△ABC中,asinAsinB+bcos*2A＝根号2a

asinAsinB+b（1-2sinA^2）=根号2乘以asinA(asinB-2bsinA)=根号2乘以a-b根据正弦定理asinB=bsinA得bcosA^2=根号（2)ab/a=根号（2)/co

是要证明呢,还是忘了写题目呢?

△ABC的三个内角ABC的对边分别为abc,asinAsinB+bcos²A=(√2)a,若c²=b²+(√3)a²,求BasinAsinB+bcos²

根据正弦定理a/sinA=b/sinB=c/sinC,原式可变形为:bsin2A+bcos2A=跟号3a,即b=根号3a.将上面的结果带入余弦定理“cosC=(a^2+b^2-c^2)/(2·a·b)

asinAsinB+bcosA＝√2a,sinAsinB+sinBcosA＝√2sinA,sinAsinB+sinB-sinAsinB＝√2sinA,sinB=√2sinA,sinB/sinA=√2,

cosC=√3/3　　b=√3a,　　c²=a²+b²-2abcosC　　c²=a²+3a²-2a²　　c²=2a

asinAsinB+bcos²A=√2a正弦定理sin²AsinB+sinBcos²A=√2sinAsinB(sin²A+cos²A)=√2sinAs

(1)根据正弦定理a=2rsinA,b=2rsinB其中r为外接圆的直径代入得2rsinAsinAsinB+2rsinB（cosA）^2=√2*2rsinA[(sinA)^2+(cosA)^2]sin

根据正弦定理,原函数asinAsinB+bcos2A=a×根号2等价于sin²AsinB+sinBcos2A=sinA*√2.①cos2A=1-2sin²A,等式①等价于sinBs

1、正弦定理：a/sinA=b/sinB=c/sinC得出：a*sinB=b*sinAasinAsinB+bcos^2A=b*sin^2A+bcos^2A=b=√2a即b/a=√2a2、余弦定理：2a

sinA=asinBasinAsinB+bcos²A=√2absin²a+bcos²A=√2ab=√2ab/a=√2(2)b²=2a²c²=

∵a/sinA=b/sinB=c/sinC=2R∴a=2RsinAb=2RsinBc=2RsinC代入已知式子得,2Rsin²AsinB+2RsinBcos²A=4RsinA即2R

右边有没有少一个a啊?再问：落写了一个a,呜呜····能解吗，拜托啦再答：a/sinA=b/sinBasinB=bsinAasinAsinB=bsin2A原式为bsin2A+bcos2A=a根号2b=

sinA=asinBasinAsinB+bcos^2A=√2absin²a+bcos²A=√2ab=√2ab^2=2a^2c^2=b^2+√3a^2=2a^2+√3a^2c=a√[

∵asinAsinB+bcos2A=2a，∴sin2AsinB+sinBcos2A=2sinA，∴sinB=2sinA，∴b=2a，∴ba=2，故选：D．

√2再问：详细过程，谢谢再答：由正弦定理可知：a/sinA=b/sinB∴asinB=bsinA∴asinAsinB=bsin²A∴题目中的条件等式可化为：bsin²A+bcos&

原式化为,aSinA*SinB+b(1-Sin^2A)=√2*a(原式为√2A,错的）或aSinA*SinB+b-bSin^2A=√2*a（1）由三角形正弦定理SinA/a=SinB/b=R,(R为外

/>原式化为,aSinA*SinB+b(1-Sin^2A)=√2*a(原式为√2A,错的）或aSinA*SinB+b-bSin^2A=√2*a（1）由三角形正弦定理SinA/a=SinB/b=R,(R

asinAsinB+bcos²A＝√2a,sin²AsinB+sinBcos²A＝√2sinA,sin²AsinB+sinB-sin²AsinB＝√2