已知函数fx=cos(x-pai/4),设gx=fx乘f(x pai/2),求函数gx在区间[-pai/6,pai/3]上的最大值和最小值.
∵f(x)=cos(x-π/4)，∴f(x π/4)=cos[(x π/4)-π/4]=cosx，∴g(x)=cos(x-π/4)cosx=(1/2)[cos(-π/4) cos(2x-π/4)]=√2/4 (1/2)cos(2x-π/4)。 ∵-π/6≦x≦π/3，∴-π/3≦2x≦2π/3，∴-π/3-π/4≦2x-π/4≦2π/3-π/4，-7π/12≦2x-π/4≦5π/12，∴cos(-7π/12)≦cos(2x-π/4)≦cos0，∴cos(6π/12 π/12)≦cos(2x-π/4)≦1，∴-sin(π/12)≦cos(2x-π/4)≦1，∴-(√6 √2)/4≦cos(...全部
∵f(x)=cos(x-π/4)，∴f(x π/4)=cos[(x π/4)-π/4]=cosx，∴g(x)=cos(x-π/4)cosx=(1/2)[cos(-π/4) cos(2x-π/4)]=√2/4 (1/2)cos(2x-π/4)。
∵-π/6≦x≦π/3，∴-π/3≦2x≦2π/3，∴-π/3-π/4≦2x-π/4≦2π/3-π/4，-7π/12≦2x-π/4≦5π/12，∴cos(-7π/12)≦cos(2x-π/4)≦cos0，∴cos(6π/12 π/12)≦cos(2x-π/4)≦1，∴-sin(π/12)≦cos(2x-π/4)≦1，∴-(√6 √2)/4≦cos(2x-π/4)≦1，∴-(√6 √2)/8≦(1/2)cos(2x-π/4)≦1/2，∴-(√6 √2)/8 √2/4≦√2/4 (1/2)cos(2x-π/4)≦1/2 √2/4，∴(√2-√6)/8≦g(x)≦(2 √2)/4。
∴g(x)在区间[-π/6，π/3]上的最大值是(2 √2)/4，最小值是(√2-√6)/8。收起
