函数y等于sin(x+兀/3)sin(x+兀/2)的最小周期T是多少
问题描述:
函数y等于sin(x+兀/3)sin(x+兀/2)的最小周期T是多少
答
y=sin(x+1/3兀)sin(x+1/2兀) =-1/2[cos(2x+5π/6)-cos(-π/6)] 所以 T=2π/w=2π/2=π
函数y等于sin(x+兀/3)sin(x+兀/2)的最小周期T是多少
y=sin(x+1/3兀)sin(x+1/2兀) =-1/2[cos(2x+5π/6)-cos(-π/6)] 所以 T=2π/w=2π/2=π