6.7.6 Fungsi Trigonometri, SPM Praktis (Kertas 1)

Soalan 15:

Buktikan identiti

\frac{2}{k o s 2 A + 1} = s e k^{2} A

Penyelesaian:

\begin{array}{l} Sebelah kiri \\ = \frac{2}{k o s 2 A + 1} \\ = \frac{2}{(2 k o s^{2} A - 1) + 1} \leftarrow k o s 2 A = 2 k o s^{2} A - 1 \\ = \frac{2}{2 k o s^{2} A} = \frac{1}{k o s^{2} A} \\ = s e k^{2} A \\ = Sebelah kanan \end{array}

Soalan 16:

Buktikan identiti

\frac{2 \tan A}{2 - s e k^{2} A} = \tan 2 A

Penyelesaian:

\begin{array}{l} Sebelah kiri \\ = \frac{2 \tan A}{2 - s e c^{2} A} \\ = \frac{2 \tan A}{2 - (\tan^{2} A + 1)} \leftarrow \tan^{2} A + 1 = s e c^{2} A \\ = \frac{2 \tan A}{1 - \tan^{2} A} \\ = \tan 2 A \\ = Sebelah kanan \end{array}

Soalan 17:

Buktikan identiti tan x + kot x = 2 kosek 2x

Penyelesaian:

Sebelah kiri,

tan x + kot x

\begin{array}{l} = \frac{\sin x}{k o s x} + \frac{k o s x}{\sin x} \\ = \frac{\sin^{2} x + k o s^{2} x}{k o s x \sin x} \\ = \frac{1}{k o s x \sin x} \leftarrow \sin^{2} x + k o s^{2} x = 1 \\ = \frac{1}{\frac{1}{2} \sin 2 x} \leftarrow \begin{matrix} \sin 2 x = 2 \sin x k o s x \\ \frac{1}{2} \sin 2 x = \sin x k o s x \end{matrix} \\ = \frac{2}{\sin 2 x} \\ = 2 (\frac{1}{\sin 2 x}) \\ = 2 k o s e k 2 x \\ = Sebelah kanan \end{array}

Soalan 18:

Buktikan identiti

\frac{k o s x - \sin 2 x}{k o s 2 x + \sin x - 1} = \frac{1}{\tan x}

Penyelesaian:

\begin{array}{l} Sebelah kiri \\ \frac{k o s x - \sin 2 x}{k o s 2 x + \sin x - 1} \\ = \frac{k o s x - 2 \sin x k o s x}{(1 - 2 \sin^{2} x) + \sin x - 1} \leftarrow k o s 2 x = 1 - 2 \sin^{2} x \\ = \frac{k o s x (1 - 2 \sin x)}{\sin x - 2 \sin^{2} x} \\ = \frac{k o s x (1 - 2 \sin x)}{\sin x (1 - 2 \sin x)} \\ = \frac{k o s x}{\sin x} \\ = k o t x \\ = \frac{1}{\tan x} \\ Sebelah kanan \end{array}