# Week 7 - Derivatives of Trigonometric Functions

## 1. Trigonometry

• Trigonometric Functions
• $\sin{\theta} = \frac{opp}{hyp}$, $\csc{\theta} = \frac{hyp}{opp}$
• $\cos{\theta} = \frac{adj}{hyp}$, $\sec{\theta} = \frac{hyp}{adj}$
• $\tan{\theta} = \frac{opp}{adj}$, $\cot{\theta} = \frac{adj}{opp}$
• $\sin{\theta} = \frac{y}{r}$, $\csc{\theta} = \frac{r}{y}$
• $\cos{\theta} = \frac{x}{r}$, $\sec{\theta} = \frac{r}{x}$
• $\tan{\theta} = \frac{y}{x}$, $\cot{\theta} = \frac{x}{y}$

### 1.1. Angles

• $\pi rad = 180^{\circ}$
• $1 rad = (\frac{180}{\pi})^{\circ} \approx 57.3^{\circ}$
Degree $0^{\circ}$ $30^{\circ}$ $45^{\circ}$ $60^{\circ}$ $90^{\circ}$ $120^{\circ}$ $135^{\circ}$ $150^{\circ}$ $180^{\circ}$ $270^{\circ}$ $360^{\circ}$
Radians $0$ $\frac{\pi}{6}$ $\frac{\pi}{4}$ $\frac{\pi}{3}$ $\frac{\pi}{2}$ $\frac{2\pi}{3}$ $\frac{3\pi}{4}$ $\frac{5\pi}{6}$ $\pi$ $\frac{3\pi}{2}$ $2\pi$
• $\theta = \frac{a}{r}$, $a = r\theta$

### 1.2. Some of Trigonometric Identities

• $\sin^2{x} + \cos^2{x} = 1$
• $\sin{-x} = -\sin{x}$
• $\cos{-x} = \cos{x}$
• $\sin{x+2\pi} = \sin{x}$, $\cos{x+2\pi} = \cos{x}$

## 2. Differentiate Trig Functions

• To prove: if $f(x) = \sin{x}$, then $f'(x) = \cos{x}$
• \begin{aligned} f'(x) &= \lim_{h \to 0}\frac{\sin(x+h) - \sin(x)}{h} \\ &= \lim_{h \to 0}\frac{\sin{x}\cos{h} + \cos{x}\sin{h} - \sin(x)}{h} \\ &= \lim_{h \to 0}[\sin{x}(\frac{\cos{h} - 1}{h}) + \cos{x}(\frac{\sin{h}}{h})] \\ &= \sin{x}\lim_{h \to 0}\frac{\cos{h} - 1}{h} + \cos{x}\lim_{h \to 0}\frac{\sin{h}}{h} \end{aligned}
• Recall the Squeeze Theorem, we know $\displaystyle\lim_{x\to{0}}\frac{\sin(x)}{x}=1$, then,
• \begin{aligned} \lim_{\theta \to 0}\frac{\cos{\theta} - 1}{\theta} &= \lim_{\theta \to 0}{\frac{\cos{\theta} - 1}{\theta} \cdot \frac{\cos{\theta} + 1}{\cos{\theta} + 1}} = \lim_{\theta \to 0}\frac{\cos^2{\theta} - 1}{\theta(\cos{\theta} + 1)} \\ &= \lim_{\theta \to 0}\frac{-\sin^2{\theta}}{\theta(\cos{\theta} + 1)} = -\lim_{\theta \to 0}{\frac{\sin{\theta}}{\theta} \cdot \frac{\sin{\theta}}{\cos{\theta}+1}} \\ &= -\lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} \cdot \lim_{\theta \to 0}\frac{\sin{\theta}}{\cos{\theta}+1} \\ &= - 1 \cdot (\frac{0}{1 + 1}) \\ \lim_{\theta \to 0}\frac{\cos{\theta} - 1}{\theta} &= 0 \end{aligned}
• So, \begin{aligned} f'(x) &= \sin{x}\lim_{h \to 0}\frac{\cos{h} - 1}{h} + \cos{x}\lim_{h \to 0}\frac{\sin{h}}{h} \\ &= (\sin{x}) \cdot 0 + (\cos{x}) \cdot 1 \\ &= \cos{x} \end{aligned}

### 2.1. Derivatives of Trigonometric Functions

• $\frac{d}{dx}(\sin{x}) = \cos{x}$, $\frac{d}{dx}(\csc{x}) = -\csc{x}\cot{x}$
• $\frac{d}{dx}(\cos{x}) = -\sin{x}$, $\frac{d}{dx}(\sec{x}) = \sec{x}\tan{x}$
• $\frac{d}{dx}(\tan{x}) = \sec^2{x}$, $\frac{d}{dx}(\cot{x}) = -\csc^2{x}$

## 3. Inverse Trigonometric Functions

• $y = \sin{x}$
• $y = \sin^{-1}{x} = \arcsin{x} \Leftrightarrow \sin{y} = x\ and\ -\frac{\pi}{2} \le x \le \frac{\pi}{2}$
• we can conclude this: $\arcsin{(\sin{x})} = x$
• Sample: Simplify the expression $\cos{(\tan^{-1}x)}$:
• If $y=\tan^{-1}x$, then $\tan{y}=x$, we can use a diagram to express this:
• Then $\cos{(\tan^{-1}x)} = \cos{y} = \frac{1}{\sqrt{1+x^2}}$

### 3.1. The Derivatives of Inverse Trig Functions

• $\frac{d}{dx}\arcsin{x} = ?$
• Let's say $f(x) = \arcsin{x}$
• We know $f(\sin{x}) = x$
• So use the chain rules, derivative the both sides, we got: \begin{aligned} f'(\sin{x}) \cdot \cos{x} &= 1 \\ f'(\sin{x}) &= \frac{1}{\cos{x}} \\ f'(\sin{x}) &= \frac{1}{\sqrt{1-\sin^2{x}}} \\ f'(x) &= \frac{1}{\sqrt{1-x^2}} \\ \end{aligned}
• $\frac{d}{dx}\arccos{x} = ?$:
• In a right triangle, the other two angles are $\alpha\, \beta$, we know:
• ($\sin{\alpha} = y\,\ \cos{\beta} = x$) and the hypotenuse equals 1, then we got:
• $\alpha + \beta = \frac{\pi}{2}$
• $\arcsin{y} = \alpha\,\ \arccos{y} = \beta$
• $\arcsin{y} + \arccos{y} = \frac{\pi}{2}$
• $\frac{d}{dy}\arcsin{y} + \frac{d}{dy}\arccos{y} = \frac{d}{dy} \frac{\pi}{2}$
• $\frac{d}{dy}\arcsin{y} + \frac{d}{dy}\arccos{y} = 0$
• $\frac{d}{dy}\arccos{y} = - \frac{1}{\sqrt{1-x^2}}$
• $\frac{d}{dx}\arctan{x} = ?$
• set $f(x) = \arctan{x}$,
• then $f(\tan{x}) = x$
• use the chain rules: \begin{aligned} f'(\tan{x}) \cdot \sec^2{x} &= 1 \\ f'(\tan{x}) &= \frac{1}{\sec^2{x}} \\ f'(\tan{x}) &= \frac{1}{\tan^2{x}+1} \\ f'(x) &= \frac{1}{x^2+1} \\ \end{aligned}

### 3.2. Why do Sine and Cosine Oscillate?

• If $f''(x) = - f(x)$
• For example:
• A running car, it's acceleration equals minor it's position, and the center point is 0.
• Which means:
• the velocity will be smaller and smaller when it's acceleration less than 0,
• but when the position is less than 0 the acceleration will be positive, and the car will move backwards sooner.
• So the car will be back and forward, just like the sine wave.
• To Sine and Cosine, we know:
• $f(t) = \sin{t},\ f''(t) = - \sin{t}$
• $f(t) = \cos{t},\ f''(t) = - \cos{t}$

### 3.3. Angle sum and difference identities

• $\sin{(x+y)} = \sin{x}\cos{y} + \cos{x}\sin{y}$
• $\cos{(x+y)} = \cos{x}\cos{y} - \sin{x}\sin{y}$
• $\sin{(x-y)} = \sin{x}\cos{y} - \cos{x}\sin{y}$
• $\cos{(x-y)} = \cos{x}\cos{y} + \sin{x}\sin{y}$
• $\sin{2x} = 2\sin{x}\cos{x}$
• $\cos{2x} = \cos^2{x} - \sin^2{x}$
• $\tan{(x+y)} = \frac{\tan(x) + \tan{y}}{1 - \tan(x)\tan(y)}$
• $\tan{(x-y)} = \frac{\tan(x) - \tan{y}}{1 + \tan(x)\tan(y)}$

### 3.4. Approximate sin 1

• $f(x) = \sin{x},\ f(0) = 0$
• $f'(x) = \cos{x},\ f'(0) = 1$
• $f(0+h) \approx f(0) + h \cdot f'(0)$, (when h approach to 0)
• $f(0+h) \approx 0 + h \cdot 1$
• $f(h) \approx h$ (when h approach to 0)
• pick an x = 1/32, then $\sin{\frac{1}{32} \approx \frac{1}{32}}$
• then use formula $\sin{2x} = 2\sin{x}\cos{x}$, we got $\sin{2x} = 2\sin{x} \cdot \sqrt{1-\sin^2{x}}$
• So, $\sin{(2 \cdot \frac{1}{32})} = 2\sin{(1/16)} \cdot \sqrt{1-\sin^2{(1/32)}}$
• We can keep multiple 2 to x until x = 1, then we got ($\sin{1}$)