求函数y=sin2x-根号3cos2x的周期,最大值以及单调递增区间
问题描述：
求函数y=sin2x-根号3cos2x的周期,最大值以及单调递增区间
答
y=sin2x-√3cos2x
=2(1/2sin2x-√3/2cos2x)
=2(sin2xcosπ/3-cos2xsinπ/3)
=2sin(2x-π/3)
周期T=2π/2=π
最大值2
增区间 2kπ+π/2>=2x-π/3>=2kπ-π/2
即 kπ+5π/12>=x>=kπ-π/12
减区间2kπ+3π/2>=2x-π/3>=2kπ+π/2
即 kπ+11π/12>=x>=kπ+5π/12
答
y=2(sin2xcosπ/3-cos2xsinπ/3)
=2sin(2x-π/3)
T=2π/2=π
显然最大值=2
sin递增则2kπ-π/2kπ-π/12
答
f(x)=sin2x-√3cos2x
=2(1/2*sin2x-√3/2cos2x)
=2(sin2xcosπ/3-cos2xsinπ/3)
=2sin(2x-π/3)
最小正周期T＝2π/2=π
因为-1