Question

已知向量a=(cosx,sinx),b=(-cosx,cosx),c=(-1,0) (1)若x=π/

(1)若x=π/6,求向量a,c的夹角 （2）当x∈[π/2,9π/8]时，求函数f(x)=2a*b+1的最大值

Answer 1

(1)若x=π/6,a=(√3/2,1/2),c=(-1,0)
cos<a,c>=(a●c)/|a||c|=(-√3/2)/(1*1)=-√3/2
∵<a,c>∈[0,π] ∴向量a,c的夹角为π/3

(2)
f(x)=2a*b+1=-cos²x+sinxcosx+1
=1/2*sin2x-1/2(1+cos2x)+1
=1/2sin2x-1/2*cos2x+1/2
=√2/2sin(2x-π/4)+1/2
∵x∈[π/2,9π/8]∴2x-π/4∈[3π/4,2π]
∴2x-π/4=3π/4时，f(x)取得最大值1

Answer 2

向量a=(cosx,sinx),向量b(-cosx,cosx),向量c(-1,0)

(1)
x=π/6
a=(√3/2,1/2)
|a|=1 |c|=1
cos<a,c>=a*c/(|a|*|c)=-√3/2
所以 <a,c>=150°

(2)
x∈[π/2，9π/8]
f(x)=2ab+1
=2(cosx,sinx)(-cosx,cosx)+1
=2(-cos²x+sinxcosx)+1
=-2cos²x+2sinxcosx+1
=-cos2x+sin2x
=√2sin(2x-π/4)
因为 x∈[π/2,9π/8]
所以 2x-π/4∈[3π/4,2π]
得 sin(2x-π/4)∈[-1,√2/2]
则 f(x)∈[-√2,1]
则 最大值为1