三道数学题(详细过程)1. △ABC中,tanA/tanB=a²/b²,判断三角形的形状2.在△ABC中,(a²+b²)sin(A-B)=(a²-b²)sin(A+B),判断△ABC的形状3.在△ABC中,已知sinA/sinC=sin(A-B)/sin(B-C),求证2b²=
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![三道数学题(详细过程)1. △ABC中,tanA/tanB=a²/b²,判断三角形的形状2.在△ABC中,(a²+b²)sin(A-B)=(a²-b²)sin(A+B),判断△ABC的形状3.在△ABC中,已知sinA/sinC=sin(A-B)/sin(B-C),求证2b²=](/uploads/image/z/104013-45-3.jpg?t=%E4%B8%89%E9%81%93%E6%95%B0%E5%AD%A6%E9%A2%98%EF%BC%88%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%EF%BC%891.+%E2%96%B3ABC%E4%B8%AD%2CtanA%2FtanB%3Da%26sup2%3B%2Fb%26sup2%3B%2C%E5%88%A4%E6%96%AD%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E5%BD%A2%E7%8A%B62.%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%EF%BC%88a%26sup2%3B%2Bb%26sup2%3B%29sin%28A-B%29%3D%28a%26sup2%3B-b%26sup2%3B%29sin%28A%2BB%29%2C%E5%88%A4%E6%96%AD%E2%96%B3ABC%E7%9A%84%E5%BD%A2%E7%8A%B63.%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E5%B7%B2%E7%9F%A5sinA%2FsinC%3Dsin%28A-B%29%2Fsin%28B-C%29%2C%E6%B1%82%E8%AF%812b%26sup2%3B%3D)
三道数学题(详细过程)1. △ABC中,tanA/tanB=a²/b²,判断三角形的形状2.在△ABC中,(a²+b²)sin(A-B)=(a²-b²)sin(A+B),判断△ABC的形状3.在△ABC中,已知sinA/sinC=sin(A-B)/sin(B-C),求证2b²=
三道数学题(详细过程)
1. △ABC中,tanA/tanB=a²/b²,判断三角形的形状
2.在△ABC中,(a²+b²)sin(A-B)=(a²-b²)sin(A+B),判断△ABC的形状
3.在△ABC中,已知sinA/sinC=sin(A-B)/sin(B-C),求证2b²=a²+c²
三道数学题(详细过程)1. △ABC中,tanA/tanB=a²/b²,判断三角形的形状2.在△ABC中,(a²+b²)sin(A-B)=(a²-b²)sin(A+B),判断△ABC的形状3.在△ABC中,已知sinA/sinC=sin(A-B)/sin(B-C),求证2b²=
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.