Question

三角形ABC中，(sinA)^2+(sinB)^2+(sinC)^2=2，证明ABC为直角三角形

Answer 1

(sinA)^2=(1-cos2A)/2
(sinB)^2=(1-cos2B)/2
(sinC)^2=(1-cos2C)/2
原式可化为
3-cos2A -cos2B -cos2C=4
cos2A+cos2B+cos2C=-1
2cos(A+B)cos(A-B)+2(cosC)^2-1=-1
-cosCcos(A-B)+(cosC)^2=0
∴ cosC=0
或者 cos(A-B)=cosC
1）cosC=0，∠C=90°
2) A-B=C,∠A=90°
3) A-B=-C,∠B=90°
∴△ABC为直角三角形

Answer 2

因为 (sinA)^2+(sinB)^2+(sinC)^2=2，所以
(sinA)^2 = 2-(sinB)^2-(sinC)^2 = (cosB)^2+(cosC)^2.
由和角公式：sinA = sin(B+C) = sinBcosC+cosBsinC，所以
(sinA)^2 = (sinB)^2(cosC)^2+(cosB)^2(sinC)^2+2sinBsinCcosBcosC.
所以
(cosB)^2+(cosC)^2-(sinA)^2
=[(cosB)^2-(cosB)^2(sinC)^2]+[(cosC)^2-(cosC)^2(sinB)^2]-2sinBsinCcosBcosC
=2(cosB)^2(cosC)^2-2sinBsinCcosBcosC
=2cosBcosC(cosBcosC-sinBsinC)
=2cosBcosCcos(B+C)
=-2cosAcosBcosC
=0
因此 cosA，cosB，cosC 中必有一个为0，即ABC是直角三角形。