Question

已知sin(3π-α)= √2sinβ, √3cos(-α)=- √2cos(π+β),0<α<π,0<β<π,求α

Answer 1

sin(3π-α)= √2sinβ,
sin(2π+π-α)= √2sinβ,
sin(π-α)= √2sinβ,
sinα= √2sinβ
0<α<π,0<β<π
sinα>0,则sinβ>0

sinα= √2sinβ(平方）
sin^2α= 2sin^2β...............1

√3cos(-α)=- √2cos(π+β)
√3cosα=√2cosβ(平方）
3cos^2α=2cos^2β................2
1式+2式得
sin^2α+3cos^2α=2
1+2cos^2α=2
2cos^2α=1
cos^2α=1/2
cosα=±√2/2

追问

已知sin(3π-α)= √2sinβ, √3cos(-α)=- √2cos(π+β),0<α<π,0<β<π,求α与β的值

追答

sin^2α= 2sin^2β...............1
3cos^2α=2cos^2β................2
1式*3+2式得
3=6sin^2β+2cos^2β
4sin^2β+2=3
4sin^2β=1
sin^2β=1/4
sinβ=±1/2