sin(A 2B) sinB-2cos(A B)

2R(sinA+sinC)(sinA-sinC)=(√3a-b)sinB有正弦定理2RsinA=a,2RsinC=c所以(a+c)(sinA-sinC)=(√3a-b)sinBsinA=a/2R,si

[(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=-

1）在方程两端同时乘以2R,有正弦定理得(a-b)x²+(c-a)x+(b-c)=0;由方程有两个相等的实根的(c-a)²-4(a-b)(b-c)=0;又(c-a)²=(

这个式子可以化为：b2-c2=a(√2b-a)b2-c2=√2ab-a2a2+b2-c2=√2abcosC=a2+b2-c2/2ab=√2ab/2ab=√2/2又因为在△ABC中,c在0—180度,所

正弦定理这一定理对于任意三角形ABC,都有a/sinA=b/sinB=c/sinC=2RR为三角形外接圆半径a/sinA=b/sinB=c/sinC=2R带入原式=2R[（a/2R）^2-(c-R)^

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

左边=sin(A+B)sin(B-A)+sin²C=sin(180-C)sin(B-A)+sin²C=sinCsin(B-A)+sin²C=sinC[sin(B-A)+s

左边=sin(A+B)sin(B-A)+sin²C=sin(180-C)sin(B-A)+sin²C=sinCsin(B-A)+sin²C=sinC[sin(B-A)+s

(1)（b+a）\a=sinB\（sinB-sinA）=(b+a)/a=b/(b-a)b^2-a^2=ab2sinAsianB=2sin^2C2ab=2c^2c^2=ab=b^2-a^2a^2+c^2

∠C=90,则A+B=90°,所以sinA=cosB.y=sin²A+2sinB=cos²B+2sinB=1-sin²B+2sinB设sinB=t,∵0°

2R(sinA+sinC)(sinA-sinC)=(√3a-b)sinB有正弦定理2RsinA=a,2RsinC=c所以(a+c)(sinA-sinC)=(√3a-b)sinBsinA=a/2R,si

a/sinA=b/sinB=c/sinC=2R=2√2=>a=2RsinA,b=2RsinB,c=2RsinC2√2（sin²A-sin²C）=（a-b）sinB=>4R²

因为∠C=90°所以A+B=90sinA=sin（B+C）=sin(B+90°)=-cosBsin^2(A)+2sinB=COS²B+2sinB=1-sin²B+2sinB=-（s

你的表述出现了一些问题,我想应该是求证：［sin（A/2）］^2＋［sin（B/2）］^2＋［sin（C/2）］^2＝1－2sin（A/2）sin（B/2）sin（C/2）若是这样,则方法如下：在三角

是直角三角形由正弦定理得(a+b)/a==sinB/(sinB-sinA)=b/(b-a)所以b^2-a^2=ab又因为2sinAsinB=2sin^2C,得ab=c^2所以有b^2-a^2=c^2也

C因为角C为90度所以角A、角B互余所以(sinA)^2+(sinB)^2=1所以y=-(sinB)^2+2sinB+1=-(sinB-1)^2+2又因为-1≤sinB≤1所以-2≤y≤2所以选C

sin²A-sin²B-sin²C=sinBsinCa/sinA=b/sinB=c/sinC则由sin²A-sin²B-sin²C=sinB

因为m垂直n所以m×n=0(要加向量符号)即(sinB+sinC,sinA-sinB)×(sinB-sinC,sin(B+C))=0又sin(B+C)=sin(π-A)=sinA所以原式=[(sinB

sin²A-sin²(180-A-B)=sinAsinB-sin²Bsin²A-sin²(A+B)=sinAsinB-sin²Bsin&su

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