f(x)=sin(πx/4-π/6)-2cos²πx/8+1=sin(πx/4-π/6)-（2cos²πx/8-1）=sinπx/4cosπ/6-cosπx/4sinπ/6-cosπx/4=√3/2*sinπx/4-3/2cosπx/4=√3(1/2sinπx/4-√3/2*cosπx/4)=√3sin(πx/4-π/3)T=2π...O(∩_∩)O哈哈~，第一問(wèn)我對(duì)了，可是第二問(wèn)那個(gè)對(duì)稱應(yīng)該怎么求？有點(diǎn)搞不清楚。g(x)與f(x)關(guān)于直線x=1對(duì)稱g(x)=f(2-x)=√3sin[π(2-x)/4-π/3)=√3sin(-πx/4+π/6)=-√3sin(πx/4-π/6)∵0≤x≤4/3∴0≤ πx/4≤π/3∴-π/6≤πx/4-π/6≤π/6∴-1/2≤sin(πx/4-π/6)≤1/2∴-√3/2≤-√3sin(πx/4-π/6)≤√3/2∴πx/4-π/6=-π/6,x=0時(shí)，g(x)取得最大值√3/2