Question

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=

Answer 1

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