PAS MATEMATIKA BLOG: Contoh Soal 2 (Segitiga dan Trigonometri)

$\begin{array}{ll} 6. & Bentuk \frac{\cos 3 x - \sin 6 x - \cos 9 x}{\sin 9 x - \cos 6 x - \sin 3 x} \\ senilai dengan . . . . \\ \begin{array}{lllllll} a . & - \tan 6 x & d . & 6 \cot x \\ b . & - \cot 6 x & e . & \tan 6 x \\ c . & 6 \tan x \end{array} \\ Jawab : \\ \begin{aligned} \frac{\cos 3 x - \sin 6 x - \cos 9 x}{\sin 9 x - \cos 6 x - \sin 3 x} \\ = \frac{\cos 3 x - \cos 9 x - \sin 6 x}{\sin 9 x - \sin 3 x - \cos 6 x} \\ = \frac{- 2 \sin 6 x \sin (- 3 x) - \sin 6 x}{2 \cos 6 x \sin 3 x - \cos 6 x} \\ = \frac{2 \sin 6 x \sin 3 x - \sin 6 x}{2 \cos 6 x \sin 3 x - \cos 6 x} \\ = \frac{\sin 6 x (2 \sin 3 x - 1)}{\cos 6 x (2 \sin 3 x - 1)} \\ = \tan 6 x \end{aligned} \end{array}$ .

$\begin{array}{ll} 7. & Nilai dari \\ \frac{\sin x + \sin 3 x + \sin 5 x + \sin 7 x}{\cos x + \cos 3 x + \cos 5 x + \cos 7 x} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \tan 2 x & d . & \tan 8 x \\ b . & \tan 4 x & e . & \tan 16 x \\ c . & \tan 6 x \end{array} \\ Jawab : \\ \begin{aligned} \frac{\sin x + \sin 3 x + \sin 5 x + \sin 7 x}{\cos x + \cos 3 x + \cos 5 x + \cos 7 x} \\ = \frac{\sin 7 x + \sin x + \sin 5 x + \sin 3 x}{\cos 7 x + \cos x + \cos 5 x + \cos 3 x} \\ = \frac{2 \sin 4 x \cos 3 x + 2 \sin 4 x \cos x}{2 \cos 4 x \cos 3 x + 2 \cos 4 x \cos x} \\ = \frac{2 \sin 4 x (\cos 3 x + \cos x)}{2 \cos 4 x (\cos 3 x + \cos x)} \\ = \tan 4 x \end{aligned} \end{array}$ .

$\begin{array}{ll} 8. & Nilai dari \cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ} = . . . . \\ \begin{array}{lll} a . 1 & d . - \frac{1}{2} \\ b . - 1 & c . \frac{1}{2} & e . 0 \end{array} \\ Jawab : e \\ \begin{aligned} \cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ} \\ = 2 \cos \frac{1}{2} (80^{\circ} + 40^{\circ}) \cos \frac{1}{2} (80^{\circ} - 40^{\circ}) - \cos 20^{\circ} \\ = 2 \cos 60^{\circ} \cos 20^{\circ} - \cos 20^{\circ} \\ = 2. \frac{1}{2} . \cos 20^{\circ} - \cos 20^{\circ} \\ = \cos 20^{\circ} - \cos 20^{\circ} \\ = 0 \end{aligned} \end{array}$ .

$\begin{array}{ll} 9. & Nilai dari \\ \sqrt{3} \sin 80^{\circ} \sin 160^{\circ} \sin 320^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{3}{8} & d . & \frac{3}{8} \\ b . & - \frac{1}{8} & e . & \frac{5}{8} \\ c . & \frac{1}{8} \end{array} \\ Jawab : \\ \begin{aligned} \sqrt{3} \sin 80^{\circ} \sin 160^{\circ} \sin 320^{\circ} \\ = \sqrt{3} \sin 80^{\circ} \sin 20^{\circ} (- \sin 40^{\circ}) \\ = - \sqrt{3} \sin 80^{\circ} \sin 40^{\circ} \sin 20^{\circ} \\ = - \sqrt{3} \sin 80^{\circ} (- \frac{1}{2} (\cos 60^{\circ} - \cos 20^{\circ})) \\ = - \sqrt{3} \sin 80^{\circ} (- \frac{1}{4} + \frac{\cos 20^{\circ}}{2}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{2} \sqrt{3} \sin 80^{\circ} \cos 20^{\circ} \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} (\sin 100^{\circ} + \sin 60^{\circ}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} (\sin 80^{\circ} + \frac{1}{2} \sqrt{3}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} \sin 80^{\circ} + \frac{1}{8} \sqrt{9} \\ = \frac{3}{8} \end{aligned} \end{array}$ .

$\begin{array}{ll} 10. & Nilai dari \\ \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{8} & d . & \frac{1}{2} \\ b . & - \frac{1}{4} & c . & 0 & e . & \frac{1}{3} \end{array} \\ Jawab : \\ Alternatif 1 \\ \begin{aligned} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \times \frac{2 \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = (\sin \frac{4 π}{7} - \sin 0) \frac{\cos \frac{π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{4 π}{7} \cos \frac{π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{(\sin \frac{5 π}{7} + \sin \frac{3 π}{7}) \cos \frac{4 π}{7}}{4 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{5 π}{7} \cos \frac{4 π}{7} + \sin \frac{3 π}{7} \cos \frac{4 π}{7}}{4 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{9 π}{7} + \sin \frac{π}{7} + \sin \frac{7 π}{7} + \sin (- \frac{π}{7})}{8 \sin \frac{2 π}{7}} \\ = \frac{- \sin \frac{2 π}{7} + \sin \frac{π}{7} + 0 - \sin \frac{π}{7}}{8 \sin \frac{2 π}{7}} \\ = \frac{- \sin \frac{2 π}{7}}{8 \sin \frac{2 π}{7}} \\ = - \frac{1}{8} \end{aligned} \\ Alternatif 2 \\ \begin{aligned} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \\ = \cos \frac{4 π}{7} \cos \frac{2 π}{7} \cos \frac{π}{7} \\ = \frac{1}{2} (\cos \frac{6 π}{7} + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (\cos (π - \frac{π}{7}) + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (- \cos \frac{π}{7} + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (- \cos^{2} \frac{π}{7} + \cos \frac{2 π}{7} \cos \frac{π}{7}) \\ = \frac{1}{4} (- \cos \frac{2 π}{7} - \cos 0 + \cos \frac{3 π}{7} + \cos \frac{π}{7}) \\ = \frac{1}{4} (- \cos 0 + \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7}) \\ = \frac{1}{4} (- 1 + \frac{1}{2}) \\ = \frac{1}{4} \times (- \frac{1}{2}) \\ = - \frac{1}{8} \end{aligned} \end{array}$ .
Berikut penjelasan untuk $\cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} = \frac{1}{2}$ .
$\begin{aligned} \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} \\ = \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} \times \frac{(2 \sin \frac{2 π}{7})}{(2 \sin \frac{2 π}{7})} \\ = \frac{2 \cos \frac{π}{7} \sin \frac{2 π}{7} - 2 \cos \frac{2 π}{7} \sin \frac{2 π}{7} + 2 \cos \frac{3 π}{7} \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} - \sin (- \frac{π}{7}) - (\sin \frac{4 π}{7} - \sin \frac{0 π}{7}) + \sin \frac{5 π}{7} - \sin \frac{π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} + \sin \frac{π}{7} - \sin \frac{4 π}{7} + \sin \frac{5 π}{7} - \sin \frac{π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} - \sin \frac{4 π}{7} + \sin \frac{5 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin (π - \frac{4 π}{7}) - \sin \frac{4 π}{7} + \sin (π - \frac{2 π}{7})}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{4 π}{7} - \sin \frac{4 π}{7} + \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{1}{2} ◼ \end{aligned}$ .

PAS MATEMATIKA BLOG

Contoh Soal 2 (Segitiga dan Trigonometri)

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