6.7.3 Fungsi Trigonometri, SPM Praktis (Kertas 1)

January 24, 2022August 22, 2021 by

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)

(c) kos (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{matrix} k o s A = - \frac{12}{13} \\ \sin B = \frac{3}{5} \end{matrix} \\ = \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

(c) sin 2A

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{matrix} k o s bernilai negatif di \\ sukuan II \end{matrix} \\ \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}

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