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