Question

已知a、b是两不共线的向量,且a=(cosα,sinα),b=(cos β,sin β)

Answer 1

（一）证明：∵a=(cosa,sina),b=(cosb,sinb),∴(a+b)·(a-b)=a²-b²=(cos²a+sin²a)-(cos²b+sin²b)=1-1=0.∴(a+b)⊥(a-b).由a·b=3/5.可得cosacosb+sinasinb=3/5.∵b=π/4.∴sina+cosa=(3√2)/5.两边平方得sin2a=-7/25.∴由题设可知：-π/4＜a＜0.===>0＜-a＜π/4.===>0＜tan(-a)＜1.===>-1＜tana＜0.又sin2a=2sinacosa/(sin²a+cos²a)=(2tana)/(tan²a+1)=-7/25.===>7tan²a+50tana+7=0.===>(tana+7)(7tana+1)=0.===>tana=-1/7.(tana=-7舍去).∴tana=-1/7.其中-π/4＜a＜0.∴sina=-1/√50=-(√2)/10.