已知角ab sin(a+b)=-3/5 sin(b-π/4)=12/13 求cos(a+π/4)) 答案是-56/65

已知角ab sin(a+b)=-3/5 sin(b-π/4)=12/13 求cos(a+π/4)) 答案是-56/65

cos(a+pi/4) = cos(a+b-b+pi/4) = cos(a+b)cos(b-pi/4)+sin(a+b)sin(b-pi/4) = -4/5*5/13-3/5*12/13 =-56/65
问题是如何得到cos(a+b) and cos(b-pi/4)
我尝试了一下从范围推到,但是好像不可以,你的角ab有其他限制吗?例如都是三角形的内角?

con(a+π/4)=con[(a+b)-(b-π/4)]=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)=代入数值