丨a丨=2,丨b丨=(4cosa,-4sina)且a垂直(a-b)
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![丨a丨=2,丨b丨=(4cosa,-4sina)且a垂直(a-b)](/uploads/image/f/1375130-2-0.jpg?t=%E4%B8%A8a%E4%B8%A8%3D2%2C%E4%B8%A8b%E4%B8%A8%3D%284cosa%2C-4sina%29%E4%B8%94a%E5%9E%82%E7%9B%B4%28a-b%29)
证明:∵在三角形ABC中,∴A+B+C=180度,得SINA=SIN(B+C)则A/2=90度-(B+C)/2,得COSA/2=SIN((B+C)/2)左边=Sin(B+C)+SinB+SinC则4C
证明:∵在三角形ABC中,∴A+B+C=180度,得SINA=SIN(B+C)则A/2=90度-(B+C)/2,得COSA/2=SIN((B+C)/2)左边=Sin(B+C)+SinB+SinC则4C
1.Acos(a+b)cos(a-b)=(cosa*cosb-sina*sinb)*(cosa*cosb+sina*sinb)=cosa*cosa*cosb*cosb-sina*sina*sinb*s
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)=
首先,由a·b=0并化简可得5/4*cos(a+b)=cos(a-b);然后,展开移项sin*sin=1/9cos*cos;最后可得tgA*tgB=1/9.公式自己去背,别问我!
cosA=4/5sinA=3/5tanA=3/4tan(A-B)=(tanA-tanB)/(1+tanAtanB)=(3/4-tanB)/(1+3tanB/4)=-1/3tanB=13/9(tanB)
已知向量a=(cosα,sinα),b=(cosβ,sinβ),且丨a-b丨=2倍根号5/5,若-π/2<β<0<α<π/2,且sinβ=-5/13求sinα的值【解】向量a=(cosα,sinα),
√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
cos^2a-sin^2b=(1+cos2a)/2-(1-cos2b)/2=(cos2a+cos2b)/2=cos(a+b)cos(a-b)=1/3
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
由和差化积公式:cosa+cosb=2cos{(a+b)/2}cos{(a-b)/2}得:cos2a+cos2b=2cos(a+b)cos(a-b)又由已知条件4sinasinb=根号2,4cosac
差角公式:cosa=cos[(2a+b)-(a+b)]=cos(2a+b)*cos(a+b)+sin(2a+b)*sin(a+b)因为a,
应该是cosA+cosB=2cos[(A+B)/2]cos[(A-B)/2]吧.
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
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
因为cos(A+B)cos(A-B)=(1/2)(cos2A+cos2B)=(1/2)[2(cosA)^2-1+2(cosB)^2-1]=(cosA)^2+(cosB)^2-1=1/4所以cosA^2
/>因为:cos(a+b)=4/5,a+b∈[7pi/4,2pi].,所以:sin(a+b)=-3/5因为:cos(a-b)=-4/5,a-b∈[3pi/4,pi].,所以:sin(a-b)=3/5所