三角公式

三角关系

\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}

graph LR
1[基本公式] --> 2[和差公式]
2--> 积化和差 & 和差化积
1 & 2 --> 3[倍角公式]
3 & 1 --> 降幂公式

三角公式的基础（向量证明两角差的余弦公式，再经过诱导公式推得其他式子）

\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

将三角的和差公式的不同角换为同一个角，即得到倍角公式（其中倍角余弦可以进一步根据三角函数基本关系得到不同的表达）：

\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}

根据三角基本关系和倍角公式，可以进一步得到：

\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}

\frac{\sin θ}{1 - \cos θ} = \cot \frac{θ}{2}

$1 - \cos θ = 2 \sin^{2} (θ / 2)$ , $\sin θ = 2 \sin (θ / 2) \cos (θ / 2)$