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Blue Snowflake

Jumat, 06 November 2015

TRIGONOMETRI

Assalamu'alaikum sobat..
Bagaimana dengan pelajaran trigonometri kalian? Masih bingung memahaminya. Tidak usah khawatir, ayo kita pahami bersama..


Hubungan fungsi trigonometri

TrigonometryTriangle.svg
Fungsi dasar:
\sin A = \frac{a}{c}\,
\cos A = \frac{b}{c}\,
\tan A = \frac{\sin A}{\cos A}\ = \frac{a}{b}\,
\cot A = \frac{1}{\tan A} = \frac{\cos A}{\sin A}\ = \frac{b}{a}\,
\sec A = \frac{1}{\cos A}\ = \frac{c}{b}\,
\csc A = \frac{1}{\sin A}\ = \frac{c}{a}\,
\sin {(-A)} = -\sin A
\cos {(-A)} = \cos A
\tan {(-A)} = - \tan A
\csc {(-A)} = - \csc A
\sec {(-A)} = \sec A
\cot {(-A)} = - \cot A

Identitas trigonometri

\sin^2 A + \cos^2 A = 1 \,
1 + \tan^2 A = \frac{1}{\cos^2 A} = \sec^2 A\,
1 + \cot^2 A = \frac{1}{\sin^2 A} = \csc^2 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} \,
sin(u)+sin(v)=2sin(\frac {u+v}{2})cos(\frac{u-v}{2})
sin(u)-sin(v)=2cos(\frac {u+v}{2})sin(\frac{u-v}{2})
cos(u)+cos(v)=2cos(\frac {u+v}{2})cos(\frac{u-v}{2})
cos(u)-cos(v)=-2sin(\frac{u+v}{2})sin(\frac{u-v}{2})

Perkalian

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),

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} \,

Aturan Sinus, Cosinus, dan Tangen

Aturan sinus

LabeledTriangle.svg
\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} \!

Turunan dari aturan sinus

Law of sines proof.svg
Luasan dari segitiga diatas dapat dirumuskan sebagai
L = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C\,.
Kalikan persamaan diatas dengan 2/abc maka akan menjadi
\frac{2L}{abc} = \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\,.

Aturan cosinus

Triangle with notations 2.svg
c^2 = a^2 + b^2 - 2ab\cos\gamma\ ,
a^2 = b^2 + c^2 - 2bc\cos\alpha\,
b^2 = a^2 + c^2 - 2ac\cos\beta\,

Aturan tangen

Triangle with notations 2.svg
\frac{a-b}{a+b} = \frac{\tan[\frac{1}{2}(\alpha-\beta)]}{\tan[\frac{1}{2}(\alpha+\beta)]}.
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