已知向量a=(2cosx/2,1),b=（√2sin(x/2+∏/4),-1),令f(x)=a*b1.化简f(x)2.求f(∏/12)的值
问题描述：
已知向量a=(2cosx/2,1),b=（√2sin(x/2+∏/4),-1),令f(x)=a*b
1.化简f(x)
2.求f(∏/12)的值
答
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