f(x)=sin2x-sin2(x-π 6)的最小正周期.对称轴以及单调增区间
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请采纳谢谢f(x)=(sin^4+cos^4+sin^2xcos^2x)/2-sin2x=((sin^2x+cos^2x)^2-sin^2xcos^2x)/2-sin2x=(1-sin^2xcos^2
(Ⅰ)因为f(x)=sin2x-(1-cos2x)=2sin(2x+π4)-1,所以函数f(x)的最小正周期为T=2π2=π(Ⅱ)由(Ⅰ)知,当2x+π4=2kπ−π2,即x=kπ−π8(k∈Z)时,
(1)f(x)=32sin2x−12cos2x−1=sin(2x−π6)−1当2x−π6=2kπ−π2,k∈Z,即x=kπ−π6,k∈Z时,f(x)取得最小值−2f(x)的最小正周期为π(2)由c=3
(1)∵f(x)=sin(2x+π6)+2sin2x∴f(x)=32sin2x+12cos2x+(−cos2x+1)=(32sin2x−12cos2x)+1=sin(2x−π6)+1.∵T=2π2=π
(1)f(x)=cos(2x+π3)+sin2x=cos2xcosπ3−sin2xsinπ3+1−cos2x2=12−32sin2x所以函数f(x)的最大值为1+32,最小正周期π.(2)由f(x)=
解 (1)f(x)=sin2x+2sinx•cosx+3cos2x=sin2x+2sinxcosx+3cos2xsin2x+cos2x=tan2x+2tanx+3tan2x+1=175;(2
(Ⅰ)f(x)=32sin2x−1+cos2x2−12=sin(2x−π6)−1则f(x)的最小值是-2,最小正周期是T=2π2=π;(7分)(Ⅱ)f(C)=sin(2C−π6)−1=0,则sin(2
(1)f(x)=2sinxcosx+2cos2x2cosx=sinx+cosx(cosx≠0),…(4分)由题意可得f(x)=2sin(x+π4)=0,故x+π4=kπ,即 x=kπ−π4(
f(x)=[1−cos(π2+2x)]−3cos2x(1分)=1+sin2x−3cos2x(2分)=2sin(2x−π3)+1,(3分)(1)T=2π2=π;(4分)(2)2sin(2x−π3)+1≥
(Ⅰ)f(x)=2cosx(12sinx+32cosx)−3sin2x+sinxcosx=2sinxcosx+3(cos2x−sin2x)=sin2x+3cos2x=2sin(2x+π3)∴T=π(Ⅱ
f(x)=sin2(2x-π4)=1−cos(4x−π2)2根据三角函数的性质知T=2π4=π2故答案为:π2
(I)函数f(x)=sin22x-3sin2xcos2x=1−cos4x2-32sin4x=12-sin(4x+π6),∵ω=4,∴T=2π4=π2;(II)∵2kπ+π2<4x+π6<2kπ+3π2
sinx的导数是cosxcosx的导数是-sinx结合求导的知识可知原式=2cos2x+2sin2x
(1)由已知f(x)=3sin2x+2cos2x+3=3sin2x+cos2x+4=2sin(2x+π6)+4.(3分)当x∈(0,π2)时,2x+π6∈(π6,7π6),sin(2x+π6)∈(−1
(1)∵向量a=(3sin2x,cos2x),b=(sin2x,sin2x),又∵函数f(x)=a•b+t(t∈R)∴f(x)=sin(4x−π3)+t+32∴f(x)的最小正周期是π2其单调递增区间
(Ⅰ)函数f(x)=sin2x+2sinxsin(π2−x) +3sin2(3π2−x)=sin2x+2sinxcosx+3cos2x=1+sin2x+2cos2x=2+sin2x+cos2
函数f(x)=3sin2x−2cos2x=3sin2x−cos 2 x−1=2sin(2x-π3)-1它的最小正周期为:π故答案为:π
(1)∵f(x)=sin2x+3cos2x=2sin(2x+π3),故[f(x)]max=2,[f(x)]min=2.(2)函数的最小正周期为T=2π2=π.(3)令2kπ-π2≤2x+π3≤2kπ+
f(x)=sin2x-3(2cos2x-1)=sin2x-3cos2x=2(12sin2x-32cos2x)=2sin(2x-π3),∵ω=2,∴函数f(x)的最小正周期是2π2=π,当-π2+2kπ