Question

已知0＜β＜派／4＜α＜3派／4，且cos﹙派／4－α﹚＝4／5，sin﹙3派＋β﹚＝5／13，求sin﹙α＋β﹚的值

Answer 1

把问题稍微转化一下就可以了，
由π/4<α<3π/4,可得：-π/2<α<0
又∵cos(π/4-α)=4/5 ∴sin(π/4-α)=-3/5
∵sin(3π+β)=5/13, ∴sinβ=-5/13; cosβ=12/13
∴sin(π/4+β)=sinπ/4cosβ+cosπ/4sinβ=7√2/26
cos(π/4+β)=cosπ/4cosβ-sinπ/4sinβ=17√2/26
∴sin(α＋β)=-sin[(π/4-α)-(π/4+β)]=-[sin(π/4-α)cos(π/4+β)-cos(π/4-α)sin(π/4+β)]
=-(-3/5×17√2/26-4/5×7√2/26)=79√2/130