RUMUS PELAJARAN: MATEMATIKA SIMPLE

Sabtu, 24 April 2010 at 06.03 | By: subhan

MATEMATIKA SIMPLE

Hubungan fungsi trigonometri

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

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

$1 + \cot^2 A = \csc^2 A \,$

$\tan A = \frac{\sin A}{\cos A}\,$

Penjumlahan

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

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

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

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

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

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

Rumus sudut rangkap dua

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

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

$\tan 2A = {2 \tan A \over 1 - \tan^2 A} = {2 \cot A \over \cot^2 A - 1} = {2 \over \cot A - \tan A} \,$

Rumus sudut rangkap tiga

$\sin 3A = 3 \sin A - 4 \sin^3 A \,$

$\cos 3A = 4 \cos^3 A - 3 \cos A \,$

Rumus setengah sudut

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

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

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

2 comments:

Slamet Wibowo mengatakan...: Sip, Tapi itu dah matengnya! mentahnya (Jalan/bukti) dari rumus itu tolong di post juga!
Thank's; 26 Oktober 2012 pukul 04.19
subhan mengatakan...: Insya Allah ta coba edit dulu,,,,; 10 Februari 2013 pukul 11.21

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