Question

已知π/4<α<3π/4,0<β<π/4,cos(π/4+α)=-3/5,sina(3π/4+β)=5/13,求sina(α+β)的值

Answer 1

π/4<α<3π/4 ，cos(π/4+α)=-3/5 …… sin(π/4+α)=4/5

0<β<π/4， sin(3π/4+β)=5/13…… cos(3π/4+β)= - 12/13

sina(α+β)= - sin（α+β+π）=- sin（(π/4+α)+(3π/4+β）)

= - (sin(π/4+α)*cos(3π/4+β) + cos(π/4+α)*sin(3π/4+β) )

=63/65

Answer 2

π/4<α<3π/4 cos(π/4+α)=-3/5 就有 sin(π/4+α)=4/5

0<β<π/4 sin(3π/4+β)=5/13 cos(3π/4+β)= - 12/13

sina(α+β)= - sin（α+β+π）=- sin（(π/4+α)+(3π/4+β）)

= - (sin(π/4+α)*cos(3π/4+β) + cos(π/4+α)*sin(3π/4+β) )

=自己算吧