在△ABC中,A、B、C为三内角,tan C=-(cos A-cos B)/(sin A-sin B),sin(B-A)=cos C,求A、C的值.
问题描述:
在△ABC中,A、B、C为三内角,tan C=-(cos A-cos B)/(sin A-sin B),sin(B-A)=cos C,求A、C的值.
不用正弦定理或余弦定理!
答
∵tan C=sinC/cosC又tan C=-(cos A-cos B)/(sin A-sin B),∴sinC/cosC=-(cos A-cos B)/(sin A-sin B),交叉相乘得:sinAsinC-sinBsinC=cosBcosC-cosAcosCcosAcosC+sinAsinC=cosBcosC+sinBsinCcos(A-C)=cos(B-C)-π...