设f(x)=4sin(2x-三分之派) 根号3

对于A，f(x+π2)＝−sinx，g(x)＝sinx，若f(x+π2)＝g(x)，只需sinx=0，即x=kπ，k∈Z，故∃x∈R，f(x+π2)＝g(x)，即A正确；对于B，f(x−π2)＝sin

f(x)=(1/2^0)·sin(x/2)+(2^0)·cos(2x)f‘(x)=(1/2)·cos(x/2)+(-2)·sin(2x)=(1/2^1)·cos(x/2)+(-2^1)·sin(2x)

f(x)=sin(2x+π/2)=cos2xg(x)=f(x)+f(π/4-x)=cos2x+cos(π/2-2x)=cos2x+sin2x=√2sin(2x+π/4)单增区间2x+π/4∈[2kπ-

你啊,要好好学习了!还没有悬赏分?把对称轴即x=∏／8代入原式子,即sin(∏／4+φ)=1或者-1,再用(-π

f(x)=sin(2x＋π3)＋sin(2x－π/3)f(x)=sin(2x)cos(π/3)＋co(2x)sin(π/3)＋sin(2x)cos(π/3)－cos(2x)sin(π/3)f(x)=2

1,f(x)=sin[(π/2)x+π/6]-2[sin(π/4)x]^2=(√3/2)sin(π/2)x+(1/2)cos(π/2)x+cos(π/2)x-1=√3sin[(π/2)x+π/3]-1

（1）F(X)=SIN(X+π/6)+2SIN^2(x/2)=SIN(X+π/6)+1-COSX=SIN(X+π/6)+1-SIN(π/2-X)=2COS[(X+π/6+π/2-X)/2]*SIN[(

1）由三角函数和差化积公式：f(x)=2sin(x+x+π/3)/2cos(x-x-π/3)/2=2sin(x+π/6)cos(π/6)=√3sin(x+π/6)f(x)的最小值为-√3.当x+π/6

cosx=1-2(sinx/2)^2f=[sin(2/x)]=1+cosx=2-2(sinx/2)^2f(x)=2-2x^2f[cos(2/x)]=2-2[cos(2/x)]^2

1、f(sin2)+f(sin(-2))=√(1-sin2)+√[1-sin(-2)]=√(1-sin2)+√(1+sin2)1-sin2=(sin1)^2+(cos1)^2-2sin1cos1=(s

f(x)=sin2(x+y/2)由于sin2x对称轴为π/4+kπ/2;故x+y/2=π/4+kπ/2x=π/4+kπ/2-y/2;将x=x=π/8代入,得y=π/4+kπ,根据y的范围可知：y=-3

1）f(x)=sin(2x+φ)一条对称轴是X=π/8则kπ+π/2=2*π/8+φ===>φ=kπ+π/4因为-π

2x+φ＝kπ＋π/2,x＝(kπ＋π/2－φ)/2(kπ＋π/2－φ)/2＝π/8当k＝0时,φ＝π/4

1.由f(x)=sin(2x+φ)一条对称轴是直线x=π/2可得：在x=π/2时,函数取极值.则2*π/2+φ=kπ+π/2(k∈Z)φ=kπ-π/2又-π

由1,3作为条件,可以得到2,由2,3作为条件,可以得到1,由1,3得到2,证明：由3可知w=2或-2,设定w=2时,由1可以得到2*π/12+t=kπ/2,k为不等于0的整数.得到t=kπ/2-π/

f(x)=(sin^4x-cos^4x-5)/(cos2x+2)=(sin^2x-cos^2x-5)/(cos2x+2)=-（cos2x+5）/（cos2x+2）=-1-3/（cos2x+2）=-1-

f(x)=sin^2x+asin^2(x/2)=sin^2x+a(1-cosx)=1-cos^2x+a-acosx1=-(cos^2x+acosx)+a+1=-(cos^2x+acosx+a^2/4)

letx=siny∫f(x)dx=∫f(siny)d(siny)=∫[y/(siny)^2]d(siny)=-∫yd[1/(siny)]=-y/siny+∫(1/siny)dy=-y/siny+ln|

f(x)=sin(2x+a)是R上的偶函数有f(x)=f(-x);sin(2x+a)=sin(-2x+a)=cos(π/2-(-2x+a))=cos(π/2+2x-a)余弦函数为R上的偶函数,a=π/