若sin(3π-α)=根号2sin(2π+β)已知sin(3π-α)=根号2 sin(π-β),根号3 cos(-α)= - 根号2 cos(π+β) 且0
问题描述：
若sin(3π-α)=根号2sin(2π+β)
已知sin(3π-α)=根号2 sin(π-β),根号3 cos(-α)= - 根号2 cos(π+β) 且0
答
解:由sin(3π-α)=√2sin(2π+β)可得: sinα=√2sinβ 由√3cos(-α)= -√2cos(π+β)可得: √3cosα=√2cosβ 所以(sinα)^2+(cosα)^2=2/3+4(sinβ)^2/3=1 所以(sinβ)^2=1/4 所以sinα*sinβ=√2(sin...