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

Revision as of 11:44, 29 September 2007

The Problem Solver's Resource

Introduction \| Other Tips and Tricks \| Methods of Proof \| You are currently viewing page 1.

Note that all measurements are in degrees, not radians.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\sin^2 A+\cos^2 A=1$ for all $A$ .

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