在△abc中,若cos^2b－sin^2a=cos^2c,试判断△abc的形状
2个回答

cos^2A=cos^2(B+C)=1-sin^2(B+C)

sin(B+C)=sinBcosC+sinCcosB

所以cos^2A+cos^2B+cos^2C=cos^2B+cos^2C-(sin^2Bcos^2C+cos^2Bsin^2C)-2(sinBcosCcosBsinC) +1=1

所以cos^2B+cos^2C-(sin^2Bcos^2C+cos^2Bsin^2C)=2(sinBcosCcosBsinC)

化简为cos^2B（1-sin^2C）+cos^2C(1-sin^2B)=

2(sinBcosCcosBsinC)

1-sin^2C=cos^2C

1-sin^2B=cos^2B

所以cos^2Bcos^2C+cos^2Ccos^2B=2(sinBcosCcosBsinC)

即为2cos^2Bcos^2C=2(sinBcosCcosBsinC)

即为cosBosC=sinBsinC

cos(B+C)=0所以B+C=90度

直角三角形

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