“我日。”

# 总结

## 反驳

Following words written in English are with no specific intention but just to avoid the censorship on my blog. I could change my mind quickly so the following words only represent my thought in the past.

# 数学

## 三角函数的变换和解三角形

$\begin{gather} 2a^2+b^2 = 2c^2 \tag{1} \\ \text{考虑余弦定理:} \\ 2ab \cos C = a^2+b^2-c^2 \tag{2} \\ \text{联立(1)(2)式:} \\ 2ab \cos C = c^2 – a^2 \to 4ab \cos C = 2c^2 – 2a^2 \\ \text{带入(1)式:} \\ 4ab \cos C = b^2 \to 4a \cos C = b \\ \text{考虑正弦定理转换:} \\ 4 \sin A \cos C = \sin A \cos C + \sin C \cos A \\ 3 \sin A \cos C = \sin C \sin A \\ 3\tan A = \tan B \\ \end{gather}$

\begin{aligned} \tan C &= \tan (\pi – A – B) = \tan (- A – B) = -\tan (A + B) \\ &= \frac{\tan A + \tan B}{\tan A \tan B – 1} \\ &= \frac{4 \tan A}{3 \tan^2 A – 1} \end{aligned} \\ \begin{aligned} \text{原式} &= \frac{1}{\tan A} + \frac{1}{\tan B} + \frac{1}{\tan C} \\ &= \frac{4}{3 \tan A} + \frac{3 \tan^2 A – 1}{4 \tan A} \\ &= \frac{16 + 9 \tan^2 A – 3}{12 \tan A} \\ &= \frac{13}{12 \tan A} + \frac{3 \tan A}{4} \geq \frac{\sqrt{13}}{2} \end{aligned}