PAS MATEMATIKA BLOG: 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}

PAS MATEMATIKA BLOG

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

Tidak ada komentar:

Posting Komentar