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السلام عليكم ورحمة الله وبركته . . . ؟

trigonometri

Diposting oleh king the world Senin, 07 Maret 2011

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} \,

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