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# 拉格朗日恆等式 (Lagrange's Identity)

$$
\cases{
\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos\theta\\
\sin^2\theta = 1-\cos^2\theta
}\implies
\begin{array}l
\sin^2\theta &= 1-(\dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|})^2\\
&= \dfrac{|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2}{|\vec{u}|^2|\vec{v}|^2}
\end{array}\\
\therefore (|\vec{u}||\vec{v}|\sin\theta)^2 = |\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2
$$

:::info
拉格朗日恆等式 (1)
$$
(|\vec{u}||\vec{v}|\sin\theta)^2 = |\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2
$$
:::

假設
$$
\cases{
\vec{u} = \pmatrix{u_1\\u_2\\\vdots\\u_n}\\
\vec{v} = \pmatrix{v_1\\v_2\\\vdots\\v_n}\\
}\\
\begin{array}l
(|\vec{u}||\vec{v}|\sin\theta)^2 &= |\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2\\
&= (u_1^2+u_2^2+\cdots+u_n^2)(v_1^2+v_2^2+\cdots+v_n^2) - (u_1v_1+u_2v_2+\cdots+u_nv_n)^2
\end{array}
$$

$(u_1^2+u_2^2+\cdots+u_n^2)(v_1^2+v_2^2+\cdots+v_n^2) = (\sum_{i=1}^n u_i^2)(\sum_{i=1}^n v_i^2)$ 依分配律可做成以下表格 (方陣):

$$
\begin{array}{c|c|c|c|c}
& v_1^2 & v_2^2 & v_3^2 & \cdots & v_n^2\\ \hline
u_1^2 & (u_1v_1)^2 & (u_1v_2)^2 & (u_1v_3)^2 & \cdots & (u_1v_n)^2\\ \hline
u_2^2 & (u_2v_1)^2 & (u_2v_2)^2 & (u_2v_3)^2& \cdots & (u_2v_n)^2\\ \hline
u_3^2 & (u_3v_1)^2 & (u_3v_2)^2 & (u_3v_3)^2& \cdots & (u_3v_n)^2\\ \hline
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ \hline
u_n^2 & (u_nv_1)^2 & (u_nv_2)^2 & (u_nv_3)^2& \cdots & (u_nv_n)^2
\end{array}
$$

對角線上的元素為 $\sum_{i=1}^n (u_iv_i)^2$
其餘元素被對角線分割成右上方三角形區域與左下方三角形區域，
每一個右上方的元素 $(u_iv_j)^2$ 可以對角線對折後對應到左下方 $(u_jv_i)^2$ 元素，我們可將此二對應的元素寫成一組 $(u_iv_j)^2 + (u_jv_i)^2$

於是表格中原本有 $n^2$ 個元素，可重新寫成

$$
(\sum_{i=1}^n u_i^2)(\sum_{i=1}^n v_i^2) =
\sum_{i=1}^n (u_iv_i)^2 + \sum_{i=1}^{n-1}\sum_{j=i+1}^n ((u_iv_j)^2+(u_jv_i)^2)
$$

同理，

$(u_1v_1+u_2v_2+\cdots+u_nv_n)^2 = (\sum_{i=1}^n u_iv_i)^2$ 也可做成以下表格 (方陣):

$$
\begin{array}{c|c|c|c|c}
& u_1v_1 & u_2v_2 & u_3v_3 & \cdots & u_nv_n\\ \hline
u_1v_1 & (u_1v_1)^2 & u_1v_2u_2v_1 & u_1v_3u_3v_1 & \cdots & u_1v_nu_nv_1\\ \hline
u_2v_2 & u_2v_1u_1v_2 & (u_2v_2)^2 & u_2v_3u_3v_2& \cdots & u_2v_nu_nv_2\\ \hline
u_3v_3 & u_3v_1u_1v_3 & u_3v_2u_2v_3 & (u_3v_3)^2& \cdots & u_3v_nu_nv_3\\ \hline
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ \hline
u_nv_n & u_nv_1u_1v_n & u_nv_2u_nv_2 & u_nv_3u_3v_n& \cdots & (u_nv_n)^2
\end{array}
$$

對角線上的元素為 $\sum_{i=1}^n (u_iv_i)^2$
與之前相同的配對方式，我們可把右上三角形中的每個 $u_iv_ju_jv_i$ 元素與左下三角形中所對應的 $u_jv_iu_iv_j$ 元素配成一組 $u_iv_ju_jv_i + u_jv_iu_iv_j = 2u_iv_ju_jv_i$

$$
(\sum_{i=1}^n u_iv_i)^2 = \sum_{i=1}^n (u_iv_i)^2 + \sum_{i=1}^{n-1}\sum_{j=i+1}^n 2u_iv_ju_jv_i
$$

則

$$
\begin{array}l
(|\vec{u}||\vec{v}|\sin\theta)^2 &= |\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2\\
&= (u_1^2+u_2^2+\cdots+u_n^2)(v_1^2+v_2^2+\cdots+v_n^2) - (u_1v_1+u_2v_2+\cdots+u_nv_n)^2\\
&= (\sum_{i=1}^n u_i^2)(\sum_{i=1}^n v_i^2) - (\sum_{i=1}^n u_iv_i)^2\\
&=\sum_{i=1}^n (u_iv_i)^2 + \sum_{i=1}^{n-1}\sum_{j=i+1}^n ((u_iv_j)^2+(u_jv_i)^2) - \sum_{i=1}^n (u_iv_i)^2 - \sum_{i=1}^{n-1}\sum_{j=i+1}^n 2u_iv_ju_jv_i\\
&= \sum_{i=1}^{n-1}\sum_{j=i+1}^n ((u_iv_j)^2-2u_iv_ju_jv_i+(u_jv_i)^2)\\
&= \sum_{i=1}^{n-1}\sum_{j=i+1}^n (u_iv_j-u_jv_i)^2
\end{array}
$$

故

:::info
拉格朗日恆等式 (2)
$$
\begin{array}l
(|\vec{u}||\vec{v}|\sin\theta)^2 &= |\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2\\
&= (\sum_{i=1}^n u_i^2)(\sum_{i=1}^n v_i^2) - (\sum_{i=1}^n u_iv_i)^2\\
&= \sum_{i=1}^{n-1}\sum_{j=i+1}^n (u_iv_j-u_jv_i)^2
\end{array}
$$
:::

###### tags: `math` `lagrange` `identity`