函数y=sin(x-π/6)cosx的递增区间

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/07 02:22:21
函数y=sin(x-π/6)cosx的递增区间

函数y=sin(x-π/6)cosx的递增区间
函数y=sin(x-π/6)cosx的递增区间

函数y=sin(x-π/6)cosx的递增区间

y=sin(x-π/6)cosx
=[sinxcos(π/6)-cosxsin(π/6)]cosx
=(√3/2)sinx*cosx-(1/2)(cosx)²
=(√3/4)sin2x-(1/2)(cos2x+1)/2
=(√3/4)*sin2x-(1/4)*cos2x-1/4
=(1/2)*[sin2x*(√3/2)-cos2x*(1/2)]-1/4
=(1/2)*[sin2x*cos(π/6)-cos2x*sin(π/6)]-1/4
=(1/2)sin(2x-π/6)-1/4
求增区间,
∴ 2kπ-π/2≤2x-π/6≤2kπ+π/2,k∈Z
∴ 2kπ-π/3≤2x≤2kπ+2π/3,k∈Z
∴ kπ-π/6≤x≤kπ+π/3,k∈Z
∴ 增区间是【kπ-π/6,kπ+π/3】,k∈Z

Y = COS(x/2-pi/6)罪(x/2-pi/6)
= 2 ^ 0.5(COS(π/ 4)COS(x/2-pi/6)罪(π/ 4)罪(x/2-pi/6))
= 2 ^ 0.5(COS(x/2-pi/6 +π/ 4))
= 2 ^ 0.5cos( X / 2 + pi/12)
(2K-1)PI <= X / 2 + pi/12 <= 2kpi
(4K-13/6)PI <= x <=(4K-圆周率1/6),该函数的单调递增

Y = COS(x/2-π/6)罪(x/2-π/6)
= SIN(π/ 6)cos(π/ 6)的
=(1半)减(一个半平方根3)

一个常数,所以也没有单调递增的范围