PAS MATEMATIKA BLOG: Odd Semester Preparation Questions and Answers

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Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Ketujuh

$\begin{array}{ll} 31. & Nilai dari \\ \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{8} & d . & \frac{1}{2} \\ b . & - \frac{1}{4} & c . & 0 & e . & \frac{1}{3} \end{array} \\ Jawab : \\ Alternatif 1 \\ \begin{aligned} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \times \frac{2 \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = (\sin \frac{4 π}{7} - \sin 0) \frac{\cos \frac{π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{4 π}{7} \cos \frac{π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{(\sin \frac{5 π}{7} + \sin \frac{3 π}{7}) \cos \frac{4 π}{7}}{4 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{5 π}{7} \cos \frac{4 π}{7} + \sin \frac{3 π}{7} \cos \frac{4 π}{7}}{4 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{9 π}{7} + \sin \frac{π}{7} + \sin \frac{7 π}{7} + \sin (- \frac{π}{7})}{8 \sin \frac{2 π}{7}} \\ = \frac{- \sin \frac{2 π}{7} + \sin \frac{π}{7} + 0 - \sin \frac{π}{7}}{8 \sin \frac{2 π}{7}} \\ = \frac{- \sin \frac{2 π}{7}}{8 \sin \frac{2 π}{7}} \\ = - \frac{1}{8} \end{aligned} \\ Alternatif 2 \\ \begin{aligned} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \\ = \cos \frac{4 π}{7} \cos \frac{2 π}{7} \cos \frac{π}{7} \\ = \frac{1}{2} (\cos \frac{6 π}{7} + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (\cos (π - \frac{π}{7}) + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (- \cos \frac{π}{7} + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (- \cos^{2} \frac{π}{7} + \cos \frac{2 π}{7} \cos \frac{π}{7}) \\ = \frac{1}{4} (- \cos \frac{2 π}{7} - \cos 0 + \cos \frac{3 π}{7} + \cos \frac{π}{7}) \\ = \frac{1}{4} (- \cos 0 + \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7}) \\ = \frac{1}{4} (- 1 + \frac{1}{2}) \\ = \frac{1}{4} \times (- \frac{1}{2}) \\ = - \frac{1}{8} \end{aligned} \end{array}$ .
Berikut penjelasan untuk $\cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} = \frac{1}{2}$ .
$\begin{aligned} \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} \\ = \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} \times \frac{(2 \sin \frac{2 π}{7})}{(2 \sin \frac{2 π}{7})} \\ = \frac{2 \cos \frac{π}{7} \sin \frac{2 π}{7} - 2 \cos \frac{2 π}{7} \sin \frac{2 π}{7} + 2 \cos \frac{3 π}{7} \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} - \sin (- \frac{π}{7}) - (\sin \frac{4 π}{7} - \sin \frac{0 π}{7}) + \sin \frac{5 π}{7} - \sin \frac{π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} + \sin \frac{π}{7} - \sin \frac{4 π}{7} + \sin \frac{5 π}{7} - \sin \frac{π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} - \sin \frac{4 π}{7} + \sin \frac{5 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin (π - \frac{4 π}{7}) - \sin \frac{4 π}{7} + \sin (π - \frac{2 π}{7})}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{4 π}{7} - \sin \frac{4 π}{7} + \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{1}{2} ◼ \end{aligned}$ .

$\begin{array}{ll} 32. & Nilai dari \\ \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{9 π}{14} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{16} & d . & \frac{1}{2} \\ b . & \frac{1}{8} & c . & \frac{1}{4} & e . & 1 \end{array} \\ Jawab : \\ \begin{aligned} Perhat & ikan bahwa \\ \sin \frac{π}{14} & = \sin (\frac{7 π}{14} - \frac{6 π}{14}) = \sin (\frac{1}{2} π - \frac{6 π}{14}) \\ = \cos \frac{6 π}{14} \\ \sin \frac{3 π}{14} & = . . . = \cos \frac{4 π}{14} \\ \sin \frac{9 π}{14} & = . . . = \sin \frac{5 π}{14} = \cos \frac{2 π}{14} \end{aligned} \\ . . . \\ \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{9 π}{14} \\ = \cos \frac{6 π}{14} \cos \frac{4 π}{14} \cos \frac{2 π}{14} \times \frac{2 \sin \frac{2 π}{14}}{2 \sin \frac{2 π}{14}} \\ = \frac{\cos \frac{6 π}{14} \cos \frac{4 π}{14} \sin \frac{4 π}{14}}{2 \sin \frac{2 π}{14}} \\ silahkan dilanjutkan \\ . . . \\ = \frac{1}{8} \end{array}$ .

\begin{array}{ll} 33. & Nilai dari \\ \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{16} & d . & \frac{1}{16} \\ b . & \frac{1}{8} & c . & 0 & e . & \frac{1}{8} \end{array} \\ Jawab : \\ \begin{aligned} \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5} \\ = \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos (π + \frac{3 π}{5}) \\ = \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} (- \cos \frac{3 π}{5}) \\ = - \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{3 π}{5} \\ = - \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{3 π}{5} \cos \frac{4 π}{5} \\ = - \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{3 π}{5} \cos \frac{4 π}{5} \times \frac{2 \sin \frac{π}{5}}{2 \sin \frac{π}{5}} \\ = \frac{- \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{3 π}{5} (\sin π - \sin \frac{3 π}{5})}{2 \sin \frac{π}{5}} \\ = \frac{\cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{3 π}{5} \sin \frac{3 π}{5}}{2 \sin \frac{π}{5}} \\ = \frac{\cos \frac{π}{5} \cos \frac{3 π}{5} (\cos \frac{2 π}{5} \sin \frac{3 π}{5})}{2 \sin \frac{π}{5}} \\ = \frac{\cos \frac{π}{5} \cos \frac{3 π}{5} (\sin π - \sin (- \frac{π}{5}))}{4 \sin \frac{π}{5}} \\ = \frac{\cos \frac{π}{5} \cos \frac{3 π}{5} \sin \frac{π}{5}}{4 \sin \frac{π}{5}} \\ = \frac{\cos \frac{3 π}{5} \cos \frac{π}{5} \sin \frac{π}{5}}{4 \sin \frac{π}{5}} \\ = \frac{\cos \frac{3 π}{5} (\cos \frac{π}{5} \sin \frac{π}{5})}{4 \sin \frac{π}{5}} \\ = \frac{\cos \frac{3 π}{5} (\sin \frac{2 π}{5} - \sin 0)}{8 \sin \frac{π}{5}} \\ = \frac{\cos \frac{3 π}{5} \sin \frac{2 π}{5}}{8 \sin \frac{π}{5}} \\ = \frac{\sin π - \sin \frac{π}{5}}{16 \sin \frac{π}{5}} \\ = - \frac{\sin \frac{π}{5}}{16 \sin \frac{π}{5}} \\ = - \frac{1}{16} \end{aligned} \end{array}

\begin{array}{ll} 34. & Nilai dari \\ \sin 18^{\circ} \cos 36^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{6} & d . & \frac{1}{3} \\ b . & \frac{1}{5} & c . & \frac{1}{4} & e . & \frac{1}{2} \end{array} \\ Jawab : \\ \begin{aligned} \sin 18^{\circ} \cos 36^{\circ} \\ = \sin 18^{\circ} \cos 36^{\circ} \times \frac{2 \cos 18^{\circ}}{2 \cos 18^{\circ}} \\ = \frac{\cos 36^{\circ} (\sin 36^{\circ} + \sin 0^{\circ})}{4 \cos 18^{\circ}} \\ = \frac{\cos 36^{\circ} \sin 36^{\circ}}{4 \cos 18^{\circ}} \\ = \frac{\sin 72^{\circ}}{4 \cos 18^{\circ}} \\ = \frac{\sin (90^{\circ} - 18^{\circ})}{4 \cos 18^{\circ}} \\ = \frac{\cos 18^{\circ}}{4 \cos 18^{\circ}} \\ = \frac{1}{4} \end{aligned} \end{array}

\begin{array}{ll} 35. & Nilai eksak dari \sin 36^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{4} \sqrt{10 + 2 \sqrt{5}} & d . & \frac{\sqrt{5} - 1}{4} \\ b . & \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} & e . & \frac{\sqrt{5} - 1}{2} \\ c . & \frac{\sqrt{5} + 1}{4} \end{array} \\ Jawab : \\ Perhatikanlah ilustrasi gambar berikut \end{array}

. \begin{aligned} Perhatikan bahwa △ A B C sama kaki \\ dengan A D = D C = C B = 1, A C = x \\ Diketahui pula C D adalah garis bagi \\ serta A B C sebangun △ B C D \\ akibatnya : \\ perbandingan sisi yang bersesuaian \\ akan sama, maka \\ \frac{A B}{B C} = \frac{B C}{A B - A D} \\ \Leftrightarrow \frac{x}{1} = \frac{1}{x - 1} \\ \Leftrightarrow x (x - 1) = 1 \\ \Leftrightarrow x^{2} - x - 1 = 0 \\ \Leftrightarrow x = \frac{1 \pm \sqrt{5}}{2} \\ akibatnya A B = A C = \frac{1 + \sqrt{5}}{2} \\ Selanjutnya gunakan aturan sinus \\ \frac{A B}{\sin ∠ C} = \frac{B C}{\sin ∠ A} \\ \Leftrightarrow \frac{A B}{B C} = \frac{\sin ∠ C}{\sin ∠ A} \\ \Leftrightarrow \frac{(\frac{1 + \sqrt{5}}{2})}{1} = \frac{\sin 72^{\circ}}{\sin 36^{\circ}} \\ \Leftrightarrow \frac{1 + \sqrt{5}}{2} = \frac{2 \sin 36^{\circ} \cos 36^{\circ}}{\sin 36^{\circ}} \\ \Leftrightarrow \frac{1 + \sqrt{5}}{2} = 2 \cos 36^{\circ} \\ \Leftrightarrow \cos 36^{\circ} =\Leftrightarrow \frac{1 + \sqrt{5}}{4} \\ Dari fakta di atas kita akan dengan \\ mudah menentukan nilai sinusnya \\ yaitu dengan menggunakan \\ identitas trigonometri berikut : \\ \sin^{2} 36^{\circ} + \cos^{2} 36^{\circ} = 1 \\ \Leftrightarrow \sin^{2} 36^{\circ} = 1 - \cos^{2} 36^{\circ} \\ \Leftrightarrow \sin 36^{\circ} = \sqrt{1 - \cos^{2} 36^{\circ}} \\ \Leftrightarrow = \sqrt{1 - {(\frac{1 + \sqrt{5}}{4})}^{2}} \\ \Leftrightarrow = \sqrt{1 - \frac{6 + 2 \sqrt{5}}{16}} \\ \Leftrightarrow = \sqrt{\frac{10 - 2 \sqrt{5}}{16}} \\ \Leftrightarrow = \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} \end{aligned}

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Keenam

$\begin{array}{ll} 26. & Bentuk sederhana dari \\ \frac{\cos 2 x - \cos 2 y}{\sin 2 x + \sin 2 y} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \sin (x - y) & d . & \cos (x - y) \\ b . & - \tan (x - y) & e . & \tan (x - y) \\ c . & \sin (x + y) \end{array} \\ Jawab : \\ \begin{aligned} \frac{\cos 2 x - \cos 2 y}{\sin 2 x + \sin 2 y} \\ = \frac{- 2 \sin (x + y) \sin (x - y)}{2 \sin (x + y) \cos (x - y)} \\ = - \tan (x - y) \end{aligned} \end{array}$ .

$\begin{array}{ll} 27. & Nilai dari \\ 8 \cos 82, 5^{\circ} \sin 37, 5^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & 4 (\sqrt{3} + \sqrt{2}) & d . & 2 (\sqrt{3} - \sqrt{2}) \\ b . & 4 (\sqrt{3} - \sqrt{2}) & e . & \sqrt{3} - \sqrt{2} \\ c . & 2 (\sqrt{3} + \sqrt{2}) \end{array} \\ Jawab : \\ \begin{aligned} 8 \cos 82, 5^{\circ} \sin 37, 5^{\circ} \\ = 4 \times 2 \cos 82, 5^{\circ} \sin 37, 5^{\circ} \\ = 4 \times (\sin (82, 5^{\circ} + 37, 5^{\circ}) - \sin (82, 5^{\circ} - 37, 5^{\circ})) \\ = 4 \times (\sin 120^{\circ} - \sin 45^{\circ}) \\ = 4 \times (\sin (180^{\circ} - 60^{\circ}) - \sin 45^{\circ}) \\ = 4 \times (\sin 60^{\circ} - \sin 45^{\circ}) \\ = 4 \times (\frac{1}{2} \sqrt{3} - \frac{1}{2} \sqrt{2}) \\ = 2 (\sqrt{3} - \sqrt{2}) \end{aligned} \end{array}$ .

$\begin{array}{ll} 28. & Bentuk lain dari \\ - 2 \cos 5 A . \cos 7 A \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \cos 6 A - \cos A \\ b . & - \cos 6 A + \cos A \\ c . & \cos 12 A - \cos 2 A \\ d . & - \cos 12 A + \cos 2 A \\ e . & - \cos 12 A - \cos 2 A \end{array} \\ Jawab : \\ \begin{aligned} - 2 \cos 5 A . \cos 7 A \\ = - (2 \cos 5 A . \cos 7 A) \\ = - (\cos 12 A + \cos (- 2 A)) \\ = - (\cos 12 A + \cos 2 A) \\ = - \cos 12 A - \cos 2 A \end{aligned} \end{array}$ .

$\begin{array}{ll} 29. & Bentuk sederhana dari \\ 4 \sin (\frac{1}{4} π + x) \cos (\frac{1}{4} π - x) \\ adalah . . . . \\ \begin{array}{lllllll} a . & 2 + 2 \sin 2 x & d . & 2 + 2 \sin x \\ b . & 2 + \sin 2 x & e . & 2 + \sin x \\ c . & 2 \sin 2 x \end{array} \\ Jawab : \\ \begin{aligned} 4 \sin (\frac{1}{4} π + x) \cos (\frac{1}{4} π - x) \\ = 2 (\sin (\frac{1}{2} π) + \sin (2 x)) \\ = 2 (1 + \sin 2 x) \\ = 2 + 2 \sin 2 x \end{aligned} \end{array}$ .

$\begin{array}{ll} 30. & Nilai dari \\ \sqrt{3} \sin 80^{\circ} \sin 160^{\circ} \sin 320^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{3}{8} & d . & \frac{3}{8} \\ b . & - \frac{1}{8} & e . & \frac{5}{8} \\ c . & \frac{1}{8} \end{array} \\ Jawab : \\ \begin{aligned} \sqrt{3} \sin 80^{\circ} \sin 160^{\circ} \sin 320^{\circ} \\ = \sqrt{3} \sin 80^{\circ} \sin 20^{\circ} (- \sin 40^{\circ}) \\ = - \sqrt{3} \sin 80^{\circ} \sin 40^{\circ} \sin 20^{\circ} \\ = - \sqrt{3} \sin 80^{\circ} (- \frac{1}{2} (\cos 60^{\circ} - \cos 20^{\circ})) \\ = - \sqrt{3} \sin 80^{\circ} (- \frac{1}{4} + \frac{\cos 20^{\circ}}{2}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{2} \sqrt{3} \sin 80^{\circ} \cos 20^{\circ} \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} (\sin 100^{\circ} + \sin 60^{\circ}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} (\sin 80^{\circ} + \frac{1}{2} \sqrt{3}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} \sin 80^{\circ} + \frac{1}{8} \sqrt{9} \\ = \frac{3}{8} \end{aligned} \end{array}$ .

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Kelima

$\begin{array}{ll} 21. & Nilai \cos \frac{5}{12} π - \cos \frac{1}{12} π adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{2} \sqrt{6} & d . & \frac{1}{2} \sqrt{2} \\ b . & - \frac{1}{2} \sqrt{3} & c . & - \frac{1}{2} \sqrt{2} & e . & \frac{1}{2} \sqrt{6} \end{array} \\ Jawab : \\ \begin{aligned} \cos \frac{5}{12} π - \cos \frac{1}{12} π \\ = - 2 \sin (\frac{\frac{5}{12} π + \frac{1}{12} π}{2}) \sin (\frac{\frac{5}{12} π - \frac{1}{12} π}{2}) \\ = - 2 \sin (\frac{\frac{6}{12} π}{2}) \sin (\frac{\frac{4}{12} π}{2}) \\ = - 2 \sin (\frac{1}{4} π) \sin (\frac{1}{6} π) \\ = - 2 (\frac{1}{2} \sqrt{2}) (\frac{1}{2}) \\ = - \frac{1}{2} \sqrt{2} \end{aligned} \end{array}$ .

$\begin{array}{ll} 22. & Bentuk \sin (2 x - \frac{3}{2} π) - \sin (4 x + \frac{1}{2} π) \\ senilai dengan . . . . \\ \begin{array}{lllllll} a . & - 2 \sin 3 x . \sin x & d . & 2 \sin 3 x . \sin x \\ b . & - 2 \cos 3 x . \sin x & e . & 2 \cos 3 x . \sin x \\ c . & 2 \sin 2 (x - π) \end{array} \\ Jawab : \\ \begin{aligned} \sin (2 x - \frac{3}{2} π) - \sin (4 x + \frac{1}{2} π) \\ = 2 \cos (\frac{2 x - \frac{3}{2} π + 4 x + \frac{1}{2} π}{2}) \\ \times \sin (\frac{2 x - \frac{3}{2} π - (4 x + \frac{1}{2} π)}{2}) \\ = 2 \cos (3 x - \frac{1}{2} π) \sin (- x - π) \\ = 2 \cos (\frac{1}{2} π - 3 x) (- \sin (π + x)) \\ = 2 (\sin 3 x) (- (- \sin x)) \\ = 2 \sin 3 x . \sin x \end{aligned} \end{array}$ .

$\begin{array}{ll} 23. & Bentuk \frac{\cos 3 x - \sin 6 x - \cos 9 x}{\sin 9 x - \cos 6 x - \sin 3 x} \\ senilai dengan . . . . \\ \begin{array}{lllllll} a . & - \tan 6 x & d . & 6 \cot x \\ b . & - \cot 6 x & e . & \tan 6 x \\ c . & 6 \tan x \end{array} \\ Jawab : \\ \begin{aligned} \frac{\cos 3 x - \sin 6 x - \cos 9 x}{\sin 9 x - \cos 6 x - \sin 3 x} \\ = \frac{\cos 3 x - \cos 9 x - \sin 6 x}{\sin 9 x - \sin 3 x - \cos 6 x} \\ = \frac{- 2 \sin 6 x \sin (- 3 x) - \sin 6 x}{2 \cos 6 x \sin 3 x - \cos 6 x} \\ = \frac{2 \sin 6 x \sin 3 x - \sin 6 x}{2 \cos 6 x \sin 3 x - \cos 6 x} \\ = \frac{\sin 6 x (2 \sin 3 x - 1)}{\cos 6 x (2 \sin 3 x - 1)} \\ = \tan 6 x \end{aligned} \end{array}$ .

$\begin{array}{ll} 24. & Nilai dari \\ \frac{\sin x + \sin 3 x + \sin 5 x + \sin 7 x}{\cos x + \cos 3 x + \cos 5 x + \cos 7 x} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \tan 2 x & d . & \tan 8 x \\ b . & \tan 4 x & e . & \tan 16 x \\ c . & \tan 6 x \end{array} \\ Jawab : \\ \begin{aligned} \frac{\sin x + \sin 3 x + \sin 5 x + \sin 7 x}{\cos x + \cos 3 x + \cos 5 x + \cos 7 x} \\ = \frac{\sin 7 x + \sin x + \sin 5 x + \sin 3 x}{\cos 7 x + \cos x + \cos 5 x + \cos 3 x} \\ = \frac{2 \sin 4 x \cos 3 x + 2 \sin 4 x \cos x}{2 \cos 4 x \cos 3 x + 2 \cos 4 x \cos x} \\ = \frac{2 \sin 4 x (\cos 3 x + \cos x)}{2 \cos 4 x (\cos 3 x + \cos x)} \\ = \tan 4 x \end{aligned} \end{array}$ .

$\begin{array}{ll} 25. & Bentuk sederhana dari \\ \frac{\cos A + \sin A}{\cos A - \sin A} - \frac{\cos A - \sin A}{\cos A + \sin A} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \tan A & d . & 2 \cos 2 A \\ b . & 2 \tan A & e . & 2 \tan 2 A \\ c . & 2 \sin 2 A \end{array} \\ Jawab : \\ \begin{aligned} \frac{\cos A + \sin A}{\cos A - \sin A} - \frac{\cos A - \sin A}{\cos A + \sin A} \\ = \frac{(\cos A + \sin A)^{2} - (\cos A - \sin A)^{2}}{(\cos A - \sin A) (\cos A + \sin A)} \\ = \frac{(\cos^{2} A + 2 \cos A \sin A + \sin^{2} A) - (\cos^{2} A - 2 \cos A \sin A + \sin^{2} A)}{\cos^{2} A - \sin^{2} A} \\ = \frac{4 \cos A \sin A}{\cos^{2} A - \sin^{2} A} \\ = \frac{2 \sin 2 A}{\cos 2 A} \\ = 2 \tan 2 A \end{aligned} \end{array}$ .

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Keempat

$\begin{array}{ll} 16. & Himpunan penyelesaian dari persamaan \\ 3 \tan (2 x - \frac{1}{3} π) = - \sqrt{3} \\ untuk 0 \leq x \leq π adalah . . . . \\ a . {\frac{1}{12} π, \frac{7}{12} π} \\ b . {\frac{2}{12} π, \frac{9}{12} π} \\ c . {\frac{3}{12} π, \frac{7}{12} π} \\ d . {\frac{3}{12} π, \frac{9}{12} π} \\ e . {\frac{5}{12} π, \frac{7}{12} π} \\ Jawab : \\ \begin{array}{ll} \begin{aligned} 3 \tan (2 x - \frac{1}{3} π) = - \sqrt{3} \\ \tan (2 x - \frac{1}{3} π) = - \frac{\sqrt{3}}{3} \\ (kuadran IV, karena Y negatif, X positif) \\ \tan (2 x - \frac{1}{3} π) = - \tan \frac{1}{6} π, menjadi \\ \tan (2 x - \frac{1}{3} π) = \tan (2 π - \frac{1}{6} π) = \tan \frac{11}{6} π \\ (2 x - \frac{1}{3} π) = \frac{11}{6} π \\ \Leftrightarrow 2 x = \frac{1}{3} π + \frac{11}{6} π + k . π = \frac{13}{6} π + k . π \\ \Leftrightarrow x = \frac{13}{12} π + \frac{k . π}{2} \\ k = 0 \Rightarrow x = \frac{13}{12} π = \frac{1}{12} π (mm) \\ k = 1 \Rightarrow x = \frac{13}{12} π + \frac{1}{2} π = \frac{19}{12} π = \frac{7}{12} π (mm) \\ k = 2 \Rightarrow x = \frac{13}{12} π + π tidak memenuhi \end{aligned} \\ HP = {\frac{1}{12} π, \frac{7}{12} π} \end{array} \end{array}$ .

$\begin{array}{ll} 17. & Salah satu nilai x yang memenuhi \\ persamaan \cos x + \sin x = \frac{1}{2} \sqrt{6} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{24} π & d . & \frac{1}{8} π \\ b . & \frac{1}{15} π & c . & \frac{1}{12} π & e . & \frac{1}{6} π \end{array} \\ Jawab : \\ \begin{array}{ll} Diketahui bahwa \\ \sin x + \cos x = \frac{1}{2} \sqrt{6} (ingat : a = 1, b = 1) \\ \sin x + \cos x = k \cos (x - θ) = \frac{1}{2} \sqrt{6} \\ {\begin{cases} k & = \sqrt{1^{2} + 1^{2}} = \sqrt{2} \\ \tan & θ = \frac{a}{b} = \frac{1}{1} = 1 \Rightarrow θ = 45^{\circ} = \frac{1}{4} π \end{cases} \\ sudut θ di kuadran I, karena a, b > 0 \\ selanjutnya \\ \begin{aligned} \sin x + \cos x = k \cos (x - θ) = \frac{1}{2} \sqrt{6} \\ \Leftrightarrow \sqrt{2} \cos (x - \frac{1}{4} π) = \frac{1}{2} \sqrt{6} \\ \Leftrightarrow \cos (x - \frac{1}{4} π) = \frac{\frac{1}{2} \sqrt{6}}{\sqrt{2}} = \frac{1}{2} \sqrt{3} \\ \Leftrightarrow \cos (x - \frac{1}{4} π) = \cos \frac{1}{6} π \\ \Leftrightarrow x - \frac{1}{4} π = \pm \frac{1}{6} π + k .2 π \\ \Leftrightarrow x = \frac{1}{4} π \pm \frac{1}{6} π + k .2 π \\ \Leftrightarrow x_{1} = \frac{5}{12} π + k .2 π atau \\ x_{2} = \frac{1}{12} π + k .2 π \\ k = 0 \Rightarrow x_{1} = \frac{5}{12} π (memenuhi) \\ \Rightarrow x_{2} = \frac{1}{12} π (memenuhi) \\ Langkah berikutnya tidak diperlukan \\ karena jawaban sudah kita dapatkan \\ yaitu : \frac{1}{12} π \end{aligned} \end{array} \end{array}$ .

$\begin{array}{ll} 18. & Himpunan penyelesaian persamaan \\ \cos x^{\circ} - \sqrt{3} \sin x^{\circ} = 1 \\ untuk 0 \leq x < 360 adalah . . . . \\ \begin{array}{lllllll} a . & {0, 240} & d . & {180, 240} \\ b . & {150, 270} & c . & {180, 300} & e . & {210, 270} \end{array} \\ Jawab : \\ \begin{array}{ll} Diketahui dari soal bahwa \\ \cos x^{\circ} - \sqrt{3} \sin x^{\circ} = 1, \\ lalu kita ubah posisinya menjadi \\ - \sqrt{3} \sin x + \cos x = 1 (ingat : a = - \sqrt{3}, b = 1) \\ - \sqrt{3} \sin x + \cos x = k \cos (x - θ) = 1 \\ {\begin{cases} k & = \sqrt{{(- \sqrt{3})}^{2} + {(1)}^{2}} = \sqrt{4} = 2 \\ \tan & θ = \frac{a}{b} = \frac{- \sqrt{3}}{1} = - \sqrt{3} \Rightarrow θ = 300^{\circ} \end{cases} \\ sudut θ di kuadran IV, karena a < 0, b > 0 \\ selanjutnya \\ \begin{aligned} - \sqrt{3} \sin x + \cos x = k \cos (x - θ) = 1 \\ \Leftrightarrow 2 \cos (x - 300^{\circ}) = 1 \\ \Leftrightarrow \cos (x - 300^{\circ}) = \frac{1}{2} \\ \Leftrightarrow \cos (x - 300^{\circ}) = \cos 60^{\circ} \\ \Leftrightarrow x - 300^{\circ} = \pm 60^{\circ} + k {.360}^{\circ} \\ \Leftrightarrow x = 300^{\circ} \pm 60^{\circ} + k {.360}^{\circ} \\ k = 0 \Rightarrow x_{1} = 300^{\circ} + 60^{\circ} = 360^{\circ} = 0^{\circ} (mm) \\ atau \\ x_{2} = 300^{\circ} - 60^{\circ} = 240^{\circ} (mm) \\ k = 1 \Rightarrow x = 300^{\circ} \pm 60^{\circ} + 360^{\circ} (tm) \end{aligned} \\ HP = {0^{\circ}, 240^{\circ}} \end{array} \end{array}$

$\begin{array}{ll} 19. & Diketahui α - β = \frac{π}{3} dan \sin α \sin β = \frac{1}{4} \\ dengan α dan β adalah sudut lancip \\ Nilai dari \cos (α + β) adalah . . . . \\ \begin{array}{lllllll} a . & 1 & d . & \frac{1}{4} \\ b . & \frac{3}{4} & c . & \frac{1}{2} & e . & 0 \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa \\ ∙ α - β = \frac{π}{3} dan \sin α \sin β = \frac{1}{4} \\ ∙ dengan α dan β sudut lancip \\ akibatnya semua sudut dikuadran I \\ sehingga {\begin{cases} \sin & = + \\ \cos & = + \\ \tan & = + \end{cases} \\ ditanya \cos (α + β), maka \\ sebagai langkah awal kita adalah : \\ \cos (α - β) = \cos (\frac{π}{3}) \\ \Leftrightarrow \cos α \cos β + \sin α \sin β = \frac{1}{2} \\ \Leftrightarrow \cos α \cos β + \frac{1}{4} = \frac{1}{2} \\ \Leftrightarrow \cos α \cos β = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \\ Selanjutnya nilai dari \\ \cos (α + β) \\ = \cos α \cos β - \sin α \sin β \\ = \frac{1}{4} - \frac{1}{4} = 0 \end{aligned} \end{array}$ .

$\begin{array}{ll} 20. & Nilai \sin 75^{\circ} - \sin 165^{\circ} adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{4} \sqrt{2} & d . & \frac{1}{2} \sqrt{2} \\ b . & \frac{1}{4} \sqrt{3} & c . & \frac{1}{4} \sqrt{6} & e . & \frac{1}{2} \sqrt{6} \end{array} \\ Jawab : \\ \begin{aligned} \sin 75^{\circ} - \sin 165^{\circ} \\ = 2 \cos (\frac{75^{\circ} + 165^{\circ}}{2}) \sin (\frac{75^{\circ} - 165^{\circ}}{2}) \\ = 2 \cos \frac{240^{\circ}}{2} \sin (- \frac{90^{\circ}}{2}) \\ = 2 \cos 120^{\circ} \sin (- 45^{\circ}) \\ = 2 (- \cos 60^{\circ}) (- \sin 45^{\circ}) \\ = 2 (- \frac{1}{2}) (- \frac{1}{2} \sqrt{2}) \\ = \frac{1}{2} \sqrt{2} \end{aligned} \end{array}$ .

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Ketiga

$\begin{array}{ll} 11. & Grafik fungsi trigonometri pada gambar \\ berikut adalah . . . . \end{array}$ .

. \begin{array}{ll} a . y = 2 \cos x \\ b . y = \cos 2 x \\ c . y = \cos \frac{1}{2} x \\ d . y = 2 \cos 2 x \\ e . y = 2 \cos \frac{1}{2} x \\ Jawab : \\ \begin{aligned} Dari grafik tampak jelas bahwa \\ gambar di atas adalah garfik \\ fungsi cosinus di geser ke \\ atas dan ke bawah \\ dengan amplitudo 2 \\ dan periodenya \frac{360^{\circ}}{2} = 180^{\circ} = π, \\ maka \\ bentuk persamaan grafik fungsinya \\ y = 2 \cos 2 x \end{aligned} \end{array}

$\begin{array}{ll} 12. & Grafik fungsi trigonometri pada gambar \\ berikut adalah . . . . \end{array}$ .

. \begin{array}{ll} a . y = 2 \sin (2 x + π) \\ b . y = \sin (2 x - \frac{1}{2} π) \\ c . y = 2 \sin (2 x - \frac{1}{2} π) \\ d . y = \sin (2 x + \frac{1}{2} π) \\ e . y = 2 \sin (x + \frac{1}{2} π) \\ Jawab : \\ \begin{aligned} Dari grafik tampak jelas bahwa \\ gambar di atas adalah garfik \\ fungsi cosinus di geser ke kiri \\ dengan amplitudo 2 \\ dan periodenya \frac{360^{\circ}}{1} = 360^{\circ} = 2 π, \\ maka \\ bentuk persamaan grafik fungsinya \\ y = 2 \sin (x + k x) \\ dengan + k adalah \\ besar geseran ke kiri \frac{1}{2} π atau 90^{\circ} \\ Jadi, y = 2 \sin (x + \frac{1}{2} π)^{\circ} \end{aligned} \end{array}

\begin{array}{ll} 13. & Himpunan penyelesaian dari persamaan \\ \sin x = \sin \frac{2}{10} π untuk 0^{\circ} \leq x \leq 360^{\circ} \\ adalah . . . . \\ a . {\frac{22}{10} π, \frac{8}{10} π} \\ b . {\frac{2}{10} π, \frac{28}{10} π} \\ c . {\frac{2}{10} π, \frac{8}{10} π} \\ d . {\frac{22}{10} π, \frac{28}{10} π} \\ e . {\frac{12}{10} π, \frac{8}{10} π} \\ Jawab : \\ \begin{array}{ll} \begin{aligned} \sin x = \sin \frac{2}{10} π \\ \Leftrightarrow x_{1} = \frac{2}{10} π + k .2 π atau \\ \Leftrightarrow x_{2} = (π - \frac{2}{10} π) + k .2 π = \frac{8}{10} π + k .2 π \\ k = 0 \Rightarrow x_{1} = \frac{2}{10} π (mm) atau \\ x_{2} = \frac{8}{10} π (mm) \\ k = 1 \Rightarrow x_{1, 2} = . . . . + 2 π (tidak memenuhi) \end{aligned} \\ HP = {\frac{2}{10} π, \frac{8}{10} π} \end{array} \end{array}

\begin{array}{ll} 14. & Himpunan penyelesaian dari persamaan \\ \tan (2 x - \frac{1}{4} π) = \tan \frac{1}{4} π untuk 0^{\circ} \leq x \leq 360^{\circ} \\ adalah . . . . \\ a . {\frac{1}{3} π, π, \frac{5}{3} π, \frac{7}{3} π} \\ b . {\frac{1}{4} π, \frac{3}{5} π, \frac{5}{4} π, \frac{8}{5} π} \\ c . {\frac{1}{4} π, \frac{3}{4} π, \frac{6}{4} π, \frac{7}{4} π} \\ d . {\frac{2}{4} π, \frac{3}{4} π, π, \frac{7}{4} π} \\ e . {\frac{1}{4} π, \frac{3}{4} π, \frac{5}{4} π, \frac{7}{4} π} \\ Jawab : \\ \begin{array}{ll} \begin{aligned} \tan (2 x - \frac{1}{4} π) = \tan \frac{1}{4} π \\ \Leftrightarrow 2 x - \frac{1}{4} π = \frac{1}{4} π + k . π \\ \Leftrightarrow 2 x = \frac{2}{4} π + k . π \\ \Leftrightarrow x = \frac{1}{4} π + k . \frac{π}{2} \\ k = 0 \Rightarrow x = \frac{1}{4} π (mm) \\ k = 1 \Rightarrow x = \frac{1}{4} π + \frac{π}{2} = \frac{3}{4} π (mm) \\ k = 2 \Rightarrow x = \frac{1}{4} π + π = \frac{5}{4} π (mm) \\ k = 3 \Rightarrow x = \frac{1}{4} π + \frac{3 π}{2} = \frac{7}{4} π (mm) \\ k = 4 \Rightarrow x = \frac{1}{4} π + 2 π = \frac{9}{4} π (tidak memenuhi) \end{aligned} \\ HP = {\frac{1}{4} π, \frac{3}{4} π, \frac{5}{4} π, \frac{7}{4} π} \end{array} \end{array}

\begin{array}{ll} 15. & Himpunan penyelesaian dari persamaan \\ \cos 2 x - 2 \cos x = - 1 untuk 0 < x < 2 π \\ adalah . . . . \\ a . {0, \frac{1}{2} π, \frac{3}{2} π, 2 π} \\ b . {0, \frac{1}{2} π, \frac{2}{3} π, 2 π} \\ c . {0, \frac{1}{2} π, π, \frac{3}{2} π} \\ d . {0, \frac{1}{2} π, \frac{2}{3} π} \\ e . {0, \frac{1}{2} π, π} \\ Jawab : \\ \begin{array}{ll} \begin{aligned} \cos 2 x - 2 \cos x = - 1 \\ \Leftrightarrow \cos 2 x - 2 \cos x + 1 = 0 \\ \Leftrightarrow (2 \cos^{2} x - 1) - 2 \cos x + 1 = 0 \\ \Leftrightarrow 2 \cos x (\cos x - 1) = 0 \\ \Leftrightarrow \cos x = 0 atau \cos x = 1 \\ \Leftrightarrow \cos x = \cos \frac{1}{2} π atau \cos x = \cos 0 \\ \Leftrightarrow x_{1, 2} = \pm \frac{1}{2} π + k .2 π atau x_{3} = k .2 π \\ maka \\ k = 0 \Rightarrow x_{1} = - \frac{1}{2} π (tm) atau \\ x_{2} = \frac{1}{2} π (mm) \\ x_{3} = 0 (mm) \\ k = 1 \Rightarrow x_{1} = \frac{3}{2} π (mm) \\ x_{2} = \frac{5}{2} π (tm) \\ x_{3} = 2 π (mm) \end{aligned} \end{array} \end{array}

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Kedua

$\begin{array}{ll} 6. & Nilai dari \frac{\sin 49^{\circ}}{\cos 41^{\circ}} - \frac{\cos 17^{\circ}}{\sin 73^{\circ}} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - 1 & d . & 0, 143 \\ b . & - 0, 321 & c . & 0 & e . & 0, 321 \end{array} \\ Jawab : \\ \begin{aligned} \frac{\sin 49^{\circ}}{\cos 41^{\circ}} - \frac{\cos 17^{\circ}}{\sin 73^{\circ}} \\ = \frac{\sin 49^{\circ}}{\cos (90^{\circ} - 49^{\circ})} - \frac{\cos 17^{\circ}}{\sin (90^{\circ} - 17^{\circ})} \\ = \frac{\sin 49^{\circ}}{\sin 49^{\circ}} - \frac{\cos 17^{\circ}}{\cos 17^{\circ}} \\ = 1 - 1 \\ = 0 \end{aligned} \end{array}$ .

$\begin{array}{ll} 7. & Nilai dari \\ p = r \sin α \cos β \\ q = r \sin α \sin β \\ s = r \cos α \\ maka pernyataan berikut yang \\ tepat adalah . . . . \\ \begin{array}{lllllll} a . & p^{2} + t^{2} + s^{2} = r^{2} \\ b . & p^{2} - t^{2} + s^{2} = r^{2} \\ c . & p^{2} + t^{2} - s^{2} = r^{2} \\ d . & - p^{2} + t^{2} + s^{2} = r^{2} \\ e . & - p^{2} - t^{2} + s^{2} = r^{2} \end{array} \\ Jawab : \\ \begin{aligned} Saat \\ p^{2} + q^{2} maka hasilnya adalah \\ \begin{array}{lll} p^{2} & = r^{2} \sin^{2} α \cos^{2} β \\ q^{2} & = r^{2} \sin^{2} α \sin^{2} β & + \\ = r^{2} \sin^{2} α (\cos^{2} β + \sin^{2} β) \\ = r^{2} \sin^{2} α (1) \\ = r^{2} \sin^{2} α \end{array} \\ Dan saat \\ p^{2} + q^{2} + s^{2} akan diperoleh hasil \\ \begin{array}{lll} p^{2} + q^{2} & = r^{2} \sin^{2} α \\ s^{2} & = r^{2} \cos^{2} α & + \\ = r^{2} \sin^{2} α + r^{2} \cos^{2} α \\ = r^{2} (\sin^{2} α + \cos^{2} α) \\ = r^{2} (1) \\ = r^{2} \end{array} \end{aligned} \end{array}$ .

$\begin{array}{ll} 8. & Nilai dari \\ \frac{\cos (90^{\circ} + θ) \sec (2 π - θ) \tan (π - θ)}{\sec (θ - 2 π) \sin (540^{\circ} + θ) \cot (θ - 90^{\circ})} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - 1 & d . & - \tan θ \\ b . & 0 & c . & 1 & e . & \tan θ \end{array} \\ Jawab : \\ \begin{aligned} Ingat kembali sudut-sudut \\ yang berelasi dari kudran selain I \\ ke kuadran I beserta tandanya \\ \frac{\cos (90^{\circ} + θ) \sec (2 π - θ) \tan (π - θ)}{\sec (θ - 2 π) \sin (540^{\circ} + θ) \cot (θ - 90^{\circ})} \\ = \frac{(- \sin θ) . \sec θ . (- \tan θ)}{\sec θ . (- \sin θ) . (- \tan θ)} \\ = 1 \end{aligned} \end{array}$ .

$\begin{array}{ll} 9. & Diketahui bahwa \\ \sin θ + \cos θ = \frac{1}{2} \\ maka nilai dari \\ \sin^{3} θ + \cos^{3} θ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{2} & d . & \frac{5}{8} \\ b . & \frac{3}{4} & c . & \frac{9}{15} & e . & \frac{11}{16} \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa \\ \sin θ + \cos θ = \frac{1}{2} \\ \Leftrightarrow {(\sin θ + \cos θ)}^{2} = \frac{1}{4} \\ \Leftrightarrow \sin^{2} θ + \cos^{2} θ + 2 \sin θ \cos θ = \frac{1}{4} \\ \Leftrightarrow 1 + 2 \sin θ \cos θ = \frac{1}{4} \\ \Leftrightarrow 2 \sin θ \cos θ = - \frac{3}{4} \\ \Leftrightarrow \sin θ \cos θ = - \frac{3}{8} \\ Selanjutnya \\ \sin^{3} θ + \cos^{3} θ \\ = (\sin θ + \cos θ) (\sin^{2} θ - \sin θ \cos θ + \cos^{2} θ) \\ = (\frac{1}{2}) (1 - (- \frac{3}{8})) \\ = \frac{1}{2} \times \frac{11}{8} \\ = \frac{11}{16} \end{aligned} \end{array}$ .

$\begin{array}{ll} 10. & Jika diketahui \frac{3}{2} π < x < 2 π \\ dan \tan x = m, \\ maka nilai dari \sin x \cos x adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{m^{2} + 1} & d . & - \frac{m}{m^{2} - 1} \\ b . & - \frac{m}{m^{2} + 1} & c . & \frac{m}{m^{2} + 1} & e . & \frac{m}{m^{2} - 1} \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa \frac{3}{2} π < x < 2 π \\ ini daerah Kwadran IV, akibatnya adalah nilai \\ {\begin{cases} \sin x & = - \\ \cos x & = + \\ \tan x & = - \end{cases} \\ Selanjutnya ada pernyataan \tan x = m \\ ini artinya \tan x = \frac{m}{1} \\ Perhatikanlah ilustrasi gambar berikut \end{aligned} \end{array}$ .

. \begin{aligned} maka nilai dari \\ \sin x \cos x (ingat yang diminta di Kwadran IV) \\ = (- \frac{m}{\sqrt{m^{2} + 1}}) \times (+ \frac{1}{\sqrt{m^{2} + 1}}) \\ = - \frac{m}{m^{2} + 1} \end{aligned}

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Pertama

$\begin{array}{ll} 1. & Nilai 105^{\circ} jika dinyatakan ke radian \\ adalah . . . . radian \\ a . \frac{1}{3} π \\ b . \frac{5}{6} π \\ c . \frac{5}{12} π \\ d . \frac{7}{12} π \\ e . \frac{9}{12} π \\ Jawab : \\ \begin{aligned} Diketah & ui bahwa \\ 180^{\circ} & = π r a d i a n \\ 1^{\circ} & = \frac{π}{180} r a d i a n \\ 105 \times 1^{\circ} & = 105 \times \frac{π}{180} r a d i a n \\ 105^{\circ} & = \frac{7}{12} π r a d i a n \end{aligned} \end{array}$ .

$\begin{array}{ll} 2. & Nilai \tan 240^{\circ} adalah . . . . \\ a . \sqrt{3} \\ b . \frac{1}{3} \sqrt{3} \\ c . - \frac{1}{3} \sqrt{3} \\ d . \frac{1}{2} \sqrt{3} \\ e . - \sqrt{3} \\ Jawab : \\ \begin{aligned} \tan 240^{\circ} & = \tan (180^{\circ} + 60^{\circ}) \\ = \tan 60^{\circ} \\ = \sqrt{3} \\ catatan & : ingat sudut berelasi \end{aligned} \end{array}$ .

$\begin{array}{ll} 3. & Perhatikanlah gambar berikut \end{array}$ .

. \begin{array}{ll} Pada gambar di atas perbandingan \\ \sin θ adalah . . . . \\ a . \sqrt{\frac{a^{2} - d^{2}}{f^{2} + g^{2}}} \\ b . \sqrt{\frac{a^{2} + b^{2}}{f^{2} + g^{2}}} \\ c . \sqrt{\frac{a^{2} - b^{2}}{f^{2} - g^{2}}} \\ d . \sqrt{\frac{a^{2} + b^{2}}{f^{2} - g^{2}}} \\ e . \sqrt{\frac{a^{2} - b^{2}}{f^{2} + g^{2}}} \\ Jawab : \\ \begin{aligned} Dari so & al diketahui bahwa \\ \sin θ & = \frac{c}{e} = \frac{\sqrt{a^{2} - b^{2}}}{\sqrt{f^{2} + g^{2}}} \\ = \sqrt{\frac{a^{2} - b^{2}}{f^{2} + g^{2}}} \end{aligned} \end{array}

\begin{array}{ll} 4. & Nilai dari (\cos^{2} 17^{\circ} - \sin^{2} 73^{\circ}) adalah . . . . \\ \begin{array}{lllllll} a . & 0 & d . & 1 \\ b . & \frac{1}{3} & c . & \frac{2}{\sqrt{3}} & e . & \frac{1}{2} \sqrt{3} \end{array} \\ Jawab : \\ \begin{aligned} (\cos^{2} 17^{\circ} - \sin^{2} 73^{\circ}) \\ = (\cos^{2} 17^{\circ} - {(\sin 73^{\circ})}^{2}) \\ = (\cos^{2} 17^{\circ} - {(\sin (90^{\circ} - 17^{\circ}))}^{2}) \\ = (\cos^{2} 17^{\circ} - \cos^{2} 17^{\circ}) \\ = 0 \end{aligned} \end{array}

\begin{array}{ll} 5. & Jika diketahui \\ \frac{x \csc^{2} 30^{\circ} \sec^{2} 45^{\circ}}{8 \cos^{2} 45^{\circ} \sin^{2} 60^{\circ}} = \tan^{2} 60^{\circ} - \tan^{2} 30^{\circ}, \\ maka nilai x adalah . . . . \\ \begin{array}{lllllll} a . & - 2 & d . & 1 \\ b . & - 1 & c . & 0 & e . & 2 \end{array} \\ Jawab : \\ \begin{aligned} \frac{x \csc^{2} 30^{\circ} \sec^{2} 45^{\circ}}{8 \cos^{2} 45^{\circ} \sin^{2} 60^{\circ}} = \tan^{2} 60^{\circ} - \tan^{2} 30^{\circ} \\ \frac{x (4) (\frac{4}{2})}{8 (\frac{2}{4}) (\frac{3}{4})} = 3 - (\frac{1}{3}) \\ \frac{8 x}{3} = \frac{8}{3} \\ x = 1 \end{aligned} \end{array}