一、求下列各式的值

一、
1)
lg12-2lg6+lg3=lg(12÷6^2×3)=lg1=0
2)
原式=(sin(α)cos(α))/(-sin(α)÷cos(α)·cos(α))=-cos(α)
二、
1)
f(-π/6)=√2cos(-π/6-π/12)
=√2cos(-π/4)
=√2·(√2/2)
=1
2)
f(θ+π/3)=√2cos(θ+π/3-π/12)
=√2cos(θ+π/4)
=√2(cos(θ)cos(π/4)-sin(θ)sin(π/4))
=√2(3/5·√2/2+4/5·√2/2)
=7/5