.obsidian/app.json

@@ -2,7 +2,6 @@
   "vimMode": true,
   "promptDelete": false,
   "pdfExportSettings": {
-    "includeName": true,
     "pageSize": "Letter",
     "landscape": false,
     "margin": "0",

education/math/MATH1060 (trig)/Identities.md

@@ -1,41 +0,0 @@
-# Trigonometric Identities
-
-All of the following only apply when the denominator is not equal to zero.
-
-$$ tan \theta = \frac{y}{x} $$
-Because the following are inverses of their counterparts, you only need to remember the equivalents for $sin$, $cos$, and $tan$, then just find the inverse by taking $1/v$. 
-
-| Base Identity                 | Inverse Identity               | Alternate Identities                          | Alternate Inverse Identities                                          |
-| ----------------------------- | ------------------------------ | --------------------------------------------- | --------------------------------------------------------------------- |
-| $$ sin\theta = y $$           | $$ csc\theta = \frac{1}{y} $$  |                                               | $$ csc\theta = \frac{1}{sin\theta} $$                                 |
-| $$ cos\theta = x $$           | $$ sec \theta = \frac{1}{x} $$ |                                               | $$ sec\theta = \frac{1}{cos\theta} $$                                 |
-| $$ tan\theta = \frac{y}{x} $$ | $$ cot\theta = \frac{x}{y} $$  | $$ tan\theta = \frac{sin\theta}{cos\theta} $$ | $$ cot\theta = \frac{1}{tan\theta} = \frac{cos\theta}{sin{\theta}} $$ |
-
-$$ cot \theta = \frac{x}{y} $$
-$$ sec\theta = \frac{1}{cos\theta}$$
-$$ csc\theta = \frac{1}{sin\theta}$$
-# Pythagorean Identities
-The Pythagorean identity expresses the Pythagorean theorem in terms of trigonometric functions. It's a basic relation between the sine and cosine functions.
-$$ sin^2 \theta + cos^2 \theta = 1 $$
-There are more forms that are useful, but they can be derived from the above formula:
-$$ 1 + tan^2\theta = sec^2\theta $$
-$$ cot^2 \theta + 1 = csc^2\theta $$
-# Even and Odd Identities
-- A function is even if $f(-x) = f(x)$.
-- A function is odd if $f(-x) = -f(x)$
-- Cosine and secant are **even**
-- Sine, tangent, cosecant, and cotangent are **odd**.
-## Examples
-### Even and Odd Functions
-> If $cot\theta = -\sqrt{3}$, what is $cot(-\theta)$? 
-
-$cot$ is an odd function, and so $cot(-\theta) = \sqrt{3}$
-### Simplifying Using Identities
-> Simplify $\frac{sin\theta}{cos\theta}$
-
-1. The above equation can be split into two components
-$$ \frac{sin\theta}{cos\theta} = \frac{sin\theta}{1} * \frac{1}{csc\theta} $$
-2. Referring to the list of trig identities, we know that $\frac{1}{csc\theta}$ is equal to $sin\theta$. 
-$$ \frac{sin\theta}{1} * \frac{1}{csc\theta} = sin\theta * sin\theta $$
-3. Simplifying further, we get:
-$$ sin^2\theta $$

education/math/MATH1060 (trig)/The Unit Circle.md

@@ -26,6 +26,19 @@ Finding a reference angle:
 | 2        | $180\degree - \theta$ |
 | 3        | $\theta - 180\degree$ |
 | 4        | $360\degree - \theta$ |
+## Other Trigonometric Functions
+All of the following only apply when the denominator is not equal to zero.
+
+$$ tan \theta = \frac{y}{x} $$
+Because the following are inverses of their counterparts, you only need to remember the equivalents for $sin$, $cos$, and $tan$, then just find the inverse by taking $1/v$. 
+$$ sec \theta = \frac{1}{x} $$
+$$ csc = \frac{1}{y} $$
+$$ cot \theta = \frac{x}{y} $$
+
+## The Pythagorean Identity
+The Pythagorean identity expresses the Pythagorean theorem in terms of trigonometric functions. It's a basic relation between the sine and cosine functions.
+$$ sin^2 \theta + cos^2 \theta = 1 $$
+
 # Definitions
 
 | Term             | Description                                                                   |

education/software development/ECE1400/Chapter 5 Exercises.md

@@ -1,103 +0,0 @@
-> 2. The following program fragments illustrate the logical operators. Show the output produced by each, assuming that `i`, `j`, and `k` are `int` variables.
-
-a. `i = 10; j = 5;`
-```c
-printf("%d", !i < j);
-
-// Expected output: `1`, because `!i` evaluates to 0, and 0 is less than 5, so that expression evaluates to true, or 1.
-```
-
-b. `i = 2; j = 1;`
-```c
-printf("%d", !!i + !j);
-
-// Expected output: `1`, because !!2 evaluates to 1, and !j evaluates to 0
-```
-
-c. `i = 5; j = 0; k = -5;`
-```c
-printf("%d", i && j || k);
-
-// Expected output: `1`, because i && j should evaluate to 0, but `0 || 1` should evalulate to true.
-```
-
-d. `i = 1; j = 2; k = 3;`
-```c
-printf("%d", i < j || k);
-
-// Expected output: `1`
-```
-
-> 4. Write a single expression whose value is either `-1`, `0`, or `1` depending on whether `i` is less than, equal to, or greater than `j`, respectively.
-
-```c
-/*
-	If i < j, the output should be -1.
-	If i == j, the output should be zero
-	If i > j, the output should be 1.
-*/
-(i > j) - (i < j)
-```
-
-> 6. Is the following `if` statement legal?
-```c
-if (n == 1-10)
-	printf("n is between 1 and 10\n");
-```
-
-Yes the statement is *legal*, but it does not produce the intended effect. It would not produce an output when `n = 5`, because `1-10` evaluates to `-9`, and `-9 != 5`.
-
-> 10. What output does the following program fragment produce? (Assume that `i` is an integer variable.)
-```c
-int i = 1;
-switch (i % 3) {
-	case 0: printf("zero");
-	case 1: printf("one");
-	case 2: printf("two");
-}
-```
-
-The program would print `onetwo` because each case is missing a `break` statement.
-
-> 11. The following table shows the telephone area codes in the state of Georgia along with the largest city in each area:
-
-| Area code | Major city |
-| --------- | ---------- |
-| 229       | Albany     |
-| 404       | Atlanta    |
-| 470       | Atlanta    |
-| 478       | Macon      |
-| 678       | Atlanta    |
-| 706       | Columbus   |
-| 762       | Columbus   |
-| 770       | Atlanta    |
-| 912       | Savannah   |
-> Write a switch statement whose controlling expression is the variable `area_code`. If the value of `area_code` is not in the table, the `switch` statement will print the corresponding city name. Otherwise, the `switch` statement will display the message `"Area code not recognized."` Use the techniques discussed in section 5.3 to make the `switch` as simple as possible.
-```c
-int area_code;
-
-switch (area_code) {
-	case 404:
-	case 470:
-	case 678:
-	case 770:
-		printf("Atlanta");
-		break;
-	case 706:
-	case 762:
-		printf("Columbus");
-		break;
-	case 229:
-		printf("Albany");
-		break;
-	case 478:
-		printf("Macon");
-		break;
-	case 912:
-		printf("Savannah");
-		break;
-	default:
-		printf("Area code not recognized.");
-		break;
-}
-```

education/software development/ECE1400/Chapter.md

@@ -0,0 +1,6 @@
+> 2. The following program fragments illustrate the logical operators. Show the output produced by each, assuming that `i`, `j`, and `k` are `int` variables.
+
+a. `i = 10; j = 5;`
+```
+printf("%d", !i < j);
+```