Bab 16 Fungsi Trigonometri

Posted on June 14, 2016 by user

5.5.1 Rumus bagi sin (A ± B), kos (A ± B), tan (A ± B), sin 2A, kos 2A, tan 2A
(Contoh Soalan)

Contoh 2:

Buktikan setiap identity trigonometri yang berikut.

$\begin{array}{l} (a) \frac{1 + k o s 2 x}{\sin 2 x} = k o t x \\ (b) k o t A sek 2 A = k o t A + \tan 2 A \\ (c) \frac{s i n x}{1 - k o s x} = k o t \frac{x}{2} \end{array}$

Penyelesaian:

(a)

$\begin{array}{l} S e b e l a h k i r i \\ = \frac{1 + k o s 2 x}{\sin 2 x} \\ = \frac{1 + (2 k o s^{2} x - 1)}{2 \sin x k o s x} \\ = \frac{2 k o s^{2} x}{2 \sin x k o s x} \\ = \frac{k o s x}{\sin x} \\ = k o t x = S e b e l a h k a n a n \end{array}$

(b)

$\begin{array}{l} S e b e l a h k a n a n \\ = k o t A + \tan 2 A \\ = \frac{k o s A}{\sin A} + \frac{\sin 2 A}{k o s 2 A} \\ = \frac{k o s A k o s 2 A + \sin A \sin 2 A}{\sin A k o s 2 A} \\ = \frac{k o s A (k o s^{2} A - \sin^{2} A) + \sin A (2 \sin A k o s A)}{\sin A k o s 2 A} \end{array}$

$\begin{array}{l} = \frac{k o s^{3} A - k o s A \sin^{2} A + 2 \sin^{2} A k o s A}{\sin A k o s 2 A} \\ = \frac{k o s^{3} A + k o s A \sin^{2} A}{\sin A k o s 2 A} \\ = \frac{k o s A (k o s^{2} A + \sin^{2} A)}{\sin A k o s 2 A} \\ = \frac{k o s A}{\sin A k o s 2 A} \leftarrow \sin^{2} A + k o s^{2} A = 1 \\ = (\frac{k o s A}{\sin A}) (\frac{1}{k o s 2 A}) \\ = k o t A sek 2 A \\ = S e b e l a h k i r i \end{array}$

(c)

$\begin{array}{l} S e b e l a h k i r i \\ = \frac{s i n x}{1 - k o s x} \\ = \frac{2 s i n \frac{x}{2} k o s \frac{x}{2}}{1 - (1 - 2 s i n^{2} \frac{x}{2})} \leftarrow \begin{array}{l} \sin x = 2 s i n \frac{x}{2} k o s \frac{x}{2}, \\ k o s x = 1 - 2 \sin^{2} \frac{x}{2} \end{array} \\ = \frac{2 s i n \frac{x}{2} k o s \frac{x}{2}}{2 s i n^{2} \frac{x}{2}} = \frac{k o s \frac{x}{2}}{s i n \frac{x}{2}} \\ = k o t \frac{x}{2} = S e b e l a h k a n a n \end{array}$

Contoh 3:

(a) Diberi bahawa

$\sin P = \frac{3}{5} dan \sin Q = \frac{5}{13},$ dengan keadaan P ialah satu sudut tirus dan Q ialah satu sudut cakah, tanpa menggunakan sifir atau kalkulator, cari nilai kos (P + Q).

(b) Diberi bahawa

$\sin A = - \frac{3}{5} dan \sin B = \frac{12}{13},$ dengan keadaan A dan B adalah sudut-sudut dalam sukuan III dan sukuan IV masing-masing, tanpa menggunakan sifir atau kalkulator, cari nilai sin (A – B).

Penyelesaian:

(a)

$\begin{array}{l} \sin P = \frac{3}{5}, k o s P = \frac{4}{5} \\ \sin Q = \frac{5}{13}, k o s Q = - \frac{12}{13} \\ k o s (P + Q) \\ = k o s A k o s B - \sin A \sin B \\ = (\frac{4}{5}) (- \frac{12}{13}) - (\frac{3}{5}) (\frac{5}{13}) \\ = - \frac{48}{65} - \frac{15}{65} \\ = - \frac{63}{65} \end{array}$

(b)

$\begin{array}{l} \sin A = - \frac{3}{5}, k o s A = - \frac{4}{5} \\ \sin B = - \frac{5}{13}, k o s B = \frac{12}{13} \\ s i n (A - B) \\ = s i n A k o s B - k o s A \sin B \\ = (- \frac{3}{5}) (\frac{12}{13}) - (- \frac{4}{5}) (- \frac{5}{13}) \\ = - \frac{36}{65} - \frac{20}{65} \\ = - \frac{56}{65} \end{array}$