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User:Temperal/The Problem Solver's Resource1

< User:Temperal
Revision as of 16:08, 29 September 2007 by Azjps (talk | contribs) (nice idea; hope you don't mind if I edit)



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

Page 1: Trigonometric Formulas

Note that all measurements are in degrees, not radians.

Basic Facts

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

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

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

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

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

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

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

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

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

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

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

The area of a triangle can be found by

$\frac 12ab\sin C$

Back to intro | Continue to page 2



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