已知cot^2α =2tan^2β +1那么sin^2β+2sin^2α

已知cot^2α =2tan^2β +1那么sin^2β+2sin^2α
已知cot^2α =2tan^2β +1那么sin^2β+2sin^2α

已知cot^2α =2tan^2β +1那么sin^2β+2sin^2α
sin^2β+2sin^2α
＝sin^2β+2sin^2α/(sin^2α+cos^2α)　　（上下同除sin^2α)
=sin^2β+2 /(1+cot^2α)
=sin^2β+2 /(1+2tan^2β +1)
=sin^2β+1 /(tan^2β +1)
=sin^2β+1 /(sec^2β )
=sin^2β+cos^2β
=1

由已知条件可得(cosα)^2/(sinα)^2 = 2 (sinβ)^2/(cosβ)^2 +1, 等号两边同加上1，可得
1/(sinα)^2 = 2/(cosβ)^2 , 即 (cosβ)^2 = 2(sinα)^2,
所以(sinβ)^2 + 2(sinα)^2 =(sinβ)^2 + (cosβ)^2 =1.

cos（α－β）=1/5,所以sin（α-β）大于0 然后根据 cos（α－β）^2sin2α＝2sinαcosα cos2α＝cos2α－sin2α＝2cos2α－1＝1－2sin2

cot^2α =2tan^2β +1
cot^2α +1/tan^2β +1=2
（cos^2α /sin^2α ）+1/（sin^2β/cos^2β）+1=2
1/sin^2α /1/cos^2β=2
cos^2β=2sin^2α
cos^2β+sin^2β=2sin^2α +sin^2β
2sin^2α +sin^2β=1