Bab 16 Fungsi Trigonometri

Posted on June 15, 2016 by user

5.7.3 Fungsi Trigonometri, SPM Praktis (Kertas 1)

Soalan 7:

Diberi bahawa

$\sin A = \frac{5}{13} dan kos B = \frac{4}{5}$ , dengan keadaan A ialah sudut cakah dan B ialah
sudut tirus. Cari

(a) tan A

(b) sin (A + B)

Penyelesaian:

(a)

$\tan A = - \frac{5}{12}$

(b)

$\begin{array}{l} \sin (A + B) = \sin A k o s B + k o s A \sin B \\ \sin (A + B) \\ = (\frac{5}{13}) (\frac{4}{5}) + (- \frac{12}{13}) (\frac{3}{5}) \leftarrow \begin{array}{l} k o s A = - \frac{12}{13} \\ \sin B = \frac{3}{5} \end{array} \\ = \frac{4}{13} - \frac{36}{65} \\ = - \frac{16}{65} \end{array}$

(c)

$\begin{array}{l} k o s (A - B) = k o s A k o s B + \sin A \sin B \\ k o s (A - B) = (- \frac{12}{13}) (\frac{4}{5}) + (\frac{5}{13}) (\frac{3}{5}) \\ k o s (A - B) = - \frac{33}{65} \end{array}$

Soalan 8:

Jika sin A = p, dan 90° < A < 180°, ungkap dalam sebutan p

(a) tan A

(b) cos A

Penyelesaian:

$\begin{array}{l} Guna Teori Pythagoras, \\ Sisi bersebelahan \\ = \sqrt{1^{2} - p^{2}} \\ = \sqrt{1 - p^{2}} \end{array}$

(a)

$\tan A = - \frac{p}{\sqrt{1 - p^{2}}} \leftarrow \begin{array}{l} tan bernilai negatif \\ di sukuan II \end{array}$

(b)

$\begin{array}{l} k o s A = - \sqrt{1 - p^{2}} \leftarrow \begin{array}{l} k o s bernilai negatif di \\ sukuan II \end{array} \\ \sin A = 2 \sin A k o s A \\ \sin A = 2 (p) (- \sqrt{1 - p^{2}}) \\ \sin A = - 2 p \sqrt{1 - p^{2}} \end{array}$

(c)

$\begin{array}{l} \sin A = 2 \sin A k o s A \\ \sin A = 2 (p) (- \sqrt{1 - p^{2}}) \\ \sin A = - 2 p \sqrt{1 - p^{2}} \end{array}$