Question

设向量a=(根号3sinx,cosx),向量b=(cosx,cosx),记f(x)=向量a×向量b

(1)写出函数f(x)的最小正周期及递增区间；
(2)若f(x0)=11/10，x0属于[π/4，π/2]，求cos2x0的值

Answer 1

设向量a = (√3sinx，cosx)，向量b = (cosx，cosx)，记ƒ(x) = 向量a • 向量b
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写出函数ƒ(x)的最小正周期及递增区间。
ƒ(x) = (√3sinx)(cosx) + (cosx)(cosx) = √3sinxcosx + cos²x
= (√3/2)sin(2x) + (1 + cos(2x))/2
= √[(√3/2)² + (1/2)²]sin(2x + π/6) + 1/2
= sin(2x + π/6) + 1/2
最小正周期T = (2π)/2 = π
- 1 ≤ sin(2x + π/6) ≤ 1
2kπ - π/2 ≤ 2x + π/6 ≤ 2kπ + π/2 或 2kπ + π/2 ≥ 2x + π/6 ≥ 2kπ + 3π/2
2kπ - 2π/3 ≤ 2x ≤ 2kπ + π/3 或 2kπ + π/3 ≥ 2x ≥ 2kπ + 4π/3
kπ - π/3 ≤ x ≤ kπ + π/6 或 kπ + π/6 ≥ x ≥ kπ + 2π/3
递增区间：[kπ - π/3，kπ + π/6]
递减区间：[kπ + π/6，kπ + 2π/3]
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若ƒ(x₀) = 11/10，x₀属于[π/4，π/2]，求cos(2x₀)的值。
π/4 ≤ x₀ ≤ π/2
π/2 ≤ 2x₀ ≤ π
在[π/2，π]里cos(2x₀) < 0
ƒ(x₀) = sin(2x₀ + π/6) + 1/2
= sin(2x₀)cos(π/6) + cos(2x₀)sin(π/6) + 1/2
= (√3/2)sin(2x₀) + (1/2)cos(2x₀) + 1/2
= (√3/2)√[1 - cos²(2x₀)] + (1/2)cos(2x₀) + 1/2 = 11/10
解得cos(2x₀) = (3 ± 4√3)/10
由于(3 + 4√3)/10 > 0，舍掉
所以cos(2x₀) = (3 - 4√3)/10

Answer 2

=√3/2sin(2α-π/3)+2
f(x)=a*b=√3sinxcosx+cos^2x
=1/2sin2x+1/2cos2x+1/2
=√2sin(2x+π/4)+1/2
T=2π/2=π
x在[kπ-3π/8,kπ+π/8]递增
f(x)==√2sin(2x+π/4)+1/2=11/10
sin(2x+π/4)=3√2/10
x属于[π/4，π/2]，
2x+π/4属于[π/4，π/2]，
cos(2x+π/4)= -2√22/10
cos2x=cos[(2x+π/4)-π/4]
=√2/2[cos(2x+π/4)+sin(2x+π/4)]
=√2/2(3√2/10-2√22/10)
=(3-√11)/10