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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∴...

(Ⅰ)f(x)=3cos2x+2sin(3π2+x)sin(π-x)=3cos2x-2cosxsinx=3cos2x-sin2x=2(32cos2x-12sin2x)=2cos(2x+π6),∴T=2π2=π,令2x+π6=kπ(k∈Z),即x=kπ2-π12(k∈Z),∴函数f(x)的对称轴方程为x=kπ2-π12(k∈Z),(Ⅱ)∵f(x)=2cos(2x+π6)...

f(x)=(sinx+3cosx+sinx)cosx--3sin2x=2sinxcosx+3cos2x-3sin2x=sin2x+3cos2x=2sin(2x+π3),(Ⅰ)令2kπ-π2≤2x+π3≤2kπ+π2,k∈Z,解得:kπ-5π12≤x≤kπ+π12,k∈Z,则函数f(x)的单调递增区间为[kπ-5π12,kπ+π12],k∈Z;(Ⅱ)∵x∈[0,5π12],∴2x...

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+...

(1)f(x)=2cosxsin(x+π3)?3sin2x+sinxcosx+2=2sin(2x+π3)+2∴最小正周期T=2π2=π,当2kπ-π2≤2x+π3≤2kπ+π2时,即kπ-5π12≤x≤kπ+π12,函数单调增∴函数的单调增区间为:[kπ-5π12,kπ+π12](k∈Z)(2)由函数y=sinx纵坐标不变,横坐标扩大2倍得到y=...

(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...

求到2x+π/3∈[π/3,4π/3]这一步 2x+π/3=π时,sin最大=1 2x+π/3=4π/3时,sin最小=-√3/2

f(x)=(sinx)^2+2√3sinxcosx+sin(x+π/4)sin(x-π/4) =1/2(1-cos2x)+√3sin2x+sin(x+π/4)sin[-π/2+(x+π/4)] =1/2-1/2cos2x+√3sin2x-sin(x+π/4)cos(x+π/4) =1/2-1/2cos2x+√3sin2x-1/2sin(2x+π/2) =1/2-1/2cos2x+√3sin2x-1/2cos2x =√3sin2x-cos2x+1/2 ...

(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...

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