1+cosα-sinβ+sinαsinβ=0,1-cosα-cosβ+sinαcosβ=0.求sinα

Question

1+cosα-sinβ+sinαsinβ=0,1-cosα-cosβ+sinαcosβ=0.求sinα

Answer 1

已知1+cosα-sinβ+sinαsinβ=0,则
1+cosα=(1-sinα)sinβ
(1+cosα)^2=[(1-sinα)sinβ]^2
1+2cosα+cos^2α=(1-sinα)^2*sin^2β.(1)
同理
由已知1-cosα-cosβ+sinαcosβ=0,得
1-cosα=(1-sinα)cosβ
(1-cosα)^2=[(1-sinα)cosβ]^2
1-2cosα+cos^2α=(1-sinα)^2*cos^2β.(2)
(1)+(2),得
2+2cos^2α=(1-sinα)^2
2+2(1-sin^2α)=1-2sinα+sin^2α
3sin^2α-2sinα-3=0
∵ -1≤sinα≤1
∴sinα=(1-√10)/3