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
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![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](/uploads/image/z/1035915-51-5.jpg?t=%E5%B7%B2%E7%9F%A5%E8%A7%92ab+sin%28a%2Bb%29%3D-3%2F5+sin%28b-%CF%80%2F4%29%3D12%2F13+%E6%B1%82cos%28a%2B%CF%80%2F4%29%29+%E7%AD%94%E6%A1%88%E6%98%AF-56%2F65)
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
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