Question

高中数学三角函数题，在线等哈~~~~~

已知四分之π＜α＜四分之3π，0＜β＜四分之π，cos(四分之π+α)=-五分之三，sin(四分之三π+β)=13分之5，求sin(α+β)的值.
请各位会的童鞋们帮帮忙哈，谢谢啦！！！

Answer 1

π/2<π/4 + α<π
3π/4<3π/4 + β<π
故 sin(π/4 + α)=√(1-9/25)=4/5
cos(3π/4 + β)=-√(1-25/169)=-12/13

sin(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)
= - 4/5 *(-12/13) - (-3/5) * 5/13
=48/65+3/13
=63/65

Answer 2

cos(π/4+α)=-3/5， sin(π/4+α)=4/5，
sin(3π/4+β)=5/13，cos(3π/4+β)=-12/13，
sin(α+β)=-sin(α+β+π)=-sin[(α+π/4)+(β+3π/4)]
=-[sin(π/4+α)cos(3π/4+β)+cos(π/4+α)sin(3π/4+β)]
=-[4/5*(-12/13)+(-3/5)*5/13]
=63/65