作业帮cos(a-b)和sin(a-b)什么关系
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/11 19:55:46
证明：输入过于麻烦,用换元法吧设A=sin²A,B=sin²B∵sin^4a/sin^2b+cos^4a/cos^2b=1即A²/B+(1-A)²/(1-B)=
最后=1具体过程我打给你了这里打不下
Sin^2A+Sin^2B-Sin^2ASin^2B+Cos^2ACos^2B=Sin^2A(1-Sin^2B)+Sin^2B+Cos^2ACos^2B=Sin^2ACos^2B+Cos^2ACos^
sin²a*cos²b-cos²a*sin²b=sin²a*(1-sin²b)-(1-sin²a)sin²b=sin&s
cos(a-b)cos(b-c)+sin(a-b)sin(b-c)=cos[(a-b)-(b-c)]=cos(a+c-2b)
A+B=π-Ccos(A+B)=cos(π-C)=-cosCcos[(A+B)/2]=cos[(π-C)/2]=cos(π/2-C/2)=sin(C/2)sin(3A+3B)=sin3*(A+B)=s
√3/2*cosa+1/2*sina=cosπ/6*cosa+sinπ/6sina=cos(π/6-a)cosa-sina=√2(√2/2cosa-√2/2sina)=√2(cosπ/4*cosa-s
吧（a+b）看成一个值-a是另一个值再用cos(a+b)=cosacosb-sinasinb就展开了
用最基本的公式：sin(A+B)=sinAcosB+cosAsinB.所以：把sin(B-y)变为-sin(y-B)再把负号提到前面,最后为sin(a+y)
Cos(a+b)cos(a-b)=[cos(a+b+a-b)+cos(a+b-a+b)]/2=(cos2a+cos2b)/2=(1-2sin²a+2cos²b-1)/2=cos&s
http://hi.baidu.com/wolf224/blog/item/296fec120385450a5aaf53c7.html
1.sin(a+b)cos(c-b)-cos(b+a)sin(b-c)=sin(a+b)cos(c-b)+cos(a+b)sin(c-b)=sin[(a+b)+(c-b)]=sin(a+c)2.sin
sin²A+sin²B=cos²C(1-cos2A)/2+(1-cos2B)/2=1-sin^2Ccos2A+cos2B=2sin^2C2cos(A+B)*cos(A-B
cos²B-cos²C=sin²Acos²B=1-sin²Bcos²C=1-sin²Csin²C-sin²B=
原式=(sin²a+cos²a)(sin²a-cos²a)=(sina+cosa)(sina-cosa)或者=(sin²a+cos²a)(s
∵cosx=(e^ix+e^-ix)/2e^ix=cosx+isinx∴cos(a+b)=[e^i(a+b)+e^-i(a+b)]/2=[(cosa+isina)(cosb+isinb)+(cosa-
sin(a-b)sin(b-r)-cos(a-b)cos(r-b)=-sin(a-b)sin(r-b)-cos(a-b)cos(r-b)=-[cos(a-b)cos(r-b)+sin(a-b)sin(
=(cosacosb-sinasinb+cosacosb+sinasinb)/(sinacosb+cosasinb+sinacosb-cosasinb)=2cosacosb/2sinacosb=cot