在△ABC中,已知2sin(A B) 2

[(sinA)^2-(sinB)^2-(sinC)^2]/(sinB*sinC)=[(a/2R)^2-(b/2R)^2-(c/2R)^2]/(b/2R*c/2R)=(a^2-b^2-c^2)/bc=-

sin²A＋sin²B＋sin²C＝sin²A＋sin²B＋sin²(A+B)＝sin²A＋sin²B＋(sinAcos

由正弦定理a/sinA=b/cosB=c/sinC令a/sinA=b/cosB=c/sinC=1/k则sinA=aksinB=bksinC=cksin^2A=sin^C+sin^B+根号3sinCsi

sin(B+C/2)=sin[B+(π-A-B)/2])=sin[π/2+(B-A)/2]=cos{π/2-[π/2+(B-A)/2]}=cos[(A-B)/2)=4/5cos(A-B)=2cos&#

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.

∵［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

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

sinA=BC/ABcosA=AC/ABSIN^2A+COS^2A=(BC^2+AC^2)/AB^2根据勾股定理,BC^2+AC^2=AB^2所以SIN^2A+COS^2A=1

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

应当是sin^2A+sin^2B【+】sin^2C=sinB*sinC+sinC*sinA+sinA*sinB吧括号中是要改的.两边同乘以22sin²A+2sin²B+2sin&s

应该是cos((A-C)/2)=2sin(B/2)2sin(B/2)cos(B/2)=cos(B/2)cos((A-C)/2)sinB=sin(π-B/2)cos((A-C)/2)=sin((A+C)

三角形ABC的形状是直角三角形,证明如下:∵a/simA=b/sinB=2R,a=sinA*2R,b=sinB*2R,(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),等式右边有

sin^2A+sin^2B=sin^2C=sin^2(A+B)=(sinAcosB+sinBcosA)^2=sin^2Acos^2B+sin^2Bcos^2A+2sinAcosAsinBcosB左边减

a²≤b²+c²-bcbc≤b²+c²-a²1/2≤(b²+c²-a²)/2bccosa≥1/2a≤60°

sin(A+B)=3/5,sin(A-B)=1/5sin(A+B)=sinAcosB+sinBcosA=3/5sin(A-B)=sinAcosB-sinBcosA=1/5两式相加相减后可得：sinAc

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

2sinAcosB=sin(A+B)+sin(A-B)=sinC+sin(A-B)=sinC所以sin(A-B)=0所以A=B所以,△ABC是等腰三角形.完毕.

a/sinA=b/sinB=c/sinC所以(sin²A-sin²B-sin²C/sinB*sinC)=(a²-b²-c²)/bc=1则(b

稍等再答：你的题目是不是搞错了？再答：Sin=SinBSinC?再问：好像是，不好意思再问：sin²A再问：平方咋个打不起？再问：sin²A再问：真的打不起平方再问：再答：应该是S

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