Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

User:Temperal/The Problem Solver's Resource1: Difference between revisions

← Older editNewer edit →
Xpmath (talk | contribs)
152 edits
Temperal (talk | contribs)
3,154 edits
additional details
Line 26: Line 26:
<math>\tan (90-A)=\cot A</math>
<math>\tan (90-A)=\cot A</math>


<math>\sec{90-A}=\csc A</math>
<math>\cos (90-A) = \sin A</math>
<math>\cot (90-A)=\tan A</math>
<math>\csc (90-A)=\sec A</math>
<math>\sin^2 A+\cos^2 A=1</math>
<math>\sec^2 A-\tan^2 A=1</math>
<math>\csc^2 A-\cot^2 A=1</math>
<math>\tan A=\frac{\sin A}{\cos A}</math>
<math>\sin^2 \frac{A}{2}=\frac{1}{2}(1-\cos A)</math>
<math>\cos^2 \frac{A}{2}=\frac{1}{2}(1+\cos A)</math>
<math>\tan \frac{A}{2}=\frac{1-\cos A}{\sin A}=\frac{\sin A}{1+\cos A}</math>
===Terminology===
<math>\cot A=\frac{1}{\tan A}</math>, but <math>\cot A\ne\tan^{-1} A}</math>.
<math>\csc A=\frac{1}{\sin A}</math>, but <math>\csc A\ne\sin^{-1} A}</math>.
<math>\sec A=\frac{1}{\sin A}</math>, but <math>\sec A\ne\cos^{-1} A}</math>.
Also:
<math>\tan^{-1} A=\atan A=\arctan A</math>
<math>\tan^{-1} A=\asin A=\arcsin A</math>
<math>\tan^{-1} A=\asin A=\arcsin A</math>
===Sum of Angle Formulas===
===Sum of Angle Formulas===
<math>\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B</math>
<math>\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B</math>
Line 54: Line 90:


===Other Formulas===
===Other Formulas===
 
====Law of Cosines==
In a triangle with sides <math>a</math>, <math>b</math>, and <math>c</math> opposite angles <math>A</math>, <math>B</math>, and <math>C</math>, respectively,
In a triangle with sides <math>a</math>, <math>b</math>, and <math>c</math> opposite angles <math>A</math>, <math>B</math>, and <math>C</math>, respectively,


<math>c^2=a^2+b^2-2bc\cos A</math>
<math>c^2=a^2+b^2-2bc\cos A</math>


and
and:
====Law of Sines====


<math>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R</math>
<math>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R</math>


====Law of Tangents===
For any <math>a</math> and <math>b</math> such that <math>\tan a,\tan b \subset \mathbb{R}</math>,
<math>\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}</math>
====Area of a Triangle====
The [[area]] of a triangle can be found by  
The [[area]] of a triangle can be found by  


Revision as of 16:19, 9 October 2007



The Problem Solver's Resource
Introduction Other Tips and Tricks Methods of Proof You are currently viewing page 1.

Trigonometric Formulas

Note that all measurements are in degrees, not radians.

Basic Facts

$\sin (-A)=-\sin A$

$\cos (-A)=\cos A$

$\tan (-A)=-\tan A$

$\sin (180-A) = \sin A$

$\cos (180-A) = -\cos A$

$\cos (360-A) = \cos A$

$\tan (180+A) = \tan A$

$\cos (90-A)=\sin A$

$\tan (90-A)=\cot A$

$\sec{90-A}=\csc A$

$\cos (90-A) = \sin A$

$\cot (90-A)=\tan A$

$\csc (90-A)=\sec A$

$\sin^2 A+\cos^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\csc^2 A-\cot^2 A=1$

$\tan A=\frac{\sin A}{\cos A}$

$\sin^2 \frac{A}{2}=\frac{1}{2}(1-\cos A)$

$\cos^2 \frac{A}{2}=\frac{1}{2}(1+\cos A)$

$\tan \frac{A}{2}=\frac{1-\cos A}{\sin A}=\frac{\sin A}{1+\cos A}$

Terminology

$\cot A=\frac{1}{\tan A}$, but $\cot A\ne\tan^{-1} A}$ (Error compiling LaTeX. Unknown error_msg).

$\csc A=\frac{1}{\sin A}$, but $\csc A\ne\sin^{-1} A}$ (Error compiling LaTeX. Unknown error_msg).

$\sec A=\frac{1}{\sin A}$, but $\sec A\ne\cos^{-1} A}$ (Error compiling LaTeX. Unknown error_msg).

Also:

$\tan^{-1} A=\atan A=\arctan A$ (Error compiling LaTeX. Unknown error_msg)

$\tan^{-1} A=\asin A=\arcsin A$ (Error compiling LaTeX. Unknown error_msg)

$\tan^{-1} A=\asin A=\arcsin A$ (Error compiling LaTeX. Unknown error_msg)

Sum of Angle Formulas

$\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B$

$\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B$

$\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

$\sin2A=2\sin A \cos A$

$\cos2A=\cos^2 A - \sin^2 A$ or $\cos2A=2\cos^2 A -1$ or $\cos2A=1- 2 \sin^2 A$

$\tan2A=\frac{2\tan A}{1-\tan^2 A}$

Pythagorean identities

$\sin^2 A+\cos^2 A=1$

$1 + \tan^2 A = \sec^2 A$

$1 + \cot^2 A = \csc^2 A$

for all $A$.

Other Formulas

==Law of Cosines

In a triangle with sides $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$, respectively,

$c^2=a^2+b^2-2bc\cos A$

and:

Law of Sines

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

=Law of Tangents

For any $a$ and $b$ such that $\tan a,\tan b \subset \mathbb{R}$, $\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}$

Area of a Triangle

The area of a triangle can be found by

$\frac 12ab\sin C$

Back to intro | Continue to page 2



Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=User:Temperal/The_Problem_Solver%27s_Resource1&oldid=17736"