# 代數方法+泰勒展開證明sin(a+b)公式
:::info
前情提要：
首先，定義如下：
$$\begin{align}
\sin(a)&=\sum_{i=0}^{\infty}{(-1)^i}\frac{a^{2i+1}}{(2i+1)!}\\
\cos(a)&=\sum_{i=0}^{\infty}{(-1)^i}\frac{a^{2i}}{(2i)!}
\end{align}$$
顯然，由Ratio Test,二者皆對於所有值皆絕對收斂。
因此，兩者的Cauchy Product也收斂。此為本證明之一大前提。
:::
以下證明 $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

$\text{Proof. Let }a,b\in \mathbb{R}.$

$\text{Since both sin and cos converge absolutely for any value, by theorem 3.50, we have}$
$$\begin{align}
\sin(a)\cos(b)&=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\overbrace{\frac{(-1)^k}{(2k+1)!}a^{2k+1}}^{\text{sin(a) part}}\cdot\overbrace{\frac{(-1)^{n-k}}{(2(n-k))!}b^{2(n-k)}}^{\text{cos(b) part}}\\
&=\sum_{n=0}^{\infty}\sum_{k=0}^{n}(-1)^n\frac{a^{2k+1}}{(2k+1)!}\cdot\frac{b^{2(n-k)}}{(2(n-k))!}\\
\cos(a)\sin(b)&=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\overbrace{\frac{(-1)^k}{(2k)!}a^{2k}}^{\text{cos(a) part}}\cdot\overbrace{\frac{(-1)^{n-k}}{(2(n-k)+1)!}b^{2(n-k)+1}}^{\text{sin(b) part}} \\
&=\sum_{n=0}^{\infty}\sum_{k=0}^{n}(-1)^n\frac{a^{2k}}{(2k)!}\cdot\frac{b^{2(n-k)+1}}{(2(n-k)+1)!}\\
\end{align}$$
$\text{That is, their cauchy product converges.}$

$\text{We now turn to analyzing }\sin(a+b).$
$$\begin{align}
\sin(a+b)&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(a+b)^{2n+1} & \text{(definition)} \\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\sum_{k=0}^{2n+1}\frac{(2n+1)!}{k!(2n+1-k)!}a^kb^{2n+1-k} & \text{(binomial theorem)} \\
&=\sum_{n=0}^{\infty}\sum_{k=0}^{2n+1}(-1)^n\frac{a^k}{k!}\frac{b^{2n+1-k}}{(2n+1-k)!} & \text{(simplification)}
\end{align}$$

$\text{Now, the monstrous looking term}$
$$\sum_{k=0}^{2n+1}(-1)^n\frac{a^k}{k!}\frac{b^{2n+1-k}}{(2n+1-k)!}$$
$\text{seems to be a pain in the ass. However, it'll no longer be, if we'd notice that}$
$$\begin{align}
&\sum_{k=0}^{2n+1}(-1)^n\frac{a^k}{k!}\frac{b^{2n+1-k}}{(2n+1-k)!}=S_{\text{even}}(n)+S_{\text{odd}}(n)\text{, for each n, where }\\
&S_{\text{even}}(n):=\sum_{k=0,2,4,\dots,2n}(-1)^n\frac{a^k}{k!}\frac{b^{2n+1-k}}{(2n+1-k)!}\\
&S_{\text{odd}}(n):=\sum_{k=1,3,5,\dots,2n+1}(-1)^n\frac{a^k}{k!}\frac{b^{2n+1-k}}{(2n+1-k)!}\\
\end{align}$$
$\text{That is, we split up the summation into even terms and odd terms.}$
$_\text{Yes. It's kind of an out-of-thin-air operation and do take time to come up with.}$
$\text{Lastly, note that, by simple algebraic maneuvering, we have}$
$$\begin{align}
S_{\text{even}}(n)&=\sum_{k=0}^{n}(-1)^n\frac{a^{2k}}{(2k)!}\cdot\frac{b^{2(n-k)+1}}{(2(n-k)+1)!}\\
S_{\text{odd}}(n)&=\sum_{k=0}^{n}(-1)^n\frac{a^{2k+1}}{(2k+1)!}\cdot\frac{b^{2(n-k)}}{(2(n-k))!}
\end{align}$$
$\text{Hence, if compare with what we have evaluated before, the coincidence is that we can rewrite}$ $$\begin{align}
\sin(a)\cos(b)&=\sum_{n=0}^{\infty}S_{\text{odd}}(n)\\
\cos(a)\sin(b)&=\sum_{n=0}^{\infty}S_{\text{even}}(n)
\end{align}$$

$\text{Hence, substituting everything back in,}$
$$\begin{align}
\sin(a+b)&=\sum_{n=0}^{\infty}S_{\text{even}}(n)+S_{\text{odd}}(n) & \text{(Theorem 3.47)}\\
&=\sum_{n=0}^{\infty}S_{\text{even}}(n)+\sum_{n=0}^{\infty}S_{\text{odd}}(n)\\
&=\cos(a)\sin(b)+\sin(a)\cos(b)
\end{align}$$

$\text{Q.E.D.}$