向量a=(sinα,cosα)向量b=(cosx,sinx)向量c=(sinx+2sinα,cosx+2cosα)

问题描述:

向量a=(sinα,cosα)向量b=(cosx,sinx)向量c=(sinx+2sinα,cosx+2cosα)
α=π/4时,求f(x)=向量b×向量c


是求最值吧
f(x)=b.c
=(cosx,sinx).(sinx+√2, cosx+√2)
=cosxsinx+√2cosx+sinxcosx+√2sinx
=2sinxcosx+√2(sinx+cosx)
令sinx+cosx=t
t=√2sin(x+π/4)∈[-√2,√2]
1+2sinxcosx=t²
f(x)=t²-1+√2t
=(t+√2/2)-3/2
所以t=-√2/2时,f(x)有最小值-3/2
t=√2时,f(x)有最大值3