# 二階數學補充-三角函數
東西很雜 寫到什麼講什麼
可能還有一點幾何
## 怪東西
### 和差化積/積化和差
以下只推導第一條 後面皆可用類似方法推得
- $\sin\alpha\cos\beta$
$=\cfrac{2\sin\alpha\cos\beta}2$
$=\cfrac{\sin\alpha\cos\beta+\cos\alpha\sin\beta+\sin\alpha\cos\beta-\cos\alpha\sin\beta}2$
$=\cfrac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}2$

- $\cos\alpha\sin\beta$
$=\cfrac{\sin(\alpha+\beta)-\sin(\alpha-\beta)}2$

- $\cos\alpha\cos\beta$
$=\cfrac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}2$

- $\sin\alpha\sin\beta$
$=\cfrac{-\cos(\alpha+\beta)+\cos(\alpha-\beta)}2$

## 寫寫考古
### EC2013-1 (複數+歐拉公式)
不想學那麼多請快轉[這裡](https://hackmd.io/4ystU6tlQy6Px79KeaXBOg#EC2013-2-%E7%A9%8D%E5%8C%96%E5%92%8C%E5%B7%AE)

我在自己的解析上寫了這麼一句話
> 請使用公式 $\sum^{n}_{k=0}\ \cos(kx)=\cfrac{1}{2}+\cfrac{\sin(\cfrac{2n+1}{2}x)}{2\sin(\cfrac{x}{2})}$

以下將嘗試證明
#### 代號與性質
- $Re(z)$:複數$z$的實部
- 令$z=a+bi$, 則$\bar{z}=a-bi$
- $z\bar{z}=|z|^2$
- $\cfrac{z_1}{z_2}=\cfrac{ac-bd+i(bc-ad)}{|z_2|^2}$
- $e^{ix}=\cos(x)+i\sin(x)\leftarrow$歐拉公式(就是那個$e^{i\pi}=-1$的來源)
- $e^{ix}-e^{-ix}$
$=(\ \cos(x)+i\sin(x)\ )-(\ \cos(-x)+i\sin(-x)\ )$
$=(\ \cos(x)+i\sin(x)\ )-(\ \cos(x)-i\sin(x)\ )$
$=2i\sin(x)$
- 積化和差:$2\sin A\cos B=\sin(A+B)-\sin(A-B)$

#### 正式開始?
$\sum^{n}_{k=0}\ \cos(kx)$
$=Re(\ \sum^{n}_{k=0}\ \cos(kx)\ )$
$=Re(\ \sum^{n}_{k=0}\ \cos(kx)+i\sin(kx)\ )$
$=Re(\ \sum^{n}_{k=0}\ e^{ixk}\ )$
$=Re(\ 1+e^{ix}+e^{2ix}+...+e^{nix}\ )$
$=Re(\ \cfrac{e^{(n+1)ix}-1}{e^{ix}-1}\ )$
$=Re(\ \cfrac{e^{\cfrac{(n+1)ix}{2}}(e^{\cfrac{(n+1)ix}{2}}-e^{-\cfrac{(n+1)ix}{2}})}{e^{\cfrac{ix}{2}}(e^{\cfrac{ix}{2}}-e^{-\cfrac{ix}{2}})}\ )$
$=Re(\cfrac{e^{\cfrac{(n+1)ix}{2}}}{e^{\cfrac{ix}{2}}}\times\cfrac{2i\sin(\cfrac{n+1}{2}x)}{2i\sin(\cfrac{x}{2})})$
$=Re(e^{\cfrac{nix}{2}}\times\cfrac{\sin(\cfrac{n+1}{2}x)}{\sin(\cfrac{x}{2})})$
$=Re(\ (\cos(\cfrac{nx}{2})+i\sin(\cfrac{nx}{2})\ )\times\cfrac{\sin(\cfrac{n+1}{2}x)}{\sin(\cfrac{x}{2})})$
$=\cos(\cfrac{nx}{2})\times\cfrac{\sin(\cfrac{n+1}{2}x)}{\sin(\cfrac{x}{2})}$
$=\cfrac{\sin(\cfrac{2n+1}{2}x)+\sin(\cfrac{x}{2})}{2\sin(\cfrac{x}{2})}$
$=\cfrac{1}{2}+\cfrac{\sin(\cfrac{2n+1}{2}x)}{2\sin(\cfrac{x}{2})}$

終於寫完了
這樣就可以代公式了
注意原題是從$k=1$開始

#### 套公式
$n=100, x=\cfrac{4\pi}{199}$代入
$S_{100}(\cfrac{4\pi}{199})=\cos(\cfrac{4\pi}{199})+\cos(2\times\cfrac{4\pi}{199})+...+\cos(100\times\cfrac{4\pi}{199})$
$=\cos(0)+\cos(\cfrac{4\pi}{199})+\cos(2\times\cfrac{4\pi}{199})+...+\cos(100\times\cfrac{4\pi}{199})-\cos(0)$
$=\cfrac{1}{2}+\cfrac{\sin(\cfrac{201}{2}\times \cfrac{4\pi}{199})}{2\sin(\cfrac{1}{2}\times\cfrac{4\pi}{199})}-1$
$=\cfrac{\sin(\cfrac{402\pi}{199})}{2\sin(\cfrac{2\pi}{199})}-\cfrac{1}{2}$
$=\cfrac{\sin(\cfrac{4\pi}{199})}{2\sin(\cfrac{2\pi}{199})}-\cfrac{1}{2}$
$=\cfrac{2\sin(\cfrac{2\pi}{199})\cos(\cfrac{2\pi}{199})}{2\sin(\cfrac{2\pi}{199})}-\cfrac{1}{2}$
$=\cos(\cfrac{2\pi}{199})-\cfrac{1}{2}$
把$\pm\cfrac{2\pi}{199}$的同位角列一列就結束ㄌ

### EC2013-2 (積化和差by范振祐)
令$A=S_{100}(\cfrac{4\pi}{199})=\cos(\cfrac{4\pi}{199})+\cos(2\times\cfrac{4\pi}{199})+...+\cos(100\times\cfrac{4\pi}{199})$
$2A\sin(\cfrac{2\pi}{199})$
$=2\sin(\cfrac{2\pi}{199})\cos(\cfrac{4\pi}{199})+2\sin(\cfrac{2\pi}{199})\cos(2\times\cfrac{4\pi}{199})+...$
$=[\sin(\cfrac{6\pi}{199})+\sin(\cfrac{-2\pi}{199})]+[\sin(\cfrac{10\pi}{199})+\sin(\cfrac{-6\pi}{199})]+...$
$=\sin(\cfrac{-2\pi}{199})+\sin(\cfrac{402\pi}{199})$
$=\sin(\cfrac{2\pi}{199})[-1+2\cos(\cfrac{2\pi}{199})]$

### EC2014 (棣美弗定理)
$C_1=-2=2(\cos\theta+i\sin\theta)\rightarrow \theta=\pi$
$C_2=2^\cfrac{1}{2}(\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2})$
$C_3=2^\cfrac{1}{4}(\cos\cfrac{\theta}{4}+i\sin\cfrac{\theta}{4})$
...
$\prod C_i=(2\times2^\cfrac{1}{2}\times2^\cfrac{1}{4}\times...)\times(\cos\theta+i\sin\theta)\times(\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2})\times(\cos\cfrac{\theta}{4}+i\sin\cfrac{\theta}{4})\times...$
$=2^2\times(\cos2\theta+i\sin2\theta)$
$=4$

### EC2015 (複數平面 單位圓)

### EC1904 (和角)

### EC1907 (疊合/斜率)

### EC1908 (極式)

### EC1910 (函數圖形)

### EE1801 (複數平面 單位圓)

### EE1803 (根與係數)

### CS1803 (tan和角) \*經典題
構造一個$\sqrt5,\sqrt5,\sqrt{10}$的等腰直角三角形

### CS1813 (和差化積)
==(1)==
和差化積 或22.5度角硬炸

==(2)==
原式$=2(\cos^8\cfrac\pi8+\sin^8\cfrac\pi8)$

### CS1701 (微分)

### CS1704 (疊合)

### CS1603 (三角函數定義)

### CS1616 (正弦定理)^

### CS1617 (積化和差)

### EE1306 (複數平面 極式)

### CS1304 (三角函數定義)

### CS1308 (等比級數)

### EE1205 (平方關係)

### EE1208 (球)*

### CS1202 (怪級數)*

### EE1101 (和差化積)

### CS1114 (疊合?)

### CS1118 (幾何)*

### EE1002 (?)*

### EE1008 (-i的立方根)

### CS1012 (?)*

### CS1013 (和差倍角公式)

### CS0802 (疊合)

### CS0803 (疊合/斜率)

### CS0814 (tan和角)
令$\tan\alpha=\cfrac13,\tan\beta=\cfrac15,\tan\gamma=\cfrac17,\tan\phi=\cfrac18$

$\tan(\alpha+\beta)=\cfrac{\cfrac13+\cfrac15}{1-\cfrac13\cfrac15}=\cfrac47$

$\tan(\gamma+\phi)=\cfrac{\cfrac17+\cfrac18}{1-\cfrac17\cfrac18}=\cfrac3{11}$

$\tan(\alpha+\beta+\gamma+\phi)$
$=\cfrac{\tan(\alpha+\beta)+\tan(\gamma+\phi)}{1-\tan(\alpha+\beta)\tan(\gamma+\phi)}$
$=\cfrac{\cfrac47+\cfrac3{11}}{1-\cfrac47\cfrac3{11}}=\cfrac{65}{65}=1$
$\rightarrow \alpha+\beta+\gamma+\phi=\pi/4$

### CS0816 (怪級數)

### CS0820 (根與係數)

### EE0705 (觀察化簡)

### EE0708 (怪級數)*

### CS0704 (?)*

### CS0604 (等比級數)*

### CS0606 (怪極限)
$=\lim(\cos\cfrac x{2}\cos\cfrac x{2^2}...\cos\cfrac x{2^n})\sin\cfrac x{2^n}/\sin\cfrac x{2^n}$
$=\lim\cfrac{\sin x}{2^n\sin\cfrac x{2^n}}$
$=\sin x\times\lim\cfrac{(\cfrac12)^n}{\sin\cfrac x{2^n}}$
$=\sin x\times\lim\cfrac{[(\cfrac12)^n]'}{\cos\cfrac x{2^n}\times(\cfrac x{2^n})'}$
$=\sin x\times\lim\cfrac1{\cos\cfrac x{2^n}\times x}$
$=\cfrac{\sin x}{x}$

### CS0612 (週期)

### CS0613 (函數圖形)*

### EE0501 (平方關係 倍角公式)
$y=2\sin^2x+4\sin x\cos x-2\cos^2x+4$
$=2\sin2x-2\cos2x+4$
$=2\sqrt2\sin(2x-\cfrac{\pi}4)+4$

### CS0501 (複數平面 極式)

### EE0406 (?)*

### CS0202 (複數平面 單位圓)
$=-\cos\cfrac{2\pi}{11}-\cos\cfrac{4\pi}{11}-\cos\cfrac{6\pi}{11}-\cos\cfrac{8\pi}{11}-\cos\cfrac{10\pi}{11}$
$=\cfrac12(-\cos\cfrac{2\pi}{11}-\cos\cfrac{4\pi}{11}-\cos\cfrac{6\pi}{11}-\cos\cfrac{8\pi}{11}-\cos\cfrac{10\pi}{11}-\cos\cfrac{12\pi}{11}-\cos\cfrac{14\pi}{11}-\cos\cfrac{16\pi}{11}-\cos\cfrac{18\pi}{11}-\cos\cfrac{20\pi}{11})$
$=\cfrac12(-\cos\cfrac{2\pi}{11}-\cos\cfrac{4\pi}{11}-\cos\cfrac{6\pi}{11}-\cos\cfrac{8\pi}{11}-\cos\cfrac{10\pi}{11}-\cos\cfrac{12\pi}{11}-\cos\cfrac{14\pi}{11}-\cos\cfrac{16\pi}{11}-\cos\cfrac{18\pi}{11}-\cos\cfrac{20\pi}{11}-\cos\cfrac{22\pi}{11})+\cfrac12$
$=\cfrac12$

###### tags: `電資二階` `Cosmos`