2cosa+2cosb+2cosc=2(2cosa-1)+(2cosb-1)+2cosc=0cos2a+cos2b+2cosc=02cos(a+b)cos(a-b)+2cosc=0cosccos(a-b)-cosc=0cosc[cos(a-b)-cosc]=0cosc[cos(a-b)+cos(a+b)]=0cosc[2cosacosb]=0cosccosacosb=0即a=90°或b=90°或c=90°.综上所述,此三角形为直角三角形.
2sinAsinB=cos(A-B)-cos(A+B)=cos(A-B)+cosC=1+cosC所以cos(A-B)=1,A=B,三角形ABC是等腰三角形.
方法一: ∵(cosa+2cosc)/(cosa+2cosb)=sinb/sinc, ∴cosasinc+2coscsinc=cosasinb+2cosbsinb ∴cosa(sinc-sinb)=sin2b-sin2c=2sin(b-c)cos(b+c)=-2sin(b-c)cosa 一、当cosa=0时,a=90°,此时三角形是直角三角形. 二、当cosa≠0时,两边同
钝角三角形
(1)(cosA)^2=(sinB)^2+(cosC)^2+sinAsinB,1-(sinA)^2=(sinB)^2+1-(sinC)^2+sinAsinB,∴(sinC)^2=(sinA)^2+(sinB)^2+sinAsinB,由正弦定理,c^2=a^2+b^2+ab,由余弦定理,cosC=-1/2,C=120°.(2)c=√3,△ABC的外接
因为a^2=b^2+c^2-2bcCos60°所以A=60°CosA=1/2CosA=-Cos(B+C)=-CosB CosC+SinB SinC=1/2SinB SinC=3/4CosB CosC=1/4Cos(B-C)=CosB CosC+SinB SinC=1所以B-C=0,因为A+B+C=180°所以A=B=C=60°所以三角形ABC是等边三角形
cosA+cosB+cosC=2cos[(A+B)/2]*cos[(A-B)/2]+cosC≤2cos[(A+B)/2]]+cosC≤2sin(C/2)+cosC=-2sin(C/2)^2+2sin(C/2)+1=-2[sin(C/2)-1/2]^2+3/2≤3/2所以构不成三角形
2sinBcosA=sinAcosC+cosAsinC=sin(A+C)=sinB2sinBcosA-sinB=0sinB(2cosA-1)=0B为三角形内角,sinB恒>0,因此只有2cosA-1=0cosA=1/2A为三角形内角,A=π/3由余弦定理得a^2=b^2+c^2-2bccosAb=2 c=1 A=π/3代入,得a^2=2^2+1^2-2*2*1*cos(π/3)=4+1-4*(1/2)=3a^2+c^2=3+1=4=b^2三角形是直角三角形.
(cosB)^2-(cosC)^2=(sinA)^2(1-(sinB)^2)-(1-(sinC)^2)=(sinA)^2(sinA)^2+(sinB)^2=(sinC)^2而:a/sinA=b/sinB=c/sinC=2R所以:a^2+b^2=c^2△ABC为直角三角形
(cosA)^2+(cosB)^2+(cosC)^2 = (1+cos2A)/2+(1+cos2B)/2+(cosC)^2 = (cos2A+cos2B)/2+(cosC)^2+1 = [2cos(A+B)cos(A-B)]/2+(cosC)^2+1 = -cosCcos(A-B)+(cosC)^2+1 =