问4道高一数学三角函数题 急1 y=cos^2(x-π/12)+sin^2(x+π/12)的周期 单调递增区间.2 y=

1.
y=[1+cos(2x-π/6)]/2+[1-cos(2x+π/6)]/2
=1+[cos(2x-π/6)-cos(2x+π/6)]/2
把里面的分解后得到y=1+sin(2x)*sin(π/6)=1+sin(2x)/2
得到周期是π, 递增区间就是sin2x的增区间,很容易拉
2. y=sinx+sinx*cos(-π/3)+cosx*sin(-π/3)
=sinx+0.5sinx-0.5cosx*根号3
=-根号3cos(x+π/3)
周期就是2π, 值域[-根号3,根号3]
3.先用积化和差
=[cos(36+72)+cos(36-72)]/2
=[cos36-cos72]/2
cos36-cos72
=-2sin54sin(-18)——这个是和差化积的公式=2sin54sin18
=2cos36sin18
=2cos36sin18cos18/cos18
=cos36sin36/cos18
=sin72/(2cos18)
=sin72/(2sin72)=1/2
那么cos36*cos72=(1/2)/2=1/4
4.用和差化积
分子=2[cos(20+40)/2]*[sin(20-40)/2]
分母=-2[sin(20+40)/2]*[sin(20-40)/2]
分子/分母=-ctn30=-根号3

1. y=cos^2(x-π/12)+sin^2(x+π/12)=[cos(2x-π/6)+1]/2+[cos(2x+π/6)+1]/2=1/2sin2x+1
所以y的周期为π,单调递增区间为[-π/4+kπ,π/4+kπ]
2.y=sinx+sin(x-π/3)=3/2sinx-根号3/2cosx=根号3sin(2x-π/6)
所以y的周期为π,值域为[-3,3]