Question

1.若α∈（π/2,π）,且3cos2α=sin（π/4-α）,则sin2α=?

Answer 1

1.

3cos2α=sin（π/4-α) ==> 3(cosα+sinα)(cosα-sinα)=√2/2(cosα-sinα)

==> cosα+sinα=√2/6 ==> cos²α+sin²α+sin2α=1/18 ==>sin2α=-17/18

2

过右焦点F2斜率为-1的直线：y=c-x

y=c-x 与y=-b/ax交于B(ac/(a-b),bc/(b-a))

∵向量F2A=向量AB,∴A是线段F2B的中点

∴ F1B//OA ∴[bc/(b-a)]/[ac/(a-b)+c]=b/a

∴b/(b-2a)=b/a ==>b=3a==>b/a=3

∴渐近线：y=±3x

3

∵1= |AB|≥|CA|-|CB|=|CB|,1=|AB|≤|CA|+|CB|=3|CB|

∴1/3≤|CB|≤1,|CA|≤2 ∴向量CA*向量CB≤|CA||CB|=2

即向量CA*向量CB的最大值为2