{"id":86,"date":"2020-11-07T04:22:55","date_gmt":"2020-11-07T04:22:55","guid":{"rendered":"https:\/\/www.perfectly-logical.com\/?p=86"},"modified":"2020-11-08T05:39:38","modified_gmt":"2020-11-08T05:39:38","slug":"pre-calculus-math-references","status":"publish","type":"post","link":"https:\/\/www.perfectly-logical.com\/index.php\/2020\/11\/07\/pre-calculus-math-references\/","title":{"rendered":"Pre-calculus Math References"},"content":{"rendered":"\n\n\n

Pre-Calculus<\/h2>\n\n\n\n

Algebra<\/h3>\n\n\n\n

Arithmetic Operations<\/h4>\n\n\n\n
\"a(b+c)=ab+ac\"<\/td>\"\frac{a}{b} + \frac{c}{d}\"<\/td><\/tr>
\"\frac{a+c}{b} = \frac{a}{b} + \frac{c}{b}\"<\/td>\"\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \centerdot \frac{d}{c} =\frac{ad}{bc}\"<\/td><\/tr><\/tbody><\/table>
Arithmetic Operations<\/figcaption><\/figure>\n\n\n\n

Exponents and Radicals<\/h4>\n\n\n\n
\"x^m +x^n = x^{m+n}\"<\/td>\"\frac{x^m}{x^n} = x^{m-n}\"<\/td><\/tr>
\"( x^m ) ^n = x^{mn}\"<\/td>\"x^{-n} = \frac{1}{x^n}\"<\/td><\/tr>
\"x^{\frac{1}{n}} = ^n \sqrt{x}\"<\/td>\"x^{\frac{m}{n}} = ^n \sqrt{x^m} = ( ^n \sqrt{x} ) ^m\"<\/td><\/tr>
\"^n \sqrt{xy} = ^n \sqrt{x} ^n \sqrt{y}\"<\/td>\"^n \sqrt{\frac{x}{y}} = \frac{ ^n \sqrt{x}}{ ^n \sqrt{y}}\"<\/td><\/tr><\/tbody><\/table>
Exponents and Radicals<\/figcaption><\/figure>\n\n\n\n

Factoring Special Polynomials<\/h4>\n\n\n\n
\"x^2 - y^2 = ( x + y ) ( x - y )\"<\/td>\"x^3 + y^3 = ( x + y ) ( x^2 - xy + y^2 )\"<\/td><\/tr>
\"x^3 - y^3 = ( x - y ) ( x^2 + xy + y^2 )\"<\/td><\/td><\/tr><\/tbody><\/table>
Factoring Special Polynomials<\/figcaption><\/figure>\n\n\n\n

Binomial Theorem<\/h4>\n\n\n\n
\"( x + y ) ^2 = x^2 + 2xy + y^2\"<\/td>\"( x - y ) ^2 = x^2 - 2xy + y^2\"<\/td><\/tr>
\"( x +y ) ^3 = x^3 + 3x^2 y + 3xy^2 + y^3\"<\/td>\"( x - y ) ^3 = x^3 - 3x^2 y + 3xy^2 - y^3\"<\/td><\/tr>
\"( x +y ) ^n = x^n + nx^{n-1} y + \frac{n(n-1)}{2} x^{n-2}\"\cdots + \binom{n}{k} x^{n-k} y^k + \cdots + nxy^{n-1} + y^n\" where \"\binom{n}{k} = \frac{n(n-1) \cdots (n-k-1)}{k!}\"<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Quadratic Formula<\/h4>\n\n\n\n

If \"ax^2 + bx + c=0\", then \"x=\frac{-b \pm \sqrt{b^2 -4ac}}{2a}\"<\/p>\n\n\n\n

Inequalities and Absolute Values<\/h4>\n\n\n\n
If a < b and b < c, the a < c.<\/td>If a < b, then a+c < b+c.<\/td><\/tr>
If a < b and c > 0, then ca < cb.<\/td>If a < b and c < 0, then ca > cb.<\/td><\/tr>
If a > 0, then<\/td>|x|=a means x=a or x=-a<\/td><\/tr>
<\/td>|x| < a means -a < x < a<\/td><\/tr>
<\/td>|x| > a means x > a or x < -a<\/td><\/tr><\/tbody><\/table>
Inequalities and Absolute Values<\/figcaption><\/figure>\n\n\n\n

Geometry<\/h3>\n\n\n\n

Geometric Formulas<\/h4>\n\n\n\n

Formulas for area A, circumference C, and volume V<\/h5>\n\n\n\n
Triangle<\/td>\"A = \frac{1}{2} b h = \frac{1}{2} a b sin \theta\" where \"\theta\" is the angle made by a and b. Also, b is the base of the triangle and h is the height of the triangle.<\/td><\/tr>
Circle<\/td>\"A = \pi r^2\"
\"C = 2 \pi r\"<\/td><\/tr>
Sector of Circle<\/td>\"A = \frac{1}{2} r^2 \theta\"
\"s = r \theta\"
(\"\theta\" in radians)<\/td><\/tr>
Sphere<\/td>\"V = \frac{4}{3} \pi r^3\"
\"A = 4 \pi r^2\"<\/td><\/tr>
Cylinder<\/td>\"V = \pi r^2 h\"<\/td><\/tr>
Cone<\/td>\"V = \frac{1}{3} \pi r^2 h\"
\"A = \pi r \sqrt{r^2 + h^2}\"<\/td><\/tr><\/tbody><\/table>
Formulas for area A, circumference C, and volume V<\/figcaption><\/figure>\n\n\n\n

Distance and Midpoint Formulas<\/h4>\n\n\n\n
Distance between points (x1<\/sub>,y1<\/sub>) and (x2<\/sub>,y2<\/sub>):<\/td>\"d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\"<\/td><\/tr>
Midpoint between points (x1<\/sub>,y1<\/sub>) and (x2<\/sub>,y2<\/sub>):<\/td>\"(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2} )\"<\/td><\/tr><\/tbody><\/table>
Distance and Midpoint Formulas<\/figcaption><\/figure>\n\n\n\n

Properties of Lines<\/h4>\n\n\n\n
Slope of a line through points (x1<\/sub>,y1<\/sub>) and (x2<\/sub>,y2<\/sub>):<\/td>\"m = \frac{y_2 - y_1}{x_2 - x_1}\"<\/td><\/tr>
Point-slope equation of a line through point (x1<\/sub>,y1<\/sub>) with slope m:<\/td>\"y - y_1 = m(x - x_1)\"<\/td><\/tr>
Slope-intercept equation of a line with slope m and y-intercept b:<\/td>\"y = mx + b\"<\/td><\/tr><\/tbody><\/table>
Properties of Lines<\/figcaption><\/figure>\n\n\n\n

Circle Equation<\/h4>\n\n\n\n
The equation of a circle with center at (h,k) and radius r:<\/td>\"(x-h)^2 + (y - k)^2 = r^2\"<\/td><\/tr><\/tbody><\/table>
Circle Equation<\/figcaption><\/figure>\n\n\n\n

Trigonometry<\/h3>\n\n\n\n

Angle Measurement<\/h4>\n\n\n\n

\"\pi\" radians = \"180^o\"<\/p>\n\n\n\n

Right Angle Trigonometry<\/h4>\n\n\n\n
\"sin(\theta) = \frac{opp}{hyp}\"<\/td>\"csc(\theta) = \frac{hyp}{opp}\"<\/td><\/tr>
\"cos(\theta) = \frac{adj}{hyp}\"<\/td>\"sec(\theta) = \frac{hyp}{adj}\"<\/td><\/tr>
\"tan(\theta) = \frac{opp}{adj}\"<\/td>\"cot(\theta) = \frac{adj}{opp}\"<\/td><\/tr><\/tbody><\/table>
Right Angle Trigonometry<\/figcaption><\/figure>\n\n\n\n

A nice thing to think of is SohCahToa. S(in)o(pp)h(yp)C(osine)a(dj)h(yp)T(angent)o(opp)a(adj)<\/p>\n\n\n\n

Fundamental Identities<\/h4>\n\n\n\n
\"csc(\theta) = \frac{1}{sin(\theta)}\"<\/td>\"sec(\theta) = \frac{1}{cos(\theta)}\"<\/td><\/tr>
\"tan(\theta) = \frac{sin(\theta)}{cos(\theta)}\"<\/td>\"cot(\theta) = \frac{cos(\theta)}{sin(\theta)}\"<\/td><\/tr>
\"cot(\theta) = \frac{1}{tan(\theta)}\"<\/td>\"sin^2 (\theta) + cos^2 (\theta) = 1\"<\/td><\/tr>
\"1 + tan^2 (\theta) = sec^2 (\theta)\"<\/td>\"1 + cot^2 (\theta) = csc^2 (\theta)\"<\/td><\/tr>
\"sin(- \theta) = - sin(\theta)\"<\/td>\"cos(- \theta) = cos(\theta)\"<\/td><\/tr>
\"tan(- \theta) = - tan(\theta)\"<\/td>\"sin(\frac{\pi}{2} - \theta) = cos(\theta)\"<\/td><\/tr>
\"cos(\frac{\pi}{2} - \theta) = sin(\theta)\"<\/td>\"tan(\frac{\pi}{2} - \theta) = cot(\theta)\"<\/td><\/tr><\/tbody><\/table>
Fundamental Identities<\/figcaption><\/figure>\n\n\n\n

The Law of Sines<\/h4>\n\n\n\n
For a triangle where a is the side opposite of angle A, b is the side opposite of angle B, and c is the side opposite of angle C:<\/td><\/tr>
\"\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}\"<\/td><\/tr><\/tbody><\/table>
The Law of Sines<\/figcaption><\/figure>\n\n\n\n

The Law of Cosines<\/h4>\n\n\n\n
For a triangle where a is the side opposite of angle A, b is the side opposite of angle B, and c is the side opposite of angle C:<\/td><\/tr>
\"a^2 = b^2 + c^2 -2bc cos(A)\"<\/td><\/tr>
\"b^2 = a^2 + c^2 -2ac cos(B)\"<\/td><\/tr>
\"c^2 = a^2 + b^2 -2ab cos(C)\"<\/td><\/tr><\/tbody><\/table>
The Law of Cosines<\/figcaption><\/figure>\n\n\n\n

Addition and Subtraction Formulas<\/h4>\n\n\n\n
\"sin(x+y) = sin(x)cos(y) + cos(x)sin(y)\"<\/td><\/tr>
\"sin(x-y) = sin(x)cos(y) - cos(x)sin(y)\"<\/td><\/tr>
\"cos(x+y) = cos(x)cos(y) - sin(x)sin(y)\"<\/td><\/tr>
\"cos(x-y) = cos(x)cos(y) + sin(x)sin(y)\"<\/td><\/tr>
\"tan(x+y) = \frac{tan(x) + tan(y)}{1 - tan(x)tan(y)}\"<\/td><\/tr>
\"tan(x-y) = \frac{tan(x) - tan(y)}{1 + tan(x)tan(y)}\"<\/td><\/tr><\/tbody><\/table>
Addition and Subtraction Formulas<\/figcaption><\/figure>\n\n\n\n

Double-Angle Formulas<\/h4>\n\n\n\n
\"sin(2x) = 2 sin(x) cos(x)\"<\/td><\/tr>
\"cos(2x) = cos ^2 (x) - sin ^2 (x) = 2 cos ^2 (x) - 1 = 1 - 2 sin ^2 (x)\"<\/td><\/tr>
\"tan(2x) = \frac{2 tan(x)}{1 - tan ^2 (x)}\"<\/td><\/tr><\/tbody><\/table>
Double-Angle Formulas<\/figcaption><\/figure>\n\n\n\n

Half-Angle Formulas<\/h4>\n\n\n\n
\"sin ^2 (x) = \frac{1 - cos(2x)}{2}\"<\/td>\"cos ^2 (x) = \frac{1 + cos(2x)}{2}\"<\/td><\/tr><\/tbody><\/table>
Half-Angle Formulas<\/figcaption><\/figure>\n","protected":false},"excerpt":{"rendered":"

Formulas used in Pre-calculus from Algebra, Geometry, Trigonometry, and more.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6,29],"tags":[14,15,8,13,12],"class_list":["post-86","post","type-post","status-publish","format-standard","hentry","category-math","category-reference","tag-calculus","tag-formulas","tag-math","tag-pre-calculus","tag-reference"],"_links":{"self":[{"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/posts\/86","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/comments?post=86"}],"version-history":[{"count":23,"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/posts\/86\/revisions"}],"predecessor-version":[{"id":109,"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/posts\/86\/revisions\/109"}],"wp:attachment":[{"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/media?parent=86"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/categories?post=86"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.perfectly-logical.com\/index.php\/wp-json\/wp\/v2\/tags?post=86"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}