Trigonometri: Perbedaan antara revisi

:<math>\sin A = \frac{a}{c} </math>

:<math>\cos A = \frac{b}{c} </math>

:<math>\tan A = \frac{\sin A}{\cos A} = \frac{a}{b} </math>

:<math>\cot A = \frac{1}{\tan A} = \frac{\cos A}{\sin A} = \frac{b}{a} </math>

:<math>\sec A = \frac{1}{\cos A} = \frac{c}{b} </math>

:<math>\csc A = \frac{1}{\sin A} = \frac{c}{a} </math>

:<math>\sin^2 A + \cos^2 A = 1 </math>

:<math>1 + \tan^2 A = \frac{1}{\cos^2 A} = \sec^2 A </math>

:<math>1 + \cot^2 A = \frac{1}{\sin^2 A} = \csc^2 A </math>

:<math>sinA\sin A = \cos (90 - A) \text{ atau } \cos \left(\frac{\pi}{2} - A \right)</math>

:<math>tanA\tan A = \cot (90 - A) \text{ atau } \cot \left(\frac{\pi}{2} - A\right)</math>

:<math>secA\sec A = \csc (90 - A) \text{ atau } \csc \left(\frac{\pi}{2} - A\right)</math>

:<math>\sin (A + B) = \sin A \cos B + \cos A \sin B </math>

:<math>\sin (A - B) = \sin A \cos B - \cos A \sin B </math>

:<math>\cos (A + B) = \cos A \cos B - \sin A \sin B </math>

:<math>\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} </math>

:<math>\tan (A - B) = \frac{\tan A - tan B}{1 +\ tan A \tan B} </math>

:<math>2 \sin A \cos B = \sin (A + B) + \sin (A - B) </math>

:<math>2 \cos A \sin B = \sin (A + B) - \sin (A - B) </math>

:<math>2 \cos A \cos B = \cos (A + B) + \cos (A - B) </math>

:<math>2 \sin A \sin B = - \cos (A + B) + \cos (A - B) </math>

:<math>\sin A + \sin B = 2 \sin \left(\frac{(A + B)}{2}\right) \cos\left( \frac{(A - B)}{2} \right)</math>

:<math>\sin A - \sin B = 2 \cos \left(\frac{(A + B)}{2}\right) \sin \left(\frac{(A - B)}{2}\right)</math>

:<math>\cos A + \cos B = 2 \cos \left( \frac{(A + B)}{2} \right) \cos \left( \frac{(A - B)}{2} \right)</math>

:<math>\cos A - \cos B = - 2 \sin \left(\frac{(A + B)}{2} \right) \sin \left(\frac{(A - B)}{2} \right)</math>

:<math>\tan A + \tan B = \tan (A + B) \cdot (1 - \tan A \tan B)</math>

:<math>\tan A - \tan B = \tan (A - B) \cdot (1 + \tan A \tan B)</math>

:<math>\sin A + \sin B + \sin C = 4 \sin \left( \frac{A}{2} \right) \cdot \sin \left(\frac{B}{2}\right) \cdot \sin \left(\frac{C}{2} \right)</math>

:<math>\cos A + \cos B + \cos C = 1 + 4 \sin \left(\frac{A}{2}\right) \cdot \sin \left(\frac{B}{2} \right) \cdot \sin \left(\frac{C}{2} \right)</math>

:<math>\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C</math>

:<math>\sin 2A = 2 \sin A \cos A </math>

:<math>\cos 2A = \cos^2 A - \sin^2 A = 1 - 2 \sin^2 A = 2 \cos^2 A - 1</math>

:<math>\tan 2A = \frac{2 \tan A}{1 - \tan^2 A} = \frac{2 \cot A}{\cot^2 A - 1} = \frac{2}{\cot A - \tan A} </math>

:<math>\sin 3A = 3 \sin A - 4 \sin^3 A </math>

:<math>\cos 3A = 4 \cos^3 A - 3 \cos A </math>

:<math>\tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} </math>

:<math>\sin \left(\frac{A}{2} \right) = \pm \sqrt{\frac{1-\cos A}{2}} </math>

:<math>\cos \left(\frac{A}{2}\right) = \pm \sqrt{\frac{1+\cos A}{2}} </math>

:<math>\tan \left(\frac{A}{2}\right) = \pm \sqrt{\frac{1-\cos A}{1+\cos A}} = \frac {\sin A}{1+\cos A} = \frac {1-\cos A}{\sin A} </math>

:Jika <math>\sin x = \sin \alpha </math>, maka <math>x = \alpha + k \cdot 360^\circ \text{ atau }x = (180^\circ - \alpha) + k \cdot 360^\circ</math> atauserta <math>x = \alpha + 2\pi k \text{ atau }x = (1802\pi - \alpha) + 2\pi k</math>

-:Jika <math>\alpha)cos +x k= \cos \alpha 360</math>, sertamaka <math>x = \pm \alpha + k 2\picdot 360^\circ</math> atauserta <math>x = (2\pi -pm \alpha) + k 2\pi k</math>

:Jika <math>\costan x = \costan \alpha </math>, maka <math>x = \pm \alpha + k 360\cdot 180^\circ</math> serta <math>x = \pm \alpha + k 2\pi k</math>

:Persamaan <math>a \tancos x + b \sin x = c </math> dapat diubah menjadi <math>k \tancos (x - \alpha) = c</math>, maka <math>xk = \alphasqrt{a^2 + k 180b^2}</math> serta, <math>x\tan \alpha = \alphafrac{b}{a}</math> serta <math>a^2 + kb^2 \pige c^2</math>