Matematik Tambahan SPM 2018, Kertas 2 (Soalan 7)

Soalan 7 (10 markah):

$\begin{array}{l} (a) Buktikan \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) = k o s 3 x \\ (b) Seterusnya, \\ (i) selesaikan persamaan s i n (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2} untuk 0 \leq x \leq 2 π \\ dan beri jawapan anda dalam bentuk pencahan termudah dalam sebutan π radian, \\ (i i) lakar graf bagi y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} untuk 0 \leq x \leq π . \end{array}$

Penyelesaian:

\begin{array}{l} (a) \\ Sebelah kiri, \\ \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) \\ = [\sin 3 x k o s \frac{π}{6} + k o s 3 x \sin \frac{π}{6}] - [\sin 3 x k o s \frac{π}{6} - k o s 3 x \sin \frac{π}{6}] \\ = 2 [k o s 3 x \sin \frac{π}{6}] \\ = 2 [k o s 3 x (\frac{1}{2})] \\ = k o s 3 x (sebelah kanan) \end{array}

\begin{array}{l} (b) (i) \\ \sin (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2}, 0 \leq x \leq 2 π \\ k o s \frac{3 x}{2} = \frac{1}{2} \\ \frac{3 x}{2} = \frac{π}{3}, (2 π - \frac{π}{3}), (2 π + \frac{π}{3}) \\ \frac{3 x}{2} = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3} \\ x = \frac{2 π}{9}, \frac{10 π}{9}, \frac{14 π}{9} \end{array}

\begin{array}{l} (b) (i i) \\ y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} untuk 0 \leq x \leq π . \\ y = k o s 3 x - \frac{1}{2} \end{array}

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