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

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m nice idea; hope you don't mind if I edit
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Get rid of "page"
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==<span style="font-size:20px; color: blue;">Page 1: Trigonometric Formulas</span>==
==<span style="font-size:20px; color: blue;">Trigonometric Formulas</span>==
Note that all measurements are in degrees, not radians.  
Note that all measurements are in degrees, not radians.  
===Basic Facts===
===Basic Facts===

Revision as of 15:08, 30 September 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

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

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

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

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

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

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

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

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

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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