If A+B+C=180°
prove that sin²A+sin²B+sin²C-2 = 2cosAcosBcosC
amath. trigo.
2008-03-02 10:54 pm
回答 (1)
2008-03-02 11:50 pm
✔ 最佳答案
L.H.S.=sinA+sinB+sinC-2
=[(1-cos2A)+(1-cos2B)+(1-cos2C)-4]/2
=-(1+cos2A+cos2B+cos2C)/2
=-[1+2cos(A+B)cos(A-B)+cos2C]/2
=-[1-2cosCcos(A-B)+cos2C]/2
=-[(cos2C+1)-2cosCcos(A-B)]/2
=-[2cos2C-2cosCcos(A-B)]/2
=-cosC[cosC-cos(A-B)]
=-cosC[-2sin(A-B+C)/2sin(C-A+B)/2]
=2cosC[sin(180-B-B)/2sin(180-A-A)/2]
=2cosCsin(90-B)sin(90-A)
=2cosAcosBcosC
收錄日期: 2021-04-25 17:18:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080302000051KK01646