cos(a-b)=12 13,sin(a b)=-3 5,求sin2x
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cos(A+B)*cos(A-B)=1/3(1/2)*(cos(2A)+cos(2B))=1/3.(积化和差)cos(2A)+cos(2B)=2/3(cosA)^2-(sinB)^2=(cos2A+1
a=1;b=1;%换成你的数据f=@(x,y)a*cos(x)+b*cos(y);ezsurf(f)
cosa=-bsin2acot(a/2)=2sinacosa(1+cosa)/sina
cos(a-b)cos(b-c)+sin(a-b)sin(b-c)=cos[(a-b)-(b-c)]=cos(a+c-2b)
√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
tan[(a+b)/2]=sin[(a+b)/2]/cos[(a+b)/2]SinC=Sin(180-a-b)=Sin(a+b)=2*[sin[(a+b)/2]*cos[(a+b)/2]]
cos^2a-sin^2b=(1+cos2a)/2-(1-cos2b)/2=(cos2a+cos2b)/2=cos(a+b)cos(a-b)=1/3
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
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
(cosa-cosb)的平方+(sina-sinb)的平方=16/13整理2-2cosacosb-2sinasinb=16/13cosacosb+sinasinb=5/13cos(a-b)=5/13
[sin(a+b)-2sinacosb]/[2sinasinb+cos(a+b)]=[cosasinb-sinacosb]/[sinasinb+cosacosb]=sin(b-a)/cos(b-a)=
应该是cosA+cosB=2cos[(A+B)/2]cos[(A-B)/2]吧.
倍角公式sin2x=2sinxcosx所以左边=[2sin(A+B)/2*cos(A+B)/2][2sin(A-B)/2*cos(A-B)/2]=sin(A+B)*sin(A-B)
原题是这样子吧:cos(a+b)cos(a-b)=1/5,则(cosa)^2-(sinb)^2=?cos(a+b)cos(a-b)=(cosacosb-sinasinb)(cosacosb+sinas
cos²B-cos²C=sin²Acos²B=1-sin²Bcos²C=1-sin²Csin²C-sin²B=
你的式子有一项好像抄错了如果原题是求证a²(cos2B-cos2C)+b²(cos2C-cos2A)+c²(cos2A-cos2B)=0的话证明如下:a²(co
∵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-