Bab 16 Fungsi Trigonometri

Posted on June 14, 2016 by user

5.5 Rumus bagi sin (A ± B), kos (A ± B), tan (A ± B), sin 2A, kos 2A, tan 2A

Rumus penambahan

\begin{array}{l} • \sin A = 2 \sin \frac{A}{2} k o s \frac{A}{2} \\ • k o s A = \sin^{2} \frac{A}{2} - k o s^{2} \frac{A}{2} \\ k o s A = 2 k o s^{2} \frac{A}{2} - 1 \\ k o s A = 1 - 2 k o s^{2} \frac{A}{2} \\ • \tan A = \frac{2 \tan \frac{A}{2}}{1 - \tan^{2} \frac{A}{2}} \end{array}

5.5.1 Pembuktian Identiti Trigonometri yang Melibatkan Sudut Majmuk dan Sudut Berganda

Contoh 1:

Buktikan setiap identity trigonometri yang berikut.

\begin{array}{l} (a) \frac{s i n (A + B) - s i n (A - B)}{k o s A k o s B} = 2 \tan B \\ (b) \frac{k o s (A + B)}{\sin A k o s B} = k o t A - \tan B \\ (c) \tan (A + 45^{o}) = \frac{\sin A + k o s A}{k o s A - \sin A} \end{array}

penyelesaian:

(a)

\begin{array}{l} (S e b e l a h K i r i) \\ = \frac{s i n (A + B) - s i n (A - B)}{k o s A k o s B} \\ = \frac{(\sin A k o s B + k o s A \sin B) - (\sin A k o s B - k o s A \sin B)}{k o s A k o s B} \\ = \frac{2 k o s A \sin B}{k o s A k o s B} \\ = \frac{2 \sin B}{k o s B} \\ = 2 \tan B = (S e b e l a h K a n a n) \end{array}

(b)

\begin{array}{l} (S e b e l a h K i r i) \\ = \frac{k o s (A + B)}{\sin A k o s B} \\ = \frac{k o s A k o s B - \sin A \sin B}{\sin A k o s B} \\ = \frac{k o s A k o s B}{\sin A k o s B} - \frac{\sin A \sin B}{\sin A k o s B} \\ = \frac{k o s A}{\sin A} - \frac{\sin B}{k o s B} \\ = k o t A - \tan B \\ = (S e b e l a h K a n a n) \end{array}

(c)

\begin{array}{l} (S e b e l a h K i r i) \\ = \tan (A + 45^{o}) \\ = \frac{t a n A + \tan 45^{o}}{1 - t a n A \tan 45^{o}} \\ = \frac{t a n A + 1}{1 - t a n A} \leftarrow \tan 45^{o} = 1 \\ = \frac{\frac{\sin A}{k o s A} + 1}{1 - \frac{\sin A}{k o s A}} \\ = \frac{\sin A + k o s A}{k o s A} \times \frac{k o s A}{k o s A - \sin A} \\ = \frac{\sin A + k o s A}{k o s A - \sin A} \\ = (S e b e l a h K a n a n) \end{array}