2x+2sinx?sin(x+π2)=3cos2x+2sinx?cosx=3cos2x+sin2x=2sin(2x+π3),…(4分)∴f(x)的最小正周期是π.…(5分)令 2x+π3=π2+2kπ,k∈Z,解得 x=π12+kπ,k∈Z,∴函数f(x)的最大值为2,此时,x值的集合为 {x|x=kπ+π12,k∈z}.…...

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