函数y=根号2sin(2x-π)cos(2(x π))是-搜答案-智慧作业帮

y=根号3cos²x+sinxcosx-根号3/2=√3（1/2cos2x+1/2)+1/2sin2x-√3/2=√3/2cos2x+√3/2+1/2sin2x-√3/2=√3/2cos2x

见图.y(天蓝线)向右平移π/8(红线),然后将每一点的横坐标加倍(纵坐标不变),得绿线.

y=sinx+cosx=√2*(√2/2sinx+√2/2cosx)=√2*[sinxcos(π/4)+sin(π/4)cosx]=√2*sin(x+π/4)

由题意可得sin(2x+π/6)

π3/4根号3-π/2--根号3

答案是这个,再问：具体过程再答：久等了~公式难打~若满意请采纳~

2cosxsin(x+π/3)=2cosx(sinxcosπ/3+cosxsinπ/3)=cosxsinx+√3cos^2xf(x)=cosxsinx+sinxcosx+√3(cos^2x-sin^2

=2sin(2x)cos[2(x)]=sin(4x)周期0.5pi再问：那他是奇还是偶函数？

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

y＝（√3/2）sin（x＋π/2）＋cos（π/6－x）＝（√3/2）cosx＋cos（π/6）cosx＋sin（π/6）sinx＝（√3/2）cosx＋（√3/2）cosx＋（1/2）sinx＝√

2sin(x+π/4）+1≥0sin(x+π/4)≥-1/22kπ-π/6≤x+π/4≤2kπ+7π/62kπ-5π/12≤x≤2kπ+11π/12[2kπ-5π/12,2kπ+11π/12],k∈Z

y=2cosxsin(x+π/3)-根号3*(sin^2)x+sinxcosx,后两项先提出一个sinx,然后括号内部分用叠加原理,得到y=2cosxsin(x+π/3)+2sinxcos(x+π/3

（1）原式=2[(1/2)sin(x/2)+（根号3/2）cos(x/2)]=2sin[(x/2)+pi/3]所以当[(x/2)+pi/3]=2kpi+pi/2时,y最大值为2解得x=4kpi+pi/

f(x)=根号下2sin(2x-π/4)+1x∈(0,π/2)则-π/4

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

y=sin(2x-π/3)+根号3cos2x=sin2xcosπ/3-cos2xsinπ/3+2sinπ/3cos2x=sin2xcosπ/3+cos2xsinπ/3=sin(2x+π/3)1.振幅=

y=根号2cosx最大值√2,最小值-√2；一个周期[0,2π].y=根号2sin（2x+π/4）=√2cos(2x+3π/4),最大值√2,最小值-√2；一个周期[π/8,5π/8].需将函数y=根

∵-1≤cosx≤1,所以2-cosx恒≥1∴原式可化为√3sinx=2Y-cosxY===>√3sinx+cosxY=2Y三角代换：sin(x+a)=2Y/√(3+y²)∴-1≤2Y/√(

2sin(x+π/4)+1>=0sin(x+π/4)>=-1/2-π/3+2kπ