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Maths › Geometry › Trigonometry ›

Formulae

General trigonometric identities

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Formulae
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Formulae

A list of important general trigonometric identities,

$\cos^2 \theta + \sin^2 \theta = 1$

$\sec^2 \theta = 1 + \tan^2 \theta$

$\csc^2 \theta = 1 + \cot^2 \theta$

$\cos(90^{\circ} - \theta) = \sin \theta$

$\sin(90^{\circ} - \theta) = \cos \theta$

$\tan(90^{\circ} - \theta) = \cot \theta$

Some more specific identities that relate to the following general diagram,

1/triangle.jpg

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$

$\cos(A-b) = \cos A \cos B + \sin A \sin B$

$\cos(A+b) = \cos A \cos B - \sin A \sin B$

$\sin(A-b) = \sin A \cos B - \cos A \sin B$

$\sin(A+b) = \sin A \cos B + \cos A \sin B$

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

$\cos 2A = 2 \cos^2 A - 1$

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

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

$\tan (A + B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$

$\tan (A - B) = \frac{\tan A - \tan B}{1+\tan A \tan B}$

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

$\tan (A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1- \tan B \tan C - \tan C \tan A - \tan A \tan B}$

If $A+B+C=180^{\circ}$

$\displaystyle \tan A + \tan B + \tan C = \tan A \tan B \tan C$
$\displaystyle \cot \tfrac{1}{2}A + \cot \tfrac{1}{2} B + \cot \tfrac{1}{2} C = \cot \tfrac{1}{2} A \cot \tfrac{1}{2} B \cot \tfrac{1}{2} C$
$\displaystyle \sin A + \sin B + \sin C = 4 \cos \tfrac{1}{2} A \cos \tfrac{1}{2} B \cos \tfrac{1}{2} C$
$\displaystyle \cos A + \cos B + \cos C -1 = 4 \sin \tfrac{1}{2} A \sin \tfrac{1}{2} B \sin \tfrac{1}{2} C$

$1+ \cos A = 2 \cos^2(\frac{1}{2}A)$

$1- \cos A = 2 \sin^2(\frac{1}{2}A)$

$\sin 2A = \frac{2 \tan A}{1+\tan^2 A}$

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

$\sin X + \sin Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}$

$\sin X - \sin Y = 2 \cos \frac{X+Y}{2} \sin \frac{X-Y}{2}$

$\cos X + \cos Y = 2 \cos \frac{X+Y}{2} \cos \frac{X-Y}{2}$

$\cos X - \cos Y = 2 \sin \frac{X+Y}{2} \sin \frac{Y-X}{2}$

$\frac{a-b}{a+b} \cot \frac{C}{2} = \tan \tfrac{1}{2}(A-B)$

$\arctan x + \arctan y \arctan \frac{x+y}{1-xy}$

$\arctan x - \arctan y \arctan \frac{x-y}{1+xy}$

where

R stands for the curcum-radius of the triangle ABC

To avoid doubt (and for those new to maths):

$\cot \theta = 1 / \tan \theta$
$\csc \theta = 1 / \sin \theta$
$\sec \theta = 1 / \cos \theta$