且3sinb=sin(2a b)化简

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

解题思路：利用和、差化积公式求解。解题过程：最终答案：略

sina+cosb=1/3,平方sin^2a+2sinacosb+cos^2b=1/9sinb-cosa=1/2,平方sin^2b-2cosasinb+cos^2a=1/4相加2-2（sinacosb

13sinb=sin(2a+b)3sin(a+b-a)=sin(a+b+a)3sin(a+b)cosa-3cos(a+b)sina=sin(a+b)cosa+sinbcos(b+a)2sin(a+b)

由tana=0.5得sin2a=2*0.5/(0.5^2+1)=4/5cos2a=(1-0.5^2)/(1+0.5^2)=3/5sin(2a+b)=sin2a*cosb+cos2a*sinb=4/5c

打开平方得：sin^2A+sin^2B-sin^C=sinA*sinB正弦定理sinA=a/2R其它也一样a2/4R2+b2/4R2-c2/4R2=ab/4R2a2+b2-c2=ab余弦定理a2+b2

sinb=sin[(a+b)-a]=sin(a+b)cosa-cos(a+b)sina=sin(a+b)cosa-sinb/sina*sina=sin(a+b)cosa-sinb2sinb=sin(a

由正弦定理a/sinA=b/sinB=c/sinCsinA+sinB=√2sinC所以a+b=√2ca+b+c=2√2+2所以√2c+c=2√2+2所以AB=c=2a+b=√2c=2√2S=1/2ab

(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

应用和差化积公式有：sin(2a+b)=sin(a+b+a)=sin(a+b)cosa+cos(a+b)sina,sinb=sin(a+b-a)=sin(a+b)cosa-cos(a+b)sina,而

∵sina=-3/5,a∈(∏,3∏/2),∴cosa=-√(1-sin²a)=-4/5∵sinb=12/13,b∈(∏/2,∏),∴cosb=-√(1-sin²b)=-5/13∴

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=√2sinb,...(1)tana=√3tanb...(2)(1)÷（2）得cosa=√2cosb/√3...(3)(1)²+（3）²得1=2sin²b+2co

4tana/2=1-tan^2a/22tana/2/(1-tan^2a/2)=1/2=tanasina=(1/2)/√(1+1/4)=√5/5cosa=√(20/25)=2√5/5sin2a=2*5*

3*sin(a+b-a)=sin(a+b+a);3*sin(a+b)cos(a)-3*cos(a+b)sin(a)=sin(a+b)cos(a)+cos(a+b)sin(a);2*sin(a+b)co

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

sin=sinAcosB+cosAsinB=根号2/2（sinA+sinB）cosB=cosA=根号2/2A=B=45°边长b=c=根号2/2ab=c=2a=2根号2

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