Related Angles in Trigonometry - Clavius

Related Angles

1. Quadrant

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2. Use of 90º, 180º, 270º and 360º

90º and 270º		180º and 360º
$\sin (90 - α) = \cos α$	$\sin (90 + α) = \cos α$	$\sin (180 - α) = \sin α$	$\sin (180 + α) = - \sin α$
$\sin (270 - α) = - \cos α$	$\sin (270 + α) = - \cos α$	$\sin (360 - α) = - \sin α$	$\sin (360 + α) = \sin α$
$\cos (90 - α) = \sin α$	$\cos (90 + α) = - \sin α$	$\cos (180 - α) = - \cos α$	$\cos (180 + α) = - \cos α$
$\cos (270 - α) = - \sin α$	$\cos (270 + α) = \sin α$	$\cos (360 - α) = \cos α$	$\cos (360 + α) = \cos α$
$\tan (90 - α) = \cot α$	$\tan (90 + α) = - \cot α$	$\tan (180 - α) = - \tan α$	$\tan (180 + α) = \tan α$
$\tan (270 - α) = \cot α$	$\tan (270 + α) = - \cot α$	$\tan (360 - α) = - \tan α$	$\tan (360 + α) = \tan α$

3. Bigger Angles

For bigger angles, we must find the remainder angles after multiplication of 360°.

Example

$\begin{aligned} \sin 1470^{o} & = \sin (360^{o} . 4 + 30^{o}) \\ \sin 1470^{o} & = \sin 30^{o} \\ \sin 1470^{o} & = \frac{1}{2} \end{aligned}$

4. Negative Angles

Negative angles can be assumed that the angle is on the 4^th quadrant.

$\sin (- α) = - \sin α$

$\cos (- α) = \cos α$

$\tan (- α) = - \tan α$

Exercise

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