sinbsinc=1 cosa
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a/sinA=b/sinB=c/sinC=2RS△ABC=(ab/2)·sinC=(bc/2)·sinA=(ac/2)·sinB=abc/(4R)故S=(ab/2)·sinC=1/2a*asinB/s
(2sinA+cosA)/(sinA-cosA)=-5上下同除cosA(2tanA+1)/(tanA-1)=-52tanA+1=-5tanA+57tanA=4tanA=4/71.(sinA+cosA)
由正弦定理得sinB=b*(sinA/a)sinC=c*(sinA/a)代入得(1/2)*a^2*[(sinBsinC)/sinA]=(1/2)*a^2*[(sinA*bc)/a^2]=(1/2)*b
三角形面积公式为:S=(1/2)abSinC=(1/2)acSinB=(1/2)bcSinA证:已知S=(1/2)a²sinBsinC/sinA由正弦定理:a/SinA=b/SinB=c/S
令k=a/sinA=b/sinBb=ksinB因为S=1/2absinC=1/2a*ksinBsinC=1/2a*(a/sinA)sinBsinC=1/2*a^2*sinBsinC/sinA
S=1/2*absinC这个公式吧,他是由bsinA是高乘以底a得来的现在只要证出1/2*absinC=1/2*a^2*(sinBsinC)/(sinA)就可以了也就是bsinC=a*(sinBsin
a/sinA=b/sinB=>b=a*sinB/sinAS=1/2absinC=1/2a*a*sinBsinC/sinA=1/2a^2sinBsinC/sinA
tgA+cosA/(1+sinA)=sinA/cosA+cosA/(1+sinA)=[sinA(1+sinA)+cos²A]/[cosA(1+sinA)]=(1+sinA)/[cosA(1+
∵[sin(A/2)]^2+sinBsinC=1,∴2[sin(A/2)]^2+2sinBsinC=2,∴2sinBsinC=1+1-2[sin(A/2)]^2=1+cosA=1+cos(180°-B
根据已知,只能推导出cosB∈[1/2,1],cosA∈[0,1],A和B的关系无法推导
sinBsinC=1/2[cos(B-C)-Cos(B+C)]=1/2[cos(B-C)+1/2]=1/2cos(B-C)=0∴B-C=90°又B+C=120B大于C,所以B=105°,C=15°
1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina=[(1+sina+cosa)²+(1+sina-cosa)²]/[(1+sina)&
cos(B+C)=cosBcosC-sinBsinC=-2分之1,∴B+C=120°∴A=60°
题目应是“在三角形ABC中,已知a平方+c平方-b平方=ac且cosA=2sinBsinc-1,试确定三角形ABC形状”首先由余弦定理的cosB=(a平方+c平方-b平方)/2ac=ac/2ac=0.
证明:(1+sinα+cosα)+2sinαcosα=(1+sinα+cosα)+2sinαcosα=(sinα+cosα)+(sinα)^+(cosα)^+2sinαcosα=(sinα+cosα)
(cosa-sina)^2=(cosa)^2-2sinacosa+(sina)^2=1-2sinacosa=1-2*1/8=3/4cosa-sina=+-√3/2
∵在△ABC中,a:b:c=1:3:5,∴设a=k,b=3k,c=5k,由正弦定理asinA=bsinB=csinC=2R,即sinA=a2R,sinB=b2R,sinC=C2R,则原式=2a2R−b
这不是分子提取一个(sina+cosa)就好了?还能继续化简吧=(sina+cosa)(1+sina+cosa)/(1+sina+cosa)=(sina+cosa)再问:(sina+cosa)
①cosBcosC-sinBsinC=cos(B+C)=cos(π-A)=-cosA则cosA=-1/2又A∈(0,π)则A=2π/3②若a=2√3则由余弦定理a²=b²+c
∵sinBsinC=cos2A/2∴1+cosA=2sinBsinC∴2sinBsinC-cosA=1,即2sinBsinC+cos(B+C)=1,即cos(B-C)=1∵在△ABC中,-π<B-C<