在△abc中 sin^2a 2=c-b 2c
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等式两边乘以4R^2用正弦定理得到a^2-c^2+b^2=ab根据余弦定理c^2=a^2+b^2-2abcosC代入第一个式子得到cosC=1/2因为C是三角形内角所以C=60度
sin^2A+sin^2B=sin^2C利用三角形正弦定理sinA/a=sinB/b=sinC/c显然a^2+b^2=c^2所以边c所对的角C为直角.
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
正弦定理:a/sinA=b/sinB=c/sinC-b?=c?/2,∴sin?A-sin?B=sin?C/2=(√3/2)?/2=3/8sin?A=(1-cos2A)/2,sin?B=(1-cos2B
∵(a2+b2)sin(A-B)=(a2-b2)sinC,∴(a2+b2)(sinAcosB-cosAsinB)=(a2-b2)(sinAcosB+cosAsinB),可得sinAcosB(a2+b2
由余弦定理得:a2=b2+c2-2bccosA,故S△ABC=a2-(b-c)2=a2-b2-c2+2bc=2bc-2bccosA.利用三角形的面积公式求出S△ABC=12bcsinA,故有S△ABC
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
由正弦定理和已知可以得到:a^2=b^2+c^2.所以三角形为直角三角形.
根据正玄定理原式=a^2+b^2=2c^2cosC=(a^2+b^2-c^2)/2ab=(a^2+b^2)/4ab>=2ab/4ab=1/2所以∠C
sin^2A+sin^2B=sin^2C=sin^2(A+B)=(sinAcosB+sinBcosA)^2=sin^2Acos^2B+sin^2Bcos^2A+2sinAcosAsinBcosB左边减
根据正弦定理:a/sinA=b/sinB=c/sinC=2R,R为该三角形外接圆半径,则:a/2R=sinAb/2R=sinBc/2R=sinC因此:sinA:sinB:sinC=a:b:c=3:2:
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
证明:由余弦定理a2=b2+c2-2bccosA,b2=a2+c2-2accosB,(3分)∴a2-b2=b2-a2-2bccosA+2accosB整理得a2-b2c2=acosB-bcosAc(6分
根据正弦定理,原式可化为sin^2Bsin^2C+sin^2Csin^2B=2sinBsinCcosBcosC2sin^2Csin^2B=2sinBsinCcosBcosCsinBsinC=cosBc
证明:原式化为a2[sin(A-B)-sin(A+B)=-b2[sin(A-B)+sin(A+B)],即a2[sin(A+B)-sin(A-B)=b2[sin(A-B)+sin(A+B)],故2a2c
2sin2CcosC-sin(2C+C)=根号3(1-cosC)2sin2CcosC-(sin2CcosC+cos2CsinC)=根号3*(2sin^2C/2)sin2CcosC-cos2CsinC=
由题意:1-sin^2A=cos^2Asin^2B+cos^2C+2sinAsinBcos(A+B)==sin^2B+cos^2C-2sinAsinBcosC=sin^2B+cosC(cosC-2si
/c=sinB/sinC&bsinB=csinC=>sinB/sinC=c/b=>b/c=c/b=>b^2=c^2i.e.b=c=>B=C=>A=180度-2B=>sinA=sin(2B)=>sin^
改了结果相同由正弦定理a/sinA=b/sinB=c/sinC(sinA)^2=(sinB)^2+(sinC)^2等价于a^2=b^2+c^2可知△ABC直角三角形A=π/2sinA=2sinBcos