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

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<math>\sin (180-A) = \sin A</math>
 
<math>\sin (180-A) = \sin A</math>
  
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<math>\cos (180-A) = -cos A</math>
 
<math>\cos (360-A) = \cos A</math>
 
<math>\cos (360-A) = \cos A</math>
  
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<math>\tan (90-A)=\cot A</math>
 
<math>\tan (90-A)=\cot A</math>
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===Sum of Angle Formulas===
 
===Sum of Angle Formulas===
 
<math>\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B</math>
 
<math>\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B</math>

Revision as of 20:54, 5 October 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

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

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

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

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

$\cos (180-A) = -cos A$ $\cos (360-A) = \cos A$

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

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

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

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

$\frac 12ab\sin C$

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