在△ABC中,(a-b)sinC (b-c)sinA
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sin²B+sin²C=sin²A+sinBsinC,正弦定理:sinA=A/2R,sinB=b/2R,sinC=c/2R,b²+c²=a²
原式可化为a^2+b^2-c^2=ab也即是a^2+b^2-c^2/2ab=1/2也即是cosC=1/2所以C=60°联立2sinC=sinA+sinB可得等边三角形
sin²A+sin²B=2sin²C由正弦定理a^2+b^2=2c^2代入余弦定理:cosC=(a^2+b^2-c^2)/(2ab)=c^2/(2ab)>0所以:cosC
sin²A=sin²B+sin²C,a/sinA=b/sinB=c/sinC=2R(a/2R)^2=(b/2R)^2+(c/2R)^2a^2=b^2+c^2,ABC是直角
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用正弦定理化作a^2-b^2+c^2=ac整理得到cosB=a^2-b^2+c^2/2ac=1/2B=π/3
sin²B+sin²C=sin²A+sinBsinC由正弦定理得到b^2+c^2=a^2+bc余弦定理得到cosA=(b^2+c^2-a^2)/2bc=1/2又在三角形中
由正弦定理,原式可化为a^2+c^2-ac=b^2即[(a^2+c^2-b^2)/2ac]=0.5即cosB=0.5∴B=π/3
∵在△ABC中,sin(A+B)=sinC∴sinC·sin(A-B)=sin²Csin(A-B)=sinC又∵sinC=sin(A+B)∴sin(A-B)=sin(A+B)sinAcosB
答:三角形ABC中,(a²+b²)sin(A-B)=(a²-b²)sin(A+B)移项合并:[sin(A-B)-sin(A+B)]a²=-[sin(A
sin方A+sin方B=sin方C根据正弦定理:a/sinA=b/sinB=c/sinC=2Ra^2/(2R)^2+b^2/(2R)^2=c^2/(2R)^2即:a^2+b^2=c^2,符合勾股定理,
sin^2A+sin^2B=sin^2C=sin^2(A+B)=(sinAcosB+sinBcosA)^2=sin^2Acos^2B+sin^2Bcos^2A+2sinAcosAsinBcosB左边减
a²[sin(A-B)-sin(A+B)]+b²[sin(A-B)+sin(A+B)]展开,得b²2sinAcosB=a²cosAsinB式①由正弦公式得a/s
这是个直角三角形用正弦定理证明a/sinA=b/sinB=c/sinC=ksinA=a/k,sinB=b/k,sinC/c/k代入sin²A=sin²B+sin²C即可得
sin^2A+sin^2B+sin^2C=(1-cosA)/2+(1-cosB)/2+(1-cos^2C)=2-cos(A+B)cos(A-B)-cos^2C=2+cosCsoc(A-B)-cos^2
(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)sin^A*(sin(A+B)-sin(A
(sina-sinb)(sina+sinb)=(sina)^2-(sinb)^2=(sina)^2-(sina)^2(sinb)^2-(sinb)^2+(sina)^2(sinb)^2=(sina)^
由题意:1-sin^2A=cos^2Asin^2B+cos^2C+2sinAsinBcos(A+B)==sin^2B+cos^2C-2sinAsinBcosC=sin^2B+cosC(cosC-2si
选C等腰三角形或者直角三角形因为B=C则推出为等腰三角形B+C=π/2则推出为直角三角形二者是或的关系再问:是(或(直角三角形)还是(等腰直角三角形)?再答:不一定是等腰直角三角形因为B=C则只能推出
你是二十一中的么如果是你是几班的(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)sin^A