已知sinα＝cos2α,α∈（二分之派,派）求tan2α,
问题描述：
已知sinα＝cos2α,α∈（二分之派,派）求tan2α,
答
sinα＝cos2α
= 1- 2(sinα)^2
2(sinα)^2+sinα-1=0
(2sinα-1)(sinα+1) =0
sinα=1/2 or -1 ( rejected)
cosα = -√3/2
tan2α = sin2α/cos2α
= 2sinαcosα/((cosα)^2 - (sinα)^2)
= 2(1/2)(-√3/2)/[ 3/4- 1/3]
= -√3