Question

高一数学求大神！！！！！！

Answer 1

解：f(x)=向量α.向量β。
=√3sin2x+(sinx+cosx)*(sinx-cosx).
=√ 3sin2x-(cos^2x-sin^2x).
=2[(√3/2)sin2x-(1/2)cos2x].
∴f(x)=2sin(2x-π/6).
(1)f(x)的最小正周期 T=2π/2=π；
(2)若f(θ)=√3, 0＜θ＜π/2, 求cos(θ+π/6).
f(θ)=2sin[2θ-π/6)=√3.
∴sin(2θ-π/6)=√3/2.
2θ-π/6=π/3
2θ=π/3+π/6
∴ θ=π/4
∵cos(θ+π/6)=cos(π/4+π/6)
=cosπ/4cosπ/6-sinπ/4sinπ/6.
=√2/2*√3/2-√2/2*1/2.
∴cos(θ+π/6)=√2/4(√3-1).