Question

求证若cos²A+cos²B+cos²C=1,则此三角形为直角三角形

Answer 1

cos^2B+cos^2C=1-cos²A=sin^2A
cos^2B+cos^2C=sin^2(B+C)
cos^2B+cos^2C=sin^2Bcos^2C+2(sinBcosCcosBsinC)+cos^2Bsin^2C
cos^2B+cos^2C-(sin^2Bcos^2C+cos^2Bsin^2C)=2(sinBcosCcosBsinC)
cos^2Bcos^2C+cos^2Ccos^2B=2(sinBcosCcosBsinC)
即cosBcosC=sinBsinC
即tanBtanC=1
所以B+C=90°
△ABC的形状是直角三角形

Answer 2

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