三角公式

\begin{aligned} \sin^{2} α + \cos^{2} α = 1 \\ 1 + \tan^{2} α = \sec^{2} α \\ 1 + \cot^{2} α = \csc^{2} α \end{aligned}

三角公式的基础（向量证明两角差的余弦公式，再经过三角基本关系推得其他式子）

\begin{array}{r} \cos (A - B) = \cos A \cos B + \sin A \sin B \\ \cos (A + B) = \cos A \cos B - \sin A \sin B \\ \sin (A + B) = \sin A \cos B + \cos A \sin B \\ \sin (A - B) = \sin A \cos B - \cos A \sin B \end{array}

\begin{array}{r} \tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \\ \tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \end{array}

\begin{array}{r} \arctan \frac{x - y}{1 + x y} = \arctan x - \arctan y \end{array}

\begin{array}{r} \sin A \sin B = \frac{1}{2} (\cos (A - B) - \cos (A + B)) \\ \cos A \cos B = \frac{1}{2} (\cos (A - B) + \cos (A + B)) \\ \cos A \sin B = \frac{1}{2} (\sin (A + B) - \sin (A - B)) \\ \sin A \cos B = \frac{1}{2} (\sin (A + B) + \sin (A - B)) \end{array}

\begin{array}{r} \sin (A + B) + \sin (A - B) = 2 \sin A \cos B \\ \sin (A + B) - \sin (A - B) = 2 \cos A \sin B \\ \cos (A + B) + \cos (A - B) = 2 \cos A \cos B \\ \cos (A + B) - \cos (A - B) = - 2 \sin A \sin B \end{array}

令 $A + B = α, A - B = β$ ，则 $A = \frac{α + β}{2}, B = \frac{α - β}{2}$
得到一般意义上的和差化积

\begin{aligned} \sin 2 A & = 2 \sin A \cos A \\ \cos 2 A & = \cos^{2} A - \sin^{2} A \\ = 2 \cos^{2} A - 1 \\ = 1 - 2 \sin^{2} A \end{aligned}

\begin{array}{r} \sin^{2} A = \frac{1}{2} (1 - \cos 2 A) \\ \cos^{2} A = \frac{1}{2} (1 + \cos 2 A) \end{array}

\begin{array}{r} \sum_{k = 1}^{m} \cos k t = \cos \frac{m + 1}{2} t \cdot \frac{\sin \frac{m t}{2}}{\sin \frac{t}{2}} \end{array}