SinA=sinCcosB+sinBcosC证明sinA=sin(B+C)证明sinA+sin(B-C)=2sinBcosCSin(A)=sin[π-(B+C)]=sinπcos(B+C)-cos(B+C)*cos(π)=sin(B+C)sinA=sin(B+C) 所以sinA+sin(B-C)=sin(B+C)+sin(B-C)=sinBcosC+sinCcosB+sinBcosC-sinCcosB=2sinBcosC
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