Tiyas Sulistiyaningsih: Rumus Trigonometri

Sabtu, 21 Desember 2013

Identitas trigonometri

$\sin^2 A + \cos^2 A = 1 \,$

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

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

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

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

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

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

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

$2 \sin A \times \cos B = \sin (A + B) + \sin (A - B),$

$2 \cos A \times \sin B = \sin (A + B) - \sin (A - B),$

$2 \cos A \times \cos B = \cos (A + B) + \cos (A - B),$

$2 \sin A \times \sin B = - \cos (A + B) + \cos (A - B),$

$\sin \frac{A}{2} = \pm \sqrt{\frac{1-\cos A}{2}} \,$

$\cos \frac{A}{2} = \pm \sqrt{\frac{1+\cos A}{2}} \,$

$\tan \frac{A}{2} = \pm \sqrt{\frac{1-\cos A}{1+\cos A}} = \frac {\sin A}{1+\cos A} = \frac {1-\cos A}{\sin A} \,$