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

(1).当α=π/4时,b•c=2sinxcosx+2sinαcosx+2cosαsinx=sin2x+(√2)(sinx+cosx)
设y=sin2x+(√2)(sinx+cosx),再令y′=2cos2x+(√2)(cosx-sinx)=2(cos²x-sin²x)+(√2)(cosx-sinx)
=2(cosx+sinx)(cosx-sinx)+(√2)(cosx-sinx)=(cosx-sinx)(2cosx+2sinx+√2)
=(cosx-sinx)[2(√2)sin(x+π/4)+√2]=2(√2)(cosx-sinx)[sin(x+π/4)+1/2]=0
于是得cosx-sinx=0,即有tanx=1,故得驻点x=π/4+kπ；
及sin(x+π/4)+1/2=0,sin(x+π/4)=-1/2,x+π/4=-π/6+2kπ,故得驻点x=-5π/12+2kπ；
因为是周期函数,极值点的分析很麻烦,详细过程太繁琐,故免去；可以肯定,x=-5π/12
是一个极小点；此时min(b•c)=sin(-5π/6)+(√2)[sin(-5π/12)+cos(-5π/12)]
=-sin(π/6)+(√2)[cos(5π/12)-sin(5π/12)]=-1/2+2cos(5π/12+π/4)=-1/2+2cos(2π/3)
=-1/2-2cos(π/3)=-1/2-1=-3/2.
(2)cos(π/3)=a•b/[︱a︱︱b︱]=(cosxcosα+sinxsinα)/(1×1)=cos(x-α)
故得x-α=π/3,x=α+π/3；∵a⊥c,
∴a•c=sinxcosα+2sinαcosα+cosxsinα+2sinαcosα=sin(x+α)+2sin2α=sin(2α+π/3)+2sin2α=0
于是得(1/2)sin2α+(√3/2)cos2α+2sin2α=0,即有(5/2)sin2α+(√3/2)cos2α=0,∴tan2α=-√3/5