User:Temperal/The Problem Solver's Resource1

< User:Temperal
Revision as of 19:35, 10 January 2009 by Temperal (talk | contribs) (Sum of Angle Formulas: hm)
Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 1.

Trigonometric Formulas

Note that all measurements are in radians.

Basic Facts

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

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

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

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

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

$\cos (2\pi-A) = \cos A$

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

$\cos (\pi/2-A)=\sin A$

$\tan (\pi/2-A)=\cot A$

$\sec{\pi/2-A}=\csc A$

$\cos (\pi/2-A) = \sin A$

$\cot (\pi/2-A)=\tan A$

$\csc (\pi/2-A)=\sec A$

The above can all be seen clearly by examining the graphs or plotting on a unit circle - the reader can figure that out themselves.

Terminology and Notation

$\cot A=\frac{1}{\tan A}$, but $\cot A\ne\tan^{-1} A}$ (Error compiling LaTeX. ! Extra }, or forgotten $.), the former being the reciprocal and the latter the inverse.

$\csc A=\frac{1}{\sin A}$, but $\csc A\ne\sin^{-1} A}$ (Error compiling LaTeX. ! Extra }, or forgotten $.).

$\sec A=\frac{1}{\sin A}$, but $\sec A\ne\cos^{-1} A}$ (Error compiling LaTeX. ! Extra }, or forgotten $.).

Speaking of inverses:

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

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

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

Sum of Angle Formulas

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

If we can prove this one, the other ones can be derived easily using the "Basic Facts" identities above. In fact, we can simply prove the addition case, for plugging $A=-B$ into the addition case gives the subtraction case.

As it turns out, there's quite a nice geometric proof of the addition case, though other methods, such as de Moivre's Theorem, exist.

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

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

The following identities can be easily derived by plugging $A=B$ into the above:

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

$\cos2A=\cos^2 A - \sin^2 A$ 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,

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


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

Invalid username
Login to AoPS