函数y=sin(x+π/18)cos(x+2π/9)的最小值
问题描述:
函数y=sin(x+π/18)cos(x+2π/9)的最小值
答
y=1/2[sin(2x+π/18+2π/9)+sin(π/18-2π/9)]
当sin(2x+π/18+2π/9)=-1,即2x+π/18+2π/9=(-π/2)+2kπ时,有最小值,最小值为
ymin=1/2(-1+sin(-1/6π))=-3/4