Bab 16 Fungsi Trigonometri

Posted on June 15, 2016 by user

5.7.4 Fungsi Trigonometri, SPM Praktis (Kertas 1)

Soalan 9:

Given that

$\sin θ = \frac{3}{5}$ , dengan keadaan θ ialah sudut tirus, tanpa menggunakan jadual atau kalkulator, cari nilai bagi

(a) sin (180º + θ),

(b) kos (180º – θ),

Penyelesaian:

$\sin θ = \frac{3}{5} k o s θ = \frac{4}{5} \tan θ = \frac{3}{4}$

(a)

sin (180º + θ)

= sin 180º kos θ + kos 180º sin θ

= (0) kos θ + (– 1) sin θ

= – sin θ

$= - \frac{3}{5}$

(b)

kos (180º – θ)

= kos 180º kos θ + sin 180º sin θ

= (– 1) kos θ + (0) sin θ

= – kos θ

$= - \frac{4}{5}$

(c)

$\begin{array}{l} \tan (360^{\circ} + θ) \\ = \frac{\tan 360^{\circ} + \tan θ}{1 - \tan 360^{\circ} \tan θ} \\ = \frac{0 + \tan θ}{1 - (0) (\tan θ)} \\ = \tan θ \\ = \frac{3}{4} \end{array}$

Soalan 10:

Buktikan setiap identiti trigonometri yang berikut.

(a) kot²x – kot² x kos²x = kos² x

$(b) \frac{sek x}{sek x - k o s x} = k o s e k^{2} x$

Penyelesaian:

(a)

Sebelah kiri:

kot²x– kot² x kos² x

= kot²x (1 – kos² x)

= kot²x (sin² x)

$\begin{array}{l} = \frac{k o s^{2} x}{s i n^{2} x} (s i n^{2} x) \\ = k o s^{2} x (Sebelah kanan) \end{array}$

(b)

$\begin{array}{l} Sebelah kiri: \frac{sek x}{sek x - k o s x} \\ = \frac{\frac{1}{k o s x}}{\frac{1}{k o s x} - k o s x} = \frac{\frac{1}{k o s x}}{\frac{1}{k o s x} - \frac{k o s^{2} x}{k o s x}} \\ = \frac{\frac{1}{k o s x}}{\frac{1 - k o s^{2} x}{k o s x}} = \frac{1}{k o s x} \times \frac{k o s x}{1 - k o s^{2} x} \\ = \frac{1}{1 - k o s^{2} x} = \frac{1}{s i n^{2} x} \\ = k o s e k^{2} x (Sebelah kanan) \end{array}$