Question

已知sin(a+3π/4)=5/13,cos(π/4-b)=4/5,且-π/4＜a＜π/4,π/4＜b＜3π/4求cos(a-b)的值

Answer 1

由-π/4＜a＜π/4, π/4＜b＜3π/4，得：
π/2＜a+3π/4＜π, -π/2＜π/4-b＜0，
由sin(a+3π/4)=5/13, cos(π/4-b)=4/5，得：
cos(a+3π/4)=-12/13, sin(π/4-b)=-3/5，
所以
cos(a-b)=cos[π+(a-b)]=cos[(a+3π/4)+(π/4-b)]
=cos(a+3π/4)*cos(π/4-b)-sin(a+3π/4)*sin(π/4-b)
=-12/13*4/5+5/13*3/5=-33/65。