已知coa(a+B)=3/5,sin(B-π/4)=5/13,a、B属于(0,π/2),那么cos(a+π/4)的值为

问题描述:

已知coa(a+B)=3/5,sin(B-π/4)=5/13,a、B属于(0,π/2),那么cos(a+π/4)的值为

coa(a+B)=3/5sin(a+b)=4/5sin(B-π/4)=5/13cos(B-π/4)=12/13cos(a+π/4)=cos[(a+b)-(b-π/4)]=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)=3/5(12/13)+(4/5)(5/13)=56/65