notes/education/math/MATH1060 (trig)/Graphing.md

# Sine/Cosine
![A graph of sine and cosine](./assets/graphsincos.png)

Given the above graph:
- At the origin, $sin(x) = 0$ and $cos(x) = 1$
- A full wavelength takes $2\pi$

# Manipulation
| Formula          | Movement                           |
| ---------------- | ---------------------------------- |
| $y = cos(x) - 1$ | Vertical shift down by 1           |
| $y = 2cos(x)$    | Vertical stretch by a factor of 2  |
| $y = -cos(x)$    | Flip over x axis                   |
| $y = cos(2x)$    | Horizontal shrink by a factor of 2 |
# Periodic Functions
A function is considered periodic if it repeats itself at even intervals, where each interval is a complete cycle, referred to as a *period*.
# Sinusoidal Functions
A function that has the same shape as a sine or cosine wave is known as a sinusoidal function.

There are 4 general functions:

| $$A * sin(B*x - C) + D$$                  | $$ y = A * cos(B*x -c) + D$$           |
| ----------------------------------------- | -------------------------------------- |
| $$ y = A * sin(B(x - \frac{C}{B})) + D $$ | $$ y = A*cos(B(x - \frac{C}{B})) + D$$ |


How to find the:
- Amplitude: $|A|$
- Period: $\frac{2\pi}{B}$
- Phase shift: $\frac{C}{|B|}$
- Vertical shift: $D$ 


$$ y = A * \sin(B(x-\frac{C}{B})) $$
# Tangent
$$ y = tan(x) $$
![Graph of tangent](assets/graphtan.png)
To find relative points to create the above graph, you can use the unit circle:

If $tan(x) = \frac{sin(x)}{cos(x})$, then:

| $sin(0) = 0$                              | $cos(0) = 1$                              | $tan(0) = \frac{cos(0)}{sin(0)} = \frac{0}{1} =0$                |
| ----------------------------------------- | ----------------------------------------- | ---------------------------------------------------------------- |
| $sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $tan(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2} = 1$ |
| $sin(\frac{\pi}{2}) = 1$                  | $cos(\frac{\pi}{2}) = 0$                  | $tan(\frac{\pi}{2}) = \frac{1}{0} = DNF$                         |
Interpreting the above table:
- When $x = 0$, $y = 0$
- When $x = \frac{\pi}{4}$, $y = 1$
- When $x = \frac{\pi}{2}$, there's an asymptote

Without any transformations applied, the period of $tan(x) = \pi$. Because $tan$ is an odd function, $tan(-x) = -tan(x)$. 
# Cotangent
$$ y = cot(x) $$
![Graph of cotangent](assets/graphcot.svg)

To find relative points to create the above graph, you can use the unit circle:

If $cot(x) = \frac{cos(x)}{sin(x)}$, then:

| $sin(0) = 0$                              | $cos(0) = 1$                              | $cot(0) = \frac{sin(0)}{cos(0)} = \frac{1}{0} = DNF$             |
| ----------------------------------------- | ----------------------------------------- | ---------------------------------------------------------------- |
| $sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cot(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2} = 1$ |
| $sin(\frac{\pi}{2}) = 1$                  | $cos(\frac{\pi}{2}) = 0$                  | $tan(\frac{\pi}{2}) = \frac{1}{0} = DNF$                         |

Without any transformations applied, the period of $cot(x) = \pi$. Because $cot$ is an odd function, $cot(-x) = -cot(x)$. 

# Features of Tangent and Cotangent
Given the form $y = A\tan(Bx - C) + D$ (the same applies for $\cot$)
- The stretching factor is $|A|$
- The period is $\frac{\pi}{|B|}$
- The domain of $tan$ is all of $x$, where $x \ne \frac{C}{B} + \frac{\pi}{2} + {\pi}{|B|}k$, where $k$ is an integer. (everywhere but the asymptotes)
- The domain of $cot$ is all of $x$, where $x \ne \frac{C}{B} + \frac{\pi}{|B|}k$, where $k$ is an integer (everywhere but the asymptotes)
- The range of both is $(-\infty, \infty)$
- The phase shift is $\frac{C}{B}$
- The vertical shift is $D$
# Examples
> Given $-2tan(\pi*x + \pi) - 1$
- $A = -2$, $B = \pi$, $C = -\pi$, $D = -1$
- Stretch: $|-2| = 2$
- Period: $\frac{\pi}{|\pi|} = 1$
- Phase shift: $\frac{-\pi}{\pi} = -1$
- Vertical shift: $