高一三角函数问题，急！！！！！！

11 |

2 | 2009-8-12 14:29:33 | 显示全部楼层 |阅读模式

设y=(3√2)sin(x/4)cos(x/4)+√6cos2(x/4)-√6/2 =3√2/2 *sin(x/2)+√6 *[1+cos(x/2)]/2-√6/2 =3√2/2 *sin(x/2)+√6/2 *cos(x/2) =√6[√3/2 *sin(x/2)+1/2 *cos(x/2)] =√6[cos(π/6)sin(x/2)+sin(π/6)cos(x/2)] =√6sin(x/2-π/6)∵x∈[-5π/6，π/6]，∴y∈[-1，(√3-3)/2]又∵y-m≤0即m≥y恒成立，∴m≥[y]max=(√3-3)/2...

使用道具举报