已知cos(π/4-a)=3/5,sin(5π/4+b)=-12/13,a∈(π/4,3π/4)b∈(0,π/4),求sin(a+b)值
问题描述:
已知cos(π/4-a)=3/5,sin(5π/4+b)=-12/13,a∈(π/4,3π/4)b∈(0,π/4),求sin(a+b)值
答
a∈(π/4,3π/4)
π/4-a∈(-π,0)
sin(π/4-a)0
sin²(π/4+b)+cos²(π/4+b)=1
cos(π/4+b)=5/13
sin(a+b)=sin[(π/4+b)-(π/4-a)]
=sin(π/4+b)cos(π/4-a)-cos(π/4+b)sin(π/4-a)
=16/65