$\begin{array}{ll}\\ 101.&\textrm{Nilai dari} \: \sin 1020^{\circ}=....\\ &\textrm{a}.\quad -1\\ &\textrm{b}.\quad \color{red}\displaystyle -\frac{1}{2}\sqrt{3}\\ &\textrm{c}.\quad \displaystyle -\frac{1}{2}\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\sqrt{3}\\\\ &\textrm{Jawab}:\qquad\color{red}\textbf{b}\\ &\begin{aligned}\sin 1020^{\circ}&=\sin \left ( 3\times 360^{\circ}-60^{\circ} \right )\\ &=\sin \left ( 0^{\circ}-60^{\circ} \right )\\ &=\sin \left ( -60^{\circ} \right )\\ &=-\sin 60^{\circ}\\ &=\color{red}-\displaystyle \frac{1}{2}\sqrt{3} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 102.&\textrm{Nilai dari} \: \cot (-1290)^{\circ}=....\\ &\textrm{a}.\quad \color{red}-\sqrt{3}\\ &\textrm{b}.\quad \displaystyle -\frac{1}{3}\sqrt{3}\\ &\textrm{c}.\quad \displaystyle \frac{1}{3}\sqrt{3}\\ &\textrm{d}.\quad \displaystyle -\frac{1}{2}\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\\ &\textrm{Jawab}:\qquad\color{red}\textrm{a}\\ &\begin{aligned}\cot (-1290)^{\circ}&=-\cot \left ( 3\times 360^{\circ}+210^{\circ} \right )\\ &=-\cot \left ( 0^{\circ}+210^{\circ} \right )\\ &=-\cot \left ( 210^{\circ} \right )\\ &=-\frac{1}{\tan 210^{\circ}}\\ &=-\frac{1}{\tan \left ( 180^{\circ}+30^{\circ} \right )}\\ &=-\frac{1}{\tan 30^{\circ}}\\ &=\color{red}-\displaystyle \sqrt{3} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 103.&\textrm{Nilai dari}\\ & \sin 240^{\circ}+\sin 225^{\circ}+\cos 315^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\sqrt{3}\qquad&&\textrm{d}.\quad \displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.\quad \color{red}\displaystyle -\frac{1}{2}\sqrt{3}&\qquad&\textrm{e}.\quad \displaystyle \frac{1}{3}\sqrt{3}\\ \textrm{c}.\quad \displaystyle -\frac{1}{2} \end{array}\\ &\textrm{Jawab}:\qquad\color{red}\textrm{b}\\ &\begin{aligned}&\sin 240^{\circ}+\sin 225^{\circ}+\cos 315^{\circ}\\ &=\sin \left ( 180^{\circ}+60^{\circ} \right )+\sin \left ( 180^{\circ}+45^{\circ} \right )+\cos \left ( 360^{\circ}-45^{\circ} \right )\\ &=-\sin 60^{\circ}+\left [ -\sin 45^{\circ} \right ]+\cos 45^{\circ}\\ &=\left ( -\frac{1}{2}\sqrt{3} \right )+\left ( -\frac{1}{2}\sqrt{2} \right )+\frac{1}{2}\sqrt{2}\\ &=\color{red}-\frac{1}{2}\sqrt{3} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 104.&\textrm{Nilai dari} \\ & \displaystyle \frac{\sin 30^{\circ}+\sin 150^{\circ}+\cos 330^{\circ}}{\tan 45^{\circ}+\cos 210^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}\\ \textrm{b}.\quad \displaystyle \frac{1-\sqrt{3}}{1+\sqrt{3}}\\ \textrm{c}.\quad \displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}\\ \textrm{d}.\quad \color{red}\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}\\ \textrm{e}.\quad \displaystyle \frac{1+2\sqrt{3}}{1-2\sqrt{3}} \end{array}\\ &\textrm{Jawab}:\qquad\color{red}\textrm{d}\\ &\begin{aligned}&\displaystyle \frac{\sin 30^{\circ}+\sin 150^{\circ}+\cos 330^{\circ}}{\tan 45^{\circ}+\cos 210^{\circ}}\\ &=\displaystyle \frac{\sin 30^{\circ}+\sin \left ( 180^{\circ}-30^{\circ} \right )+\cos \left ( 360^{\circ}-30^{\circ} \right )}{\tan 45^{\circ}+\cos \left ( 180^{\circ}+30^{\circ} \right )}\\ &=\displaystyle \frac{\sin 30^{\circ}+\sin 30^{\circ}+\cos 30^{\circ}}{\tan 45^{\circ}-\cos 30^{\circ}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}\sqrt{3}}{1-\displaystyle \frac{1}{2}\sqrt{3}}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{2}\sqrt{3}}{1-\displaystyle \frac{1}{2}\sqrt{3}}\\ &=\color{red}\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 105.&\textrm{Nilai dari} \\ &\displaystyle \frac{\sin 270^{\circ}\times \cos 135^{\circ}\times \tan 135^{\circ}}{\sin 150^{\circ}\times \cos 225^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -2\qquad&&\textrm{d}.\quad 1\\ \textrm{b}.\quad \displaystyle -\frac{1}{2}&\textrm{c}.\quad \displaystyle \frac{1}{2}\qquad&\textrm{e}.\quad 2 \end{array}\\ &\textrm{Jawab}:\qquad\color{red}\textrm{e}\\ &\begin{aligned}&\displaystyle \frac{\sin 270^{\circ}\times \cos 135^{\circ}\times \tan 135^{\circ}}{\sin 150^{\circ}\times \cos 225^{\circ}}\\ &=\displaystyle \frac{\sin 270^{\circ}\times \cos \left ( 180^{\circ}-45^{\circ} \right )\times \tan \left ( 180^{\circ}-45^{\circ} \right )}{\sin \left ( 180^{\circ}-30^{\circ} \right )\times \cos \left ( 180^{\circ}+45^{\circ} \right )}\\ &=\displaystyle \frac{-1\times \left (-\cos 45^{\circ} \right )\times \left ( - \tan 45^{\circ}\right )}{\sin 30^{\circ}\times \left ( -\cos 45^{\circ} \right )}\\ &=\displaystyle \frac{-1\times \left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\times -1}{\displaystyle \frac{1}{2}\times \left ( -\frac{1}{2}\sqrt{2} \right )}\\ &=\displaystyle \frac{-\frac{1}{2}\sqrt{2}}{-\frac{1}{4}\sqrt{2}}\\ &=\color{red}\displaystyle 2\end{aligned} \end{array}$
$\begin{array}{ll}\\ 106.&\textrm{Perhatikanlah gambar kurva berikut ini}\end{array}$.
$\begin{array}{ll}\\ .\, \: \: \quad&\begin{array}{llllll}\\ \textrm{a}.&y=-2\cos 2x\\ \textrm{b}.&y=2\cos \displaystyle \frac{3}{2}x\\ \textrm{c}.&y=-2\cos \displaystyle \frac{3}{2}x\\ \textrm{d}.&y=2\sin \displaystyle \frac{3}{2}x\\ \textrm{e}.&y=-2\sin \displaystyle \frac{3}{2}x\\\\ &(\textbf{SIMAK UI 2009 Mat Das}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Dari gambar tampak jelas bahwa}\\ &\textrm{grafik di atas atas adalah grafik}\: \textbf{fungsi cosinus}\\ &\textrm{dengan amplitudo 2 dan periodenya}\: :\: \displaystyle \frac{3}{2}\pi \\ &\textrm{Maka persamaan grafiknya adalah}:\\ &y=2\cos \displaystyle \frac{3}{2}\pi \\ &\textrm{Karena posisinya terbalik, maka}\\ &y=\color{red}-2\cos \displaystyle \frac{3}{2}\pi \\ \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 107.&\textrm{Nilai minimum jika}\\ & f(x)=\left ( 2004\cos 2005x-2006 \right )^{2}+2007\\ & \textrm{adalah}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&2005\\ \textrm{b}.&2007\\ \textrm{c}.&\color{red}2011\\ \textrm{d}.&2013\\ \textrm{e}.&\textrm{tidak ada satupun jawaban dari a sampai d}\\\\ &(\textbf{NUS Mathematics A Level}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{c}\\ &\begin{aligned}&f(x)=\left ( 2004\cos 2005x-2006 \right )^{2}+2007\\ &\textrm{Supaya bernilai minimum, }\\ &\textrm{maka nilai}\quad \color{red}\cos 500x=1,\\ & \color{black}\textrm{ingat nilai}\: \: -1\leq \cos n\pi \leq 1\\ &\textrm{maka},\\ &f_{min}=\left ( 2004.1-2006 \right )^{2}+2007\\ &=(-2)^{2}+2007\\ &=4+2007\\ &=\color{red}2011 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 108.&\textrm{Penyelesaian persamaan}\\ & \cos ^{2}x-2\cos x=4\sin x-2\sin x\cos x\\ & \textrm{adalah}\, ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\pi -\cot ^{-1}\left ( \displaystyle \frac{1}{2} \right )\\ \textrm{b}.&\pi +\tan ^{-1}\left ( \displaystyle \frac{1}{2} \right )\\ \textrm{c}.&\pi -\cot ^{-1}\left ( -1 \right )\\ \textrm{d}.&\color{red}\pi +\tan ^{-1}\left ( -\displaystyle \frac{1}{2} \right )\\ \textrm{e}.&\pi -\tan ^{-1}\left ( \displaystyle \frac{1}{4} \right ) \\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Perhatikan bahwa},\\ &\cos ^{2}x-2\cos x=4\sin x-2\sin x\cos x\\ &\cos x\left ( \cos x-2 \right )=2\sin x\left ( 2-\cos x \right )\\ &\cos x\left ( \cos x-2 \right )=-2\sin x\left ( \cos x -2\right )\\ &\left (\cos x+2\sin x \right )\left ( \cos x-2 \right )=0\\ &\left (\cos x+2\sin x \right )=0\: \: \textrm{atau} \: \: \left ( \cos x-2 \right )=0\\ &2\sin x=-\cos x\: \: \textbf{(mm)}\: \: \textrm{atau}\: \: \cos x=2\: \: \textbf{(tm)}\\ &\textrm{maka}\\ &\displaystyle \frac{\sin x}{\cos x}=-\displaystyle \frac{1}{2}\: \: \Leftrightarrow \: \: \tan x=-\displaystyle \frac{1}{2}\\ &x=\color{red}\tan ^{-1}\left ( -\displaystyle \frac{1}{2} \right )+k.\pi ,\quad k\in \mathbb{Z} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 109.&\textrm{Jika diketahui bahwa}\\ & \sin \beta -\tan \beta -2\cos \beta +2=0\\ & \textrm{dengan}\: \: 0<\beta <\displaystyle \frac{\pi }{2},\\ &\textrm{maka himpunan harga}\: \: \sin \beta =....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}\left \{ \displaystyle \frac{2}{5}\sqrt{5} \right \}\\ \textrm{b}.&\left \{ 0 \right \}\\ \textrm{c}.&\left \{ \displaystyle \frac{2}{5}\sqrt{5},0 \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5} \right \}\\ \textrm{e}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5},0 \right \}\\\\ &\textbf{(SIMAK UI 2009)} \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{a}\\ &\begin{aligned}&\sin \beta -\tan \beta -2\cos \beta +2=0\\ &\sin \beta -\displaystyle \frac{\sin \beta }{\cos \beta } -2\cos \beta +2=0\\ &\sin \beta \cos \beta -\sin \beta -2\cos^{2} \beta +2\cos \beta =0\\ &\sin \beta \left ( \cos \beta -1 \right )-2\cos \beta \left ( \cos \beta -1 \right )=0\\ &\left ( \sin \beta -2\cos \beta \right )\left ( \cos \beta -1 \right )=0\\ &\left ( \sin \beta -2\cos \beta \right )=0\: \textbf{(mm)}\\ &\qquad \textrm{atau}\quad \left ( \cos \beta -1 \right )=0\: \textbf{(tmm)}\\ &\textrm{maka},\\ &\left ( \sin \beta -2\cos \beta \right )=0\\ &\sin \beta =2\cos \beta\\ &\displaystyle \frac{\sin \beta }{\cos \beta }=2\\ &\tan \beta =2=\displaystyle \frac{2}{1},\qquad \textrm{buatlah ilustrasi} \\ &\textrm{dengan membuat segitiga siku-siku}.\\ &\textrm{Sehingga akan didapatkan nilai}\\ &\sin \beta =\color{red}\displaystyle \frac{2}{5}\sqrt{5} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 110.&\textrm{Diketahui bahwa}\\ &\sin \theta -\cos \theta =\displaystyle \frac{\sqrt{5}-\sqrt{3}}{2}\: \: \textrm{dan}\\ & \cos ^{3}\theta -\sin ^{3}\theta =\displaystyle \frac{1}{a}\left ( b\sqrt{5}-c\sqrt{3} \right ),\\ &\textrm{dengan}\: \: a,\: b,\: c\: \: \textrm{adalah bilangan asli, maka}\\ &(1) \quad b-c>0\\ &(2) \quad a-b=7\\ &(3)\quad a-3b+c=0\\ &(4)\quad a+b+c=12\\ &\begin{array}{llllll}\\ \textrm{a}.&(1),\: (2).\: \textrm{dan}\: (3)\: \textrm{benar}\\ \textrm{b}.&(1),\: \textrm{dan}\: (3)\: \textrm{benar}\\ \textrm{c}.&\color{red}(2),\: \textrm{dan}\: (4)\: \textrm{benar}\\ \textrm{d}.&\textrm{hanya}\: (4)\: \textrm{yang benar}\\ \textrm{e}.&\textrm{semuanya benar}\\\\ &(\textbf{SIMAK UI 2015 Mat IPA}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{c}\\ &\begin{aligned}&\sin \theta -\cos \theta =\displaystyle \frac{\sqrt{5}-\sqrt{3}}{2}\\ &\sin ^{2}\theta +\cos^{2}\theta -2\sin \theta \cos \theta =\displaystyle \frac{8-2\sqrt{5}}{4}\\ &1-2\sin \theta \cos \theta =\displaystyle \frac{8-2\sqrt{5}}{4}\\ &\sin \theta \cos \theta =\displaystyle \frac{\sqrt{5}-2}{4}\\ &\textrm{maka},\\ &\cos ^{3}\theta -\sin ^{3}\theta \\ &\qquad=\left ( \cos \theta -\sin \theta \right )\left ( \cos ^{2}\theta +\sin \theta \cos \theta +\sin ^{2}\theta \right )\\ &=\left ( \displaystyle \frac{\sqrt{3}-\sqrt{5}}{2} \right )\left ( 1+\displaystyle \frac{\sqrt{5}-2}{4} \right )\\ &=\displaystyle \frac{1}{8}\left ( \sqrt{3}-\sqrt{5} \right )\left ( 2+\sqrt{5} \right )\\ &=\displaystyle \frac{1}{8}\left ( 2\sqrt{3}+3\sqrt{5}-2\sqrt{5}-5\sqrt{3} \right )\\ &=\displaystyle \frac{1}{8}\left (\sqrt{5}-3\sqrt{3} \right )=\displaystyle \frac{1}{a}\left ( b\sqrt{5}-c\sqrt{3} \right ) \left\{\begin{matrix} a=8\\ b=1\\ c=3 \end{matrix}\right.\\ &\textrm{sehingga}\\ &a-b=8-1=\color{red}7\\ &a+b+c=8+1+3=\color{red}12 \end{aligned} \end{array}$
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