A |
2 |
B |
2 |
C |
2 |
⇔2cos
A+B |
2 |
A−B |
2 |
C |
2 |
C |
2 |
A+B |
2 |
A−B |
2 |
⇔cos
A+B |
2 |
A−B |
2 |
C |
2 |
C |
2 |
A+B |
2 |
C |
2 |
A−B |
2 |
⇔0=sin
C |
2 |
C |
2 |
A+B |
2 |
C |
2 |
A+B |
2 |
A−B |
2 |
⇔0=(sin
C |
2 |
A+B |
2 |
C |
2 |
A−B |
2 |
∵A、B、C是锐角,∴sin
C |
2 |
A−B |
2 |
所以上式⇔0=sin
C |
2 |
A+B |
2 |
⇔
C |
2 |
A+B |
2 |
π |
2 |
A |
2 |
B |
2 |
C |
2 |
A |
2 |
B |
2 |
C |
2 |
A+B |
2 |
A−B |
2 |
C |
2 |
C |
2 |
A+B |
2 |
A−B |
2 |
A+B |
2 |
A−B |
2 |
C |
2 |
C |
2 |
A+B |
2 |
C |
2 |
A−B |
2 |
C |
2 |
C |
2 |
A+B |
2 |
C |
2 |
A+B |
2 |
A−B |
2 |
C |
2 |
A+B |
2 |
C |
2 |
A−B |
2 |
C |
2 |
A−B |
2 |
C |
2 |
A+B |
2 |
C |
2 |
A+B |
2 |
π |
2 |