在△ABC中,若sin(2 A)=-2sin(2-B),3cosA

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

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

由和差化积公式：sinA=sin(B+C)=sinBcosC+cosBsinC=2sinBcosC,所以cosBsinC-sinBcosC=0,即sin(B-C)=0.从而B=C,因此三角形ABC是等

sinA/a=sinB/b=sinC/c=rsin^2A=sin^2B+sin^2C得出a^2*r^2=（b^2+c^2）*r^2即a^2=b^2+c^2所以ABC是直角三角形

原式可化为：a^2+b^2a^2+b^2∴2abcosC0∴cosC

∵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

锐角三角形,高中数学题做过.

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.

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°

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:

sin^2A=sin^2B+sin^2C,sinA=2sinBsinC所以sin^2A-sinA=sin^2B+sin^2C-sinA=sin^2B+sin^2C-2sinBsinC即sinA(sin

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

【1】sin方a+sin方b+sin方c=sin方a+sin方b+sin方(180-(a+b))=sin方a+sin方b+sin方(a+b)=sin方a+sin方b+(sina*cosb+cosa*s

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

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