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

(Ⅰ) f(x)=3sinxcosx+sin2x=32sin2x-12cos2x+12=sin(2x-π6)+12,∵ω=2,∴T=2π2=π,则函数f(x)的最小正周期是π; (Ⅱ)∵x∈[0,π2],∴2x-π6∈[-π6,5π6],∴sin(2x-π6)∈[-12,1],即sin(2x-π6)+12∈[0,32],则f(x)在[0,π2]上的最大值和...

(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)=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)=sinxcosx-√3cosx+√3/2=(1/2)sin2x-√3×(1+cos2x)/2+√3/2=(1/2)sin2x-(√3/2)cos2x =sin2xcosπ/3-sinπ/3cos2x=sin(2x-π/3) 1)最小正周期T=2π/2=π 2)单调增区间:2x-π/3∈[2kπ-π/2,2kπ+π/2], 解得x∈[kπ-π/12,kπ+5π/12] ∴单调增区间为[kπ-π/12,kπ+...

(1)∵f(x)=[2sin(x+π3)+sinx]cosx?3sin2x =2sinxcosx+3cos2x?3sin2x =sin2x+3cos2x=2sin(2x+π3).∴最小正周期T=2π2=π.(2)∵x 0∈[0,5π12],∴2x 0+π3∈[23,7π6],∴sin(2x 0+π3)∈[?12,1],∴f(x0)的值域为[-1,2].∵存在x 0∈[0,5π12],使...

f(x)=2sinx[a?sin(x+π2)+12sinx]-12=2asinxcosx-12(1-2sin2x)=asin2x-12cos2x,∵函数f(x)=2sinx[a?sin(x+π2)+12sinx]-12(x∈R)的图象关于直线x=π3对称,∴当x=π3时,函数f(x)取得最大值或最小值±a2+(?12)2,即asin2π3-12cos2π3=±a2+...

(Ⅰ)f(x)=sin2x+3sin2x+12(sin2x-cos2x)=1?cos2x2+3sin2x-12cos2x,=3sin2x-cos2x+12=2sin(2x-π6)+12,∴f(x)的周期为π,由-π2+2kπ≤2x-π6≤π2+2kπ得:-π6+kπ≤x≤π3+kπ,k∈Z.∴f(x)的单调递增区间为[-π6+kπ,π3+kπ]k∈Z.(Ⅱ)由f(x0)=2...

(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...

(1)f(x)=3(1?cos2x)2+12sin2x?32=12sin2x?32cos2x=sin(2x-π3)∵x∈(0,π2)∴?π3<2x?π3<2π3.∴当?2x?π3=π2时,即x=5π12时,f(x)的最大值为1.(2)由f(x)=sin(2x-π3),若x是三角形的内角,则0<x<π,∴?π3<2x?π3<5π3.令f(x)=12...

(1)∵f(x)=32sin2x+1?cos2x2?52=sin(2x?π6)?2,∴f(x)的最大值为1-2=-1,最小正周期是T=2π2=π. (2)f(C)=sin(2C?π6)?2=?1,∴sin(2C?π6)=1,∵0<C<π,∴0<2C<2π,∴?π6<2C?π6<116π,∴2C?π6=π2,∴C=π3. ∵sinB=2sinA,∴由正弦定理...

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