题目

sin（2A-B）=3/5,A∈[π/2,π],sinB=-12/13,B∈[-π/2,0],求sinA的值

提问时间：2021-01-25

答案

A∈[π/2,π],B∈[-π/2,0] 2A-B∈[π,5π/2] sin(2A-B)=3/5>0 2A-B∈[2π,5π/2] cos(2A-B)>0 cos(2A-B) =√[1-sin^2(2A-B)] =4/5 sinB=-12/13,B∈[-π/2,0] cosB =√[1-sin2(B)] =5/13 sin2A =sin[(2A-B)+B] =-33/65 sin(2A-B) =sin2AcosB-sinBcos2A =3/5 即cos2A=56/65 A∈[π/2,π] sinA>0 1-2sin2(A)=cos2A=56/65 sinA=(3√130)/130