Question

已知sin＝4/5,α∈(π/2,π),cosβ＝-5/13,β＝(π,3π/2)

求①sin(α+β) ②cos(α-β)

Answer 1

sina＝4/5,α∈(π/2,π),可得：cosa=3/5
cosβ＝-5/13,β＝(π,3π/2) 可得：sinβ=-12/13

sin(a+β)
=sinacosβ+cosasinβ
=4/5x(-5/13)+3/5x(-12/13)
=-56/65

cos(a-β)
=cosacosβ+sinasinβ
=3/5x(-5/13)+4/5x(-12/13)
=-63/65

Answer 2

1.16/65 2.-33/65