函数y=sin(3x+π/4)+2cos(3x+π/4)的最小正周期是
问题描述:
函数y=sin(3x+π/4)+2cos(3x+π/4)的最小正周期是
答
令cosa=√5/5,sina=2√5/5
y=sin(3x+π/4)+2cos(3x+π/4)
=√5[√5/5sin(3x+π/4)+2√5/5cos(3x+π/4)]
=√5[√5/5sin(3x+π/4)+2√5/5cos(3x+π/4)]
=√5[sin(3x+π/4)cosa+cos(3x+π/4)sina]
=√5sin(3x+π/4+a)
T=2π/3