已知g(x)=sin(2x 30°)求对称轴方程

1）f(x)=1/2cos2x=1/2sin(2x+π/2)=1/2sin2(x+π/4)而g(x)=1/2sin(2x)-1/4f(x)可由g(x)向左平移π/4,再向下平移1/4而得到.2）h(x

∵f（x）＝cos（2x－π/3）＋（sinx）^2－（cosx）^2＝cos（2x－π/3）－cos2x＝2sin（2x－π/6）sin（π/6）＝sin（2x－π/6）.∴g（x）＝［sin（2x

化简就可以啦,f(x)+g(x)=3sin(2x-π/3)+4sin(2x+π/3)=3（sin2x*1/2-cos2x*根号3/2）+4(sin2x*1/2+cos2x*根号3/2)=7/2sin2

f(x)=cos^2x+sinxcosx=(1+cos2x)/2+1/2*sin2x=1/2+1/2(cos2x+sin2x)=√2/2*(√2/2*cos2x+√2/2sin2x)+1/2=√2/2

f(x)=2sinxcosx-asinx-acosx令sinx+cosx=t=(√2)sin(x+π/4)则-√2≤t≤√2f(x)=t^2-1-at=t^2-at-1=(t-a/2)^2-1-a^2

f(x)=cos(x-π/)+sin^2x-cos^2x=-cosx+sin^2x-cos^2x=-2cos^2x-cosx+1最小正周期2π

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-

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

函数g(x)＝sin(x+π6)，f(x)＝2cosx•g(x)−12=32sin2x+12cos2x=sin（2x+π6）．（1）函数f（x）的最小正周期T=π，因为2x+π6=kπ，所以对称中心坐

(1)f(x)=sin(x-π/6)+cos(x-π/3),g(x)=2sin²x/2f(x)=sinxcosπ/6-cosxsinπ/6+cosxcosπ/3+sinxsinπ/3=√3s

25%x+3=3分之2*60.25x+3=40.25x=4-30.25x=1x=1÷0.25x=418-5%x=12.818-0.05x=12.80.05x=18-12.80.05x=5.2x=5.2

（1）单调递增,∴-π/2+2nπ

设arcsinx=t,则有：g(x)=sint.对于arcsinx=t,取反对数,得到：sint=x,则有：g(x)=sint=x,为本题结果.

f(x)=sin(x/2)-√3[1-cos(x/2)]+√3=2[(1/2)sin(x/2)+(√3/2)cos(x/2)]=2sin(x/2+π/3)(1)g(x)=f(x+π/3)=2sin[(

因为t(x)是奇函数且不是偶函数,所以t(-x)=-t(x)且t(-x)≠t(x)又t(x)=f(x)+g(X)=sin(x+α)+cos(x+β)=sinxcosα+cosxsinα+cosxcos

（1）f(x)=(cos^x-sin^x)/2=1/2cos2x=1/2sin2(x+π/4）,它的图像可由g(x)的图像向上平移1/4个单位,再向左平移π/4个单位得到.（2）h(x)=f(x)-g

当w=2时∵f（x）g（x）=sin2xsin（2x+π2）=sin2xcos2x=12sin4x，T=2π4=π2，故①正确；当w=1时∵f（x）+g（x）=sinx+sin（2x+π2）=sinx

f(a)=根号（3）sina=3根号（3）/5sina=3/5cosa=4/52sin^)2a/2=1-cosa=1/5f(x)>g(x)2sin(x+pi/6)>1sin(x+pi/6)>1/22n

sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)

f(x1)的值域A=[-3/2,1/2],g(x2)的值域B=[-3-m,-m]A是B的子集-3-m1/2-3/2