（1）f(x)=a·b
=(2,cos x)·(sin(x+π/6）,-2)
=2sin(x+π/6)-2cosx
=2[sin(x+π/6)-sin(π/2-x)]
=2×2cos(2π/3)sin(x-π/6)
=-2sin(x-π/6)
=2sin(x+5π/6)
令x+5π/6 ∈[2kπ-π/2,2kπ+π/2],得x∈[2kπ-4π/3,2kπ-π/3],此为单增区间；
令x+5π/6 ∈[2kπ+π/2,2kπ+3π/2],得x∈[2kπ-π/3,2kπ+2π/3],此为单减区间；
（2）由已知有
5/6=2sin(x+5π/6)
令θ为满足 sinθ=5/12 的角 解得 x=θ-5π/6；
则 cos(2x-π/3)=cos(2θ-5π/3-π/3)
=cos(2θ)
=1-2sin²θ
=1-2×(5/12)²
=47/72
说明：2[sin(x+π/6)-sin(π/2-x)]
=2×2cos(2π/3)sin(x-π/6)
这步用的是和差化积公式：
sinα-sinβ=sin[(α+β)/2+(α-β)/2] - sin[(α+β)/2-(α-β)/2]
=sin[(α+β)/2]cos[(α-β)/2] + cos[(α+β)/2]sin[(α-β)/2]
- sin[(α+β)/2]cos[(α-β)/2] + cos[(α+β)/2]sin[(α-β)/2]
=2cos[(α+β)/2]sin[(α-β)/2]
这里的证明过程又用到了另一个公式：
sin(θ±γ)=sinθcosγ ± cosθsinγ
这个应该知道吧?