Question

已知向量a=(cosx+sinx,sinx),b=(cosx-sinx,2cosx),设f(x)=ab 当x∈[-π/4,π/4]时，求函数f(x)的

最大值和最小值
有满意答案追加10分

Answer 1

向量a=(cosx+sinx,sinx),b=(cosx-sinx,2cosx),
f(x)=a●b
=(cosx+sinx)(cosx-sinx)+2sinxcosx
=cos²x-sin²x+sin2x
=cos2x+sin2x
=√2(√2/2sin2x+√2/2cos2x)
=√2sin(2x+π/4)
∵x∈[-π/4,π/4]
∴2x∈[-π/2,π/2]
∴2x+π/4∈[-π/4,3π/4]
∴sin(2x+π/4]∈[-√2/2,1]
∴√2sin(2x+π/4]∈[-1,√2]
f(x)最大值为√2，最小值为-1

Answer 2

f(x)=(cosx+sinx)*(cosx-sinx)+sinx*2cosx
=cos2x+sin2x
=√2sin(2x+π/4)
∵ x∈[-π/4，π/4]
所以∴ 2x+π/4∈[-π/4，0]
所以 当 2x+π/4=0时， 最大=f(-π/8)=0
当2x+π/4=-π/4时， 最小=f(-π/4）=-1