回覆 15# Gary 的帖子
計算第二題：
\(\displaystyle \sin A+\sin B -\cos C = \frac{3}{2}\)
\(\displaystyle\Rightarrow 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} -\left(2\cos^2\frac{C}{2}-1\right) = \frac{3}{2}\)
\(\displaystyle\Rightarrow 2\sin\frac{\pi - C}{2}\cos\frac{A-B}{2} -\left(2\cos^2\frac{C}{2}-1\right) = \frac{3}{2}\)
\(\displaystyle\Rightarrow 2\cos\frac{C}{2}\cos\frac{A-B}{2} -\left(2\cos^2\frac{C}{2}-1\right) = \frac{3}{2}\)
\(\displaystyle\Rightarrow 4\cos^2\frac{C}{2}-4\cos\frac{A-B}{2}\cos\frac{A-B}{2}+1=0\)
\(\displaystyle\Rightarrow \left(2\cos\frac{C}{2}-\cos\frac{A-B}{2}\right)^2 = \cos^2\frac{A-B}{2}-1\geq 0\)
因為 \(\displaystyle\cos^2\frac{A-B}{2}\leq1\)，得 \(\displaystyle\cos^2\frac{A-B}{2}=1\Rightarrow \cos\frac{A-B}{2}=\pm1\) 。
因為 \(\displaystyle\frac{-\pi}{2}<\frac{A-B}{2}<\frac{\pi}{2}\)，得 \(\displaystyle\cos\frac{A-B}{2}=1\) 。
\(\displaystyle\Rightarrow \left(2\cos\frac{C}{2}-1\right)^2 = 0\)
\(\displaystyle\Rightarrow \cos\frac{C}{2}=\frac{1}{2}\)
\(\displaystyle\Rightarrow \frac{C}{2}=\frac{\pi}{3}\)
\(\displaystyle\Rightarrow \angle C=\frac{2\pi}{3}\)
