Question

已知向量a=(-2cosθ,cosx),b=(2sinθ,-sinx),c=(1/2,-1)(1)若a*c=1/3,b*c=1/4(1)求cos(θ-x)

(2)设θ=π/3,求f(x)=a^2+a*b+根3的最大值及单调递增区间。

Answer 1

1. a*c=-cosθ-cosx=1/3
cos²θ+2cosθcosx+cos²x=1/9 (1)
b*c=sinθ+sinx=1/4
sin²θ+2sinθsinx+sin²x=1/16 (2)
(1)+(2) 2+2cos(θ-x)=25/144
所以cos(θ-x)=-263/288
2. f(x)=4*(1/2)²+cos²x-4*(1/2)*(√3/2)-sinxcosx+√3
=1+cos²x-(1/2)sin2x
=(1/2)(cos2x-sin2x)+3/2
=(√2/2)cos(2x+π/4)+3/2
可见f(x)最大值=(√2+3)/2
单调递增区间为2x+π/4∈[2kπ-π/2, 2kπ+π/2]
x∈[kπ-3π/8, kπ+π/8]
希望能帮到你，祝学习进步O(∩_∩)O

Answer 2

1/3=a*c=-cos(θ)-cos(x), 1/9=[cos(θ)]^2+[cos(x)]^2+2cos(θ)cos(x)
1/4=b*c=sin(θ)+sin(x),1/16=[sin(θ)]^2+[sin(x)]^2+2sin(θ)sin(x)
1/9+1/16=2+2[cos(θ)cos(x)+sin(θ)sin(x)]=2+2cos(θ-x)
cos(θ-x)=[1/9+1/16-2]/2=[25/144-2]/2=-263/288

a*a=4[cos(θ)]^2+[cos(x)]^2=1+[cos(x)]^2,
a*b=-2sin(2θ)-sin(2x)/2=-3^(1/2)-sin(2x)/2,
f(x)=[cos(x)]^2+1-3^(1/2)-sin(2x)/2+3^(1/2)
=[cos(2x)+1]/2 + 1 - sin(2x)/2
= [cos(2x)-sin(2x) +3]/2
=[2^(1/2)cos(2x+π/4) + 3]/2
<=[2^(1/2)+3]/2=f(nπ-π/8)
f(x) 的最大值为f(nπ-π/8)=[2^(1/2)+3]/2,
π+2nπ<=2x+π/4<=2π+2nπ时，cos(2x+π/4)单调递增，f(x)单调递增。
此时，
3π/4+2nπ<=2x<=7π/4+2nπ,
3π/8+nπ<=x<=7π/8+nπ.
f(x)的单调递增区间为[3π/8+nπ,7π/8+nπ], n=0, +1,-1, +2,-2,...