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已知函数F(x)=3sinxsin(x+π2)+sin2x(x∈R)....

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]

因为f(x)=[2sin(x+π3)+sinx]cosx?3sin2x=[2(sinxcosπ3+sinπ3cosx)+sinx]cosx-3sin2x=sin2x+3cos2x=2sin(2x+π3)于是(I)函数f(x)的最小正周期T=2π2=π.(II)因为x∈[0,π4]∴π3≤2x+π3≤5π6∴12≤sin(2x+π3)≤ 1即:1≤y≤2∴f(x)max=2,f(x)min=1

(I)f(x)=1?cos2x2+32sin2x+(1+cos2x)=32sin2x+12cos2x+32=sin(2x+π6)+32.∴f(x)的最小正周期T=2π2=π.由题意得2kπ?π2≤2x+π6≤2kπ+π2,k∈Z,即kπ?π3≤x≤kπ+π6,k∈Z.∴f(x)的单调增区间为[kπ?π3,kπ+π6],k∈Z.(II)先把y=sin2x图象上所有...

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

∵A+π属于(π/2,3π2) 又sin(A+π/4)=7根号2/10 ∴cos(A+π/4)=-根号2/10 sinA=sin[(A+π/4)-π/4]=4/5 f(x)=cos2x+5/2*4/5*sinx=1-2sin²x+2sinx=-2(sinx-1/2)²+3/2 当sinx=1/2时f(x)max=3/2 当sinx=-1时f(x)min=-3

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...

一、使用排除法 A、当x=0,sin2x=0,f(sin2x)=f(0)=sinx =sin0 =0 而当x=π/2时,sin2x=0,f(sin2x)=f(0)=sinx =sin(π/2)=1 出现了f(0)=0和f(0)=1,不符合函数定义,错误。 B、当x=0,sin2x=0,f(sin2x)=f(0)=x^2+x=0 而当x=π/...

(1)解法一:∵f(x)=1?cos2x2+sin2x+3(1+cos2x)2=2+sin2x+cos2x=2+2sin(2x+π4)(4分)∴当2x+π4=2kπ+π2,即x=kπ+π8(k∈Z)时,f(x)取得最大值2+2.因此,f(x)取得最大值的自变量x的集合是{x|x=kπ+π8,k∈Z}. (8分)解法二:∵f(x)=(sin2...

f(x)=cos(x)-2sin2x+1,当x=0时,最大,f(x)=1-0+1=2;第二问找出x0/2和cos2x0的关系即可。希望可以帮到你

(Ⅰ)f(x)=32sin2x-12-12cos2x+m=sin(2x-π6)+m-12,∵点M(π12,0)在函数f(x)的图象上,∴sin(2×π12-π6)+m-12=0,解得:m=12,∴f(x)=sin(2x-π6),由2kπ-π2≤2x-π6≤2kπ+π2,k∈Z,得kπ-π6≤x≤kπ+π3,k∈Z,∴函数f(x)的单调增区间为[kπ-π...

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