Question

若sin(3派/4+α)=5/13，cos(派/4－α)＝3/5，0<α<派/4＜β＜3派/4，求cos(α+β)的值？这是大题，过程

Answer 1

哪里有β？

3π/4<3π/4+α<π
所以cos(3π/4+α)<0
sin²(3π/4+α)+cos²(3π/4+α)=1
所以cos(3π/4+α)=-12/13

π/4<β<3π/4
-π/2<π/4-β<0
所以sin(π/4-β)<0
sin²(π/4-β)+cos²(π/4-β)=1
sin(π/4-β)=-4/5

cos(α+β)
=sin[π/2-(α+β)]
=sin[(3π/4+α)-(π/4-β)]
=sin(3π/4+α)cos(π/4-β)-cos(3π/4+α)sin(π/4-β)
=-33/65