PAS MATEMATIKA BLOG: Latihan Soal 10 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll} 91. & Bentuk sederhana dari \\ 4 \sin (\frac{1}{4} π + x) \cos (\frac{1}{4} π - x) \\ adalah . . . . \\ \begin{array}{lllllll} a . & 2 + 2 \sin 2 x & d . & 2 + 2 \sin x \\ b . & 2 + \sin 2 x & e . & 2 + \sin x \\ c . & 2 \sin 2 x \end{array} \\ Jawab : \\ \begin{aligned} 4 \sin (\frac{1}{4} π + x) \cos (\frac{1}{4} π - x) \\ = 2 (\sin (\frac{1}{2} π) + \sin (2 x)) \\ = 2 (1 + \sin 2 x) \\ = 2 + 2 \sin 2 x \end{aligned} \end{array}$ .
$\begin{array}{ll} 92. & Bentuk sederhana dari \\ 2 \sin (\frac{3}{4} π + x) . \cos (\frac{π}{4} + x) = . . . . \\ \begin{array}{lll} a . 1 - \sin 2 x & d . \cos 2 x \\ b . 1 + \sin 2 x & c . 1 - \cos 2 x & e . - \cos 2 x \end{array} \\ Jawab : a \\ \begin{aligned} 2 \sin (\frac{3}{4} π + x) . \cos (\frac{π}{4} + x) \\ = 2 \sin (135^{\circ} + x) . \cos (45^{\circ} + x) \\ = 2 \sin (90^{\circ} + 45^{\circ} + x) . \cos (45^{\circ} + x) \\ = 2 \cos (45^{\circ} + x) . \cos (45^{\circ} + x) \\ = 2 \cos^{2} (45^{\circ} + x) \\ = 2 {(\cos 45^{\circ} \cos x - \sin 45^{\circ} \sin x)}^{2} \\ = 2 {(\frac{1}{2} \sqrt{2} . \cos x - \frac{1}{2} \sqrt{2} . \sin x)}^{2} \\ = {(\cos x - \sin x)}^{2} \\ = \cos^{2} x + \sin^{2} x - 2 \sin x \cos x \\ = 1 - \sin 2 x \end{aligned} \end{array}$ .
$\begin{array}{ll} 93. & Bentuk sederhana dari \\ 2 \cos (x + \frac{π}{4}) . \cos (\frac{3}{4} π - x) = . . . . \\ \begin{array}{lll} a . \cos (2 x + π) & d . - \cos (2 x - \frac{π}{2}) \\ b . \cos (2 x + \frac{π}{2}) & e . \sin 2 x - 1 \\ c . \cos 2 x \end{array} \\ Jawab : e \\ \begin{aligned} 2 \cos (x + \frac{π}{4}) . \cos (\frac{3}{4} π - x) \\ = \cos (x + \frac{π}{4} + \frac{3}{4} π - x) + \cos (x + \frac{π}{4} - (\frac{3}{4} π - x)) \\ = \cos π + \cos (2 x - \frac{2}{4} π) \\ = - 1 + \cos (\frac{1}{2} π - 2 x), \\ ingat bahwa \cos (- α) = \cos α \\ = - 1 + \sin 2 x \\ = \sin 2 x - 1 \end{aligned} \end{array}$ .
$\begin{array}{ll} 94. & Nilai dari \\ \sqrt{3} \sin 80^{\circ} \sin 160^{\circ} \sin 320^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{3}{8} & d . & \frac{3}{8} \\ b . & - \frac{1}{8} & e . & \frac{5}{8} \\ c . & \frac{1}{8} \end{array} \\ Jawab : \\ \begin{aligned} \sqrt{3} \sin 80^{\circ} \sin 160^{\circ} \sin 320^{\circ} \\ = \sqrt{3} \sin 80^{\circ} \sin 20^{\circ} (- \sin 40^{\circ}) \\ = - \sqrt{3} \sin 80^{\circ} \sin 40^{\circ} \sin 20^{\circ} \\ = - \sqrt{3} \sin 80^{\circ} (- \frac{1}{2} (\cos 60^{\circ} - \cos 20^{\circ})) \\ = - \sqrt{3} \sin 80^{\circ} (- \frac{1}{4} + \frac{\cos 20^{\circ}}{2}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{2} \sqrt{3} \sin 80^{\circ} \cos 20^{\circ} \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} (\sin 100^{\circ} + \sin 60^{\circ}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} (\sin 80^{\circ} + \frac{1}{2} \sqrt{3}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} \sin 80^{\circ} + \frac{1}{8} \sqrt{9} \\ = \frac{3}{8} \end{aligned} \end{array}$ .
$\begin{array}{ll} 95. & Nilai dari \\ \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{8} & d . & \frac{1}{2} \\ b . & - \frac{1}{4} & c . & 0 & e . & \frac{1}{3} \end{array} \\ Jawab : \\ Alternatif 1 \\ \begin{aligned} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \times \frac{2 \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = (\sin \frac{4 π}{7} - \sin 0) \frac{\cos \frac{π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{4 π}{7} \cos \frac{π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{(\sin \frac{5 π}{7} + \sin \frac{3 π}{7}) \cos \frac{4 π}{7}}{4 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{5 π}{7} \cos \frac{4 π}{7} + \sin \frac{3 π}{7} \cos \frac{4 π}{7}}{4 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{9 π}{7} + \sin \frac{π}{7} + \sin \frac{7 π}{7} + \sin (- \frac{π}{7})}{8 \sin \frac{2 π}{7}} \\ = \frac{- \sin \frac{2 π}{7} + \sin \frac{π}{7} + 0 - \sin \frac{π}{7}}{8 \sin \frac{2 π}{7}} \\ = \frac{- \sin \frac{2 π}{7}}{8 \sin \frac{2 π}{7}} \\ = - \frac{1}{8} \end{aligned} \\ Alternatif 2 \\ \begin{aligned} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \\ = \cos \frac{4 π}{7} \cos \frac{2 π}{7} \cos \frac{π}{7} \\ = \frac{1}{2} (\cos \frac{6 π}{7} + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (\cos (π - \frac{π}{7}) + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (- \cos \frac{π}{7} + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (- \cos^{2} \frac{π}{7} + \cos \frac{2 π}{7} \cos \frac{π}{7}) \\ = \frac{1}{4} (- \cos \frac{2 π}{7} - \cos 0 + \cos \frac{3 π}{7} + \cos \frac{π}{7}) \\ = \frac{1}{4} (- \cos 0 + \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7}) \\ = \frac{1}{4} (- 1 + \frac{1}{2}) \\ = \frac{1}{4} \times (- \frac{1}{2}) \\ = - \frac{1}{8} \end{aligned} \end{array}$ .
Berikut penjelasan untuk $\cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} = \frac{1}{2}$ .
$\begin{aligned} \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} \\ = \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} \times \frac{(2 \sin \frac{2 π}{7})}{(2 \sin \frac{2 π}{7})} \\ = \frac{2 \cos \frac{π}{7} \sin \frac{2 π}{7} - 2 \cos \frac{2 π}{7} \sin \frac{2 π}{7} + 2 \cos \frac{3 π}{7} \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} - \sin (- \frac{π}{7}) - (\sin \frac{4 π}{7} - \sin \frac{0 π}{7}) + \sin \frac{5 π}{7} - \sin \frac{π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} + \sin \frac{π}{7} - \sin \frac{4 π}{7} + \sin \frac{5 π}{7} - \sin \frac{π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} - \sin \frac{4 π}{7} + \sin \frac{5 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin (π - \frac{4 π}{7}) - \sin \frac{4 π}{7} + \sin (π - \frac{2 π}{7})}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{4 π}{7} - \sin \frac{4 π}{7} + \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{1}{2} ◼ \end{aligned}$ .

$\begin{array}{ll} 96. & Nilai dari \\ \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{9 π}{14} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{16} & d . & \frac{1}{2} \\ b . & \frac{1}{8} & c . & \frac{1}{4} & e . & 1 \end{array} \\ Jawab : \\ \begin{aligned} Perhat & ikan bahwa \\ \sin \frac{π}{14} & = \sin (\frac{7 π}{14} - \frac{6 π}{14}) = \sin (\frac{1}{2} π - \frac{6 π}{14}) \\ = \cos \frac{6 π}{14} \\ \sin \frac{3 π}{14} & = . . . = \cos \frac{4 π}{14} \\ \sin \frac{9 π}{14} & = . . . = \sin \frac{5 π}{14} = \cos \frac{2 π}{14} \end{aligned} \\ . . . \\ \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{9 π}{14} \\ = \cos \frac{6 π}{14} \cos \frac{4 π}{14} \cos \frac{2 π}{14} \times \frac{2 \sin \frac{2 π}{14}}{2 \sin \frac{2 π}{14}} \\ = \frac{\cos \frac{6 π}{14} \cos \frac{4 π}{14} \sin \frac{4 π}{14}}{2 \sin \frac{2 π}{14}} \\ silahkan dilanjutkan \\ . . . \\ = \frac{1}{8} \end{array}$ .

\begin{array}{ll} 97. & 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} 98. & Nilai dari \sin \frac{π}{24} . \sin \frac{5 π}{24} . \sin \frac{7 π}{24} . \sin \frac{11 π}{24} \\ \begin{array}{lll} a . \frac{5}{16} & d . \frac{2}{16} \\ b . \frac{4}{16} & c . \frac{3}{16} & e . \frac{1}{16} \end{array} \\ (Olimpiade Sains PORSEMA NU 2012) \\ Jawab : e \\ \begin{aligned} \sin \frac{π}{24} . \sin \frac{5 π}{24} . \sin \frac{7 π}{24} . \sin \frac{11 π}{24} \\ = \frac{1}{4} (2 \sin \frac{11 π}{24} . \sin \frac{π}{24} .2 \sin \frac{7 π}{24} . \sin \frac{5 π}{24}) \\ = \frac{1}{4} [(\cos (\frac{10 π}{24}) - \cos (\frac{12 π}{24})) \times (\cos (\frac{2 π}{24}) - \cos (\frac{12 π}{24}))] \\ = \frac{1}{4} [(\cos 75^{\circ} - \cos 90^{\circ}) \times (\cos 15^{\circ} - \cos 90^{\circ})] \\ = \frac{1}{4} [\cos 75^{\circ} . \cos 15^{\circ}] \\ = \frac{1}{8} [\cos 90^{\circ} + \cos 60^{\circ}] \\ = \frac{1}{8} (0 + \frac{1}{2}) \\ = \frac{1}{16} \end{aligned} \end{array}

\begin{array}{ll} 99. & 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} 100. & 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

Latihan Soal 10 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

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