1.f(x) = a•b = (2cos(x/2),1)•(√2sin(x/2 + π/4),-1)
= 2√2cos(x/2)sin(x/2 + π/4) - 1
= 2√2 * (1/2) * {sin[(x/2) + (x/2 + π/4)] - sin[(x/2) - (x/2 + π/4)] - 1 (积化和差)
= √2[sin(x + π/4) - sin(-π/4)] - 1
= √2 sin(x + π/4)
2.f(π/12) = √2 sin(π/12 + π/4)
= √2 sin(π/3)
= √6/2