1 |
2 |
∴2S=absinC=c2-(a-b)2,化简得ab(sinC-2)=-(a2+b2-c2)
∵根据余弦定理,得a2+b2-c2=2abcossC
∴ab(sinC-2)=-2abcossC,整理得sinC=2-2cosC
由此可得:
sinC |
1−cosC |
2−2cosC |
1−cosC |
(2)由(1)得
sinC |
1−cosC |
4 |
5 |
∴S=
1 |
2 |
2 |
5 |
∵a+b=2,∴S=
2 |
5 |
2 |
5 |
2 |
5 |
当且仅当a=b=1时,面积S的最大值为
2 |
5 |
sinC |
1−cosC |
1 |
2 |
sinC |
1−cosC |
2−2cosC |
1−cosC |
sinC |
1−cosC |
4 |
5 |
1 |
2 |
2 |
5 |
2 |
5 |
2 |
5 |
2 |
5 |
2 |
5 |