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User:Temperal/The Problem Solver's Resource1

< User:Temperal
Revision as of 18:13, 10 January 2009 by Temperal (talk | contribs) (hm)
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:

$\tan^{-1} A=\text{atan } A=\arctan A$

$\cos^{-1} A=\text{acos } A=\arccos A$

$\sin^{-1} A=\text{asin } A=\arcsin A$

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

$\frac 12ab\sin C$

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