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幼苗
共回答了16个问题采纳率:100% 举报
1、△ABC中,sinA/sinB=a/b;则tanA/tanB=acosB/bcosA=a²/b²;
即:cosB/cosA=a/b=sinA/sinB;则sinBcosB=sinAcosA;
即:sin(2B)=sin(2A);则2B=2A或π-2B=2A;即A=B;或A+B=π/2;
故△ABC为等腰三角形(a=b),或直角三角形(C=π/2).
2、△ABC中,(a²+b²)sin(A-B)=(a²-b²)sin(A+B);
则a²[sin(A-B)-sin(A+B)]=-b²[sin(A+B)+sin(A-B)];
有-2a²cosAsinB=-2b²sinAcosB;即:a²cosA/(b²cosB)=sinA/sinB=a/b;
则acosA=bcosB;则a/b=sinA/sinB=cosB/cosA;则sinBcosB=sinAcosA;
即:sin(2B)=sin(2A);则2B=2A或π-2B=2A;即A=B;或A+B=π/2;
故△ABC为等腰三角形(a=b),或直角三角形(C=π/2).
3、证明:
∵sinA/sinC=sin(A-B)/sin(B-C)=(sinAcosB-cosAsinB)/(sinBcosC-cosBsinC);
∴sinAsinBcosC-sinAcosBsinC=sinAcosBsinC-cosAsinBsinC;
∴sinB(sinAcosC+cosAsinC)=2sinAsinCcosB,即sinBsin(A+C)=2sinAsinCcosB;
又∵△ABC中,B=π-(A+C),∴sin(A+C)=sinB;
∴sinBsinB=2sinAsinCcosB,即b²=2accosB;
又∵△ABC中cosB=(a²+c²-b²)/(2ac);故b²=2accosB=2ac(a²+c²-b²)/(2ac);
整理有:2b²=a²+c² ;证毕.
1年前
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