| a |
| sinA |
| b |
| sinB |
| c |
| sinC |
则
| 2c−a |
| b |
| 2ksinC−ksinA |
| ksinB |
| 2sinC−sinA |
| sinB |
| cosA−2cosC |
| cosB |
整理求得sin（A+B）=2sin（B+C）
又A+B+C=π
∴sinC=2sinA，即
| sinC |
| sinA |
（Ⅱ）由余弦定理可知cosB=
| a2+c2−b2 |
| 2ac |
| 1 |
| 4 |
由（Ⅰ）可知
| sinC |
| sinA |
| c |
| a |
①②联立求得c=2，a=1
sinB=
1−
|
| ||
| 4 |
∴S=
| 1 |
| 2 |
| ||
| 4 |
| cosA−2cosC |
| cosB |
| 2c−a |
| b |
| sinC |
| sinA |
| 1 |
| 4 |
| a |
| sinA |
| b |
| sinB |
| c |
| sinC |
| 2c−a |
| b |
| 2ksinC−ksinA |
| ksinB |
| 2sinC−sinA |
| sinB |
| cosA−2cosC |
| cosB |
| sinC |
| sinA |
| a2+c2−b2 |
| 2ac |
| 1 |
| 4 |
| sinC |
| sinA |
| c |
| a |
1−
|
| ||
| 4 |
| 1 |
| 2 |
| ||
| 4 |