Lanjutan Contoh Soal Bab Trigonometri (3) (K13 Revisi)

\begin{array}{ll}\\ \fbox{21}.&\textrm{Jika pada segitiga ABC diketahui}\: \: \angle ABC=60^{\circ},\: \: \textrm{AC = 8 cm},\\ &\textrm{maka luas lingkaran luar segitiga ABC adalah}\, ....\, \textrm{cm}^{2}\\ &\begin{array}{llllll}\\ \textrm{a}.&64\pi &&&\textrm{d}.&\displaystyle \frac{196}{3}\pi \\ \textrm{b}.&32\pi &\textrm{c}.&\displaystyle \frac{64}{3}\pi &\textrm{e}.&\displaystyle \frac{32}{3}\pi \\\\ &&&&&(\textbf{SIMAK UI 2009 Mat Das}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Proses penyelesaian}}\\\hline \textrm{Mencari}&\textrm{Penentuan}\\\cline{1-1} \textrm{jari-jari R}&\textrm{luas lingkaran}\\\hline \begin{aligned}2R&=\displaystyle \frac{AC}{\sin \angle B}\\ 2R&=\displaystyle \frac{8}{\sin 60^{\circ}}\\ R&=\displaystyle \frac{4}{\displaystyle \frac{1}{2}\sqrt{3}}\\ &=\displaystyle \frac{8}{\sqrt{3}} \end{aligned}&\begin{aligned}L_{\bigodot }&=\pi .R^{2}\\ &=\pi \left ( \displaystyle \frac{8}{\sqrt{3}} \right )^{2}\\ &=\displaystyle \frac{64}{3}\pi \\ &\\ & \end{aligned}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{22}.&\textrm{Diketahui harga}\: \: \sin ^{2}\theta =\displaystyle \frac{2x-7}{x+1}.\\ &\textrm{maka nilai \textit{x} yang memenuhi adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&-1\leq x\leq 8&&&\textrm{d}.&0\leq x\leq 1\\ \textrm{b}.&1\leq x\leq 8&\textrm{c}.&3\displaystyle \frac{1}{2}\leq x\leq 8&\textrm{e}.&1\leq x\leq 3\displaystyle \frac{1}{2}\\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{aligned}\textrm{Nilai}&\textrm{nya adalah}\\ 0&\leq \sin ^{2}\theta =\displaystyle \frac{2x-7}{x+1}\leq 1\quad\Leftrightarrow\quad \begin{cases} \displaystyle \frac{2x-7}{x+1} &\leq 1 \\\\ \cap &(\textrm{irisan})\\\\ \displaystyle \frac{2x-7}{x+1} &\geq 0 \end{cases}\\ \therefore \: &\: \displaystyle \frac{7}{2}\leq x\leq 8 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{23}.&\textrm{Perhatikanlah gambar kurva berikut ini}\end{array}...

\begin{array}{ll}\\ .\, \: \: \quad&\begin{array}{llllll}\\ \textrm{a}.&y=-2\cos 2x&&&\textrm{d}.&y=2\sin \displaystyle \frac{3}{2}x\\ \textrm{b}.&y=2\cos \displaystyle \frac{3}{2}x&\textrm{c}.&y=-2\cos \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\textbf{c}\\ &\begin{aligned}&\textrm{Dengan mensubstitusikan}\\ &\textrm{nilai}\: \: x=0,\: \: \textrm{maka akan mendapatkan nilai}\: \: -2\\ &\textrm{maka yang mungkin adalah jawaban \textbf{a} dan jawaban \textbf{c}}.\\ &\textrm{Dan perhatikanlah periodenya juga yaitu cuma}\: \: \displaystyle \frac{3}{2}. \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{24}.&\textrm{Penyelesaian umum untuk nilai}\: \: \theta \: \: \textrm{yang memenuhi persamaan}\\ &\sin \theta =\displaystyle \frac{1}{2},\: \: \tan \theta =\displaystyle \frac{1}{\sqrt{3}} \: \: \textrm{adalah}\, ....\\ &\begin{array}{llllll}\\ \textrm{a}.&2n\pi +\displaystyle \frac{\pi }{6}&&&\textrm{d}.&2n\pi +\displaystyle \frac{\pi }{3}\\ \textrm{b}.&2n\pi +\displaystyle \frac{\pi }{5}&\textrm{c}.&2n\pi +\displaystyle \frac{\pi }{4} &\textrm{e}.&2n\pi +\displaystyle \frac{\pi }{2} \\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{a}\\ &\begin{aligned}\textrm{Perhatikan}&\: \textrm{bahwa}\, ;\\ &\begin{cases} \sin x & =\sin \theta \\ x& = \theta +k.2\pi \\ \tan x & =\tan \theta \\ x&=\theta +k.\pi \end{cases}\\ \textrm{Karena}\quad&\sin x=\sin \theta =\displaystyle \frac{1}{2}\Rightarrow \theta =30^{\circ}=\displaystyle \frac{\pi }{6}\\ \textrm{Demikian}&\: \textrm{juga untuk nilai}\: \: \tan \theta ,\: \: \textrm{yaitu}:\\ &\tan x=\tan \theta =\displaystyle \frac{1}{\sqrt{3}}\Rightarrow \theta =30^{\circ}=\displaystyle \frac{\pi }{6}\\ \textrm{maka dari}&\: \textrm{sini akan diperoleh \textbf{penyelesaian umumnya}}\\ \textrm{yaitu}\quad &:2n\pi +\displaystyle \frac{\pi }{6} \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{25}.&\textrm{Jika diketahui}\: \: f\left ( \displaystyle \frac{6}{\sqrt{\sin ^{2}x+4}} \right )=\tan x,\: \: \pi \leq x\leq 2\pi ,\\ &\textrm{maka nilai}\: \: f(3)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&0&&&\textrm{d}.&\pi \\ \textrm{b}.&1&\textrm{c}.&\displaystyle \frac{\pi }{2}&\textrm{e}.&2\pi \\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{a}\\ &\begin{aligned}\textrm{Diketahui bahwa}\: &\\ f\left ( \displaystyle \frac{6}{\sqrt{\sin ^{2}x+4}} \right )&=\tan x,\: \: \pi \leq x\leq 2\pi\\ f(3)&=....\\ \textrm{maka selanjutnya}\, &\\ \displaystyle \frac{6}{\sqrt{\sin ^{2}x+4}}&=3\\ \sqrt{\sin ^{2}x+4}&=2\\ \sin ^{2}x+4&=4\\ \sin ^{2}x&=0\\ \sin x&=\begin{cases} x_{1}=0^{\circ} &+k.2\pi \\ x_{2}=180^{\circ} & +k.2\pi \end{cases}\\ \tan \pi =\tan 2\pi &=0\\ \therefore \quad f(3)&=0 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{26}.&\textrm{Nilai minimum jika}\: \: f(x)=\left ( 2004\cos 2005x-2006 \right )^{2}+2007\: \: \textrm{adalah}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&2005&&&\textrm{d}.&2013\\ \textrm{b}.&2007&\textrm{c}.&2011&\textrm{e}.&\textrm{tidak ada satupun jawaban dari \textbf{a} sampai \textbf{d}}\\\\ &&&&&(\textbf{NUS Mathematics A Level}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{aligned}f(x)&=\left ( 2004\cos 2005x-2006 \right )^{2}+2007\\ \textrm{Sup}&\textrm{aya bernilai minimum, maka nilai}\: \: \cos 500x=1,\: \textrm{ingat nilai}\: \: -1\leq \cos n\pi \leq 1\\ \textrm{mak}&\textrm{a},\\ f_{min}&=\left ( 2004.1-2006 \right )^{2}+2007\\ &=(-2)^{2}+2007\\ &=4+2007\\ &=2011 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{27}.&\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{d}.&\pi +\tan ^{-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{e}.&\pi -\tan ^{-1}\left ( \displaystyle \frac{1}{4} \right ) \\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{aligned}\textrm{Perhatikan bahwa},\qquad\qquad\qquad\qquad&\\ \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&=\tan ^{-1}\left ( -\displaystyle \frac{1}{2} \right )+k.\pi ,\quad k\in \mathbb{Z} \end{aligned} \end{array}..

Sumber Referensi

  1. Kanginan, M,. 2017. Matematika untuk Siswa SMA-MA/SMA-MAK Kelas X (Cetakan ke-2). Bandung: Srikandi Empat Widya Utama.
  2. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.
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Lanjutan Contoh Soal Bab Trigonometri (2) (K13 Revisi)

\begin{array}{ll}\\ \fbox{11}.&\textrm{Jika}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{memenuhi}\: \: 12\cos ^{2}x-\cos x-1=0.\\ &\textrm{maka nilai}\: \: \sec ^{2}x_{1}+\sec ^{2}x_{2}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&22&&&\textrm{d}.&25\\ \textrm{b}.&23&\textrm{c}.&24&\textrm{e}.&26\\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{aligned}12\cos ^{2}x+\cos x-1&=0\\ \left ( 3\cos x-1 \right )\left ( 4\cos x+1 \right )&=0\\ \cos x_{1}=\displaystyle \frac{1}{3}\quad \textrm{atau}\quad \cos x_{2}&=-\displaystyle \frac{1}{4}\\ &\begin{cases} \sec x_{1} & =\displaystyle \frac{1}{\cos x_{1}} \\ \Leftrightarrow &\sec ^{2}x_{1}=\displaystyle \frac{1}{\cos^{2} x_{1}}=3^{2}=9\\ \sec x_{2} & =\displaystyle \frac{1}{\cos x_{2}}\\ \Leftrightarrow &\sec ^{2}x_{2}=\displaystyle \frac{1}{\cos^{2} x_{2}}=(-4)^{2}=16 \end{cases}\\ \sec ^{2}x_{1}+\sec ^{2}x_{2}&=9+16=25 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Jika}\: \: \displaystyle \frac{6\sin \gamma \cos \gamma }{2\sin ^{2}\gamma }=\displaystyle \frac{1}{2}\: \: \textrm{dengan}\: \: 0^{\circ}<\gamma <180^{\circ},\\ &\textrm{maka nilai}\: \: \tan \gamma =....\\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&6\\ \textrm{b}.&2&\textrm{c}.&4&\textrm{e}.&8\\\\ &&&&&(\textbf{SAT Subject Test}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{aligned}\displaystyle \frac{6\sin \gamma \cos \gamma }{2\sin ^{2}\gamma }&=\displaystyle \frac{1}{2}\\ \displaystyle \frac{6\sin \gamma \cos \gamma }{2\sin \gamma \sin \gamma }&=\displaystyle \frac{1}{2}\\ \displaystyle \frac{\cos \gamma }{\sin \gamma }&=\displaystyle \frac{1}{6}\\ \cot \gamma &=\displaystyle \frac{1}{6},\: \textrm{maka}\\ \therefore \: \: \: \tan \gamma &=6 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Jika}\: \: \sin \theta =p,\: \: \textrm{maka}\: \: \displaystyle \frac{1}{2p^{2}-3+\displaystyle \frac{1}{p^{2}}}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&\sec ^{2}\theta \tan ^{2}\theta &&&\textrm{d}.&\displaystyle \frac{\tan ^{2}\theta }{2\cos ^{2}\theta -1}\\ \textrm{b}.&\sec ^{2}\theta \cos \theta &\textrm{c}.&\left ( 2\cos ^{2}-1 \right )\sec \theta &\textrm{e}.&\displaystyle \frac{1+\cot \theta }{\sin \theta \cos \theta }\\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{aligned}\displaystyle \frac{1}{2p^{2}-3+\displaystyle \frac{1}{p^{2}}}&=\displaystyle \frac{1}{\left ( p-\displaystyle \frac{1}{p} \right )\left ( 2p-\displaystyle \frac{1}{p} \right )}\\ &=\displaystyle \frac{p^{2}}{\left ( p^{2}-1 \right )\left ( 2p^{2}-1 \right )}\\ &=\displaystyle \frac{\sin ^{2}\theta }{\left ( \sin ^{2}\theta -1 \right )\left ( 2\sin ^{2}\theta -1 \right )}\\ &=\displaystyle \frac{\sin ^{2}\theta }{\left ( -\cos ^{2}\theta \right )\left ( 2\left ( 1-\cos ^{2}\theta \right ) -1 \right )}\\ &=\displaystyle \frac{\sin ^{2}\theta }{\cos ^{2}\theta }\times \displaystyle \frac{1}{2\cos ^{2}\theta -1}\\ &=\displaystyle \frac{\tan ^{2}\theta }{2\cos ^{2}\theta -1} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Jika diketahui}\: \: \displaystyle \frac{\cos ^{2}\alpha -\sin ^{2}\alpha }{\sin \alpha \cos \alpha }=p,\: \: \textrm{maka}\: \: \cot ^{2}\alpha +\tan ^{2}\alpha =....\\ &\begin{array}{llllll}\\ \textrm{a}.&p^{2}+2&&&\textrm{d}.&1-p^{2}\\ \textrm{b}.&p^{2}+1&\textrm{c}.&p^{2}&\textrm{e}.&2-p^{2}\\\\ &&&&&(\textbf{SIMAK UI 2013 Mat Das}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{a}\\ &\begin{aligned}\displaystyle \frac{\cos ^{2}\alpha -\sin ^{2}\alpha }{\sin \alpha \cos \alpha }&=p\\ \displaystyle \frac{\cos \alpha }{\sin \alpha }-\displaystyle \frac{\sin \alpha }{\cos \alpha }&=p\\ \cot \alpha -\tan \alpha &=p\qquad (\textrm{dikuadratkan})\\ \cot ^{2}\alpha -2+\tan ^{2}\alpha &=p^{2}\\ \cot ^{2}\alpha +\tan ^{2}\alpha &=p^{2}+2 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Jika diketahui}\: \: \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}.&\left \{ \displaystyle \frac{2}{5}\sqrt{5} \right \}&&&\textrm{d}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5} \right \}\\ \textrm{b}.&\left \{ 0 \right \}&\textrm{c}.&\left \{ \displaystyle \frac{2}{5}\sqrt{5},0 \right \}&\textrm{e}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5},0 \right \}\\\\ &&&&&(\textbf{SIMAK UI 2009}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\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})\quad \textrm{atau}\quad \left ( \cos \beta -1 \right )&=0\: (\textbf{tmm})\\ \textrm{maka},\qquad\qquad\quad&\\ \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 &=\displaystyle \frac{2}{5}\sqrt{5} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Diketahui panjang dua sisi sebuah segitiga adalah 10 cm dan 8 cm}.\\ &\textrm{Nilai keliling dari segitiga tersebut yang tidak mungkin adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\textrm{32 cm}&&&\textrm{d}.&\textrm{35 cm}\\ \textrm{b}.&\textrm{33 cm}&\textrm{c}.&\textrm{34 cm}&\textrm{e}.&\textrm{36 cm}\\\\ &&&&&(\textbf{SNMPTN 2010 Mat IPA}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{e}\\ &\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Misalkan sisi-sisi}&\textrm{nya}\\ &\left\{\begin{matrix} a=10\\ b=8\\ c=c \end{matrix}\right.\\ \textrm{maka}&\quad a+b>c\: (\textrm{definisi})\\ &\quad 10+8>c \Leftrightarrow 18>c\\ &\end{aligned}}\\\hline \textrm{No}.&\textrm{Keliling(K)},\: K=(a+b)+\textbf{c}&\textrm{Keterangan}\\\hline \textrm{a}&32=18+(14)&\textrm{Boleh}\\\hline \textrm{b}&33=18+(15)&\textrm{Boleh}\\\hline \textrm{c}&34=18+(16)&\textrm{Boleh}\\\hline \textrm{d}&35=18+(17)&\textrm{Boleh}\\\hline \textbf{e}&36=18+(18)&\textbf{Tidak boleh}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Jika dalam segitiga PQR diketahui bahwa QR = 10, PR = 40}\\ &\textrm{dan}\: \: \angle R=120^{\circ},\: \textrm{maka PQ =}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&10\sqrt{13}&&&\textrm{d}.&10\sqrt{17}\\ \textrm{b}.&20\sqrt{13}&\textrm{c}.&10\sqrt{21}&\textrm{e}.&\sqrt{5}\\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Diketahui bahwa}&:\\ &\begin{cases} QR &=10 \\ PR &=40 \\ \angle R &=120^{\circ} \end{cases}\\ & \end{aligned}}\\\hline \textrm{Gambar}&\textrm{Solusi}\\\hline \begin{aligned}&\textrm{perhatikanlah}\\ &\textrm{ilustrasi gambar}\\ &\textrm{di bawah} \end{aligned}&\begin{aligned}\textrm{Dengan}\: \, &\textbf{aturan cosinus}\\ PR^{2}&=PR^{2}+QR^{2}-2.PR.QR.\cos \angle R\\ &=40^{2}+10^{2}-2.40.10.\cos 120^{\circ}\\ &=1600+100-800.\left ( -\displaystyle \frac{1}{2} \right )\\ &=1700+400\\ PR^{2}&=2100\\ PR&=\sqrt{2100}\\ &=10\sqrt{21} \end{aligned}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{18}.&\textrm{Jika dalam segitiga ABC diketahui bahwa}\: \: \angle A=60^{\circ},\: \: \angle B=75^{\circ}\\ &\textrm{dan}\: \: \textrm{BC = 3},\: \textrm{maka panjang sisi AB =}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&3\sqrt{2}&&&\textrm{d}.&6\sqrt{2}\\ \textrm{b}.&3\sqrt{6}&\textrm{c}.&\sqrt{6}&\textrm{e}.&3\sqrt{3}\\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Diketahui bahwa}&:\: \textrm{pada}\: \: \triangle \textrm{ABC}\\ &\begin{cases} \angle A &=60^{\circ} \\ \angle B&=75^{\circ} \\ \angle C&=180^{\circ}-\left ( 60^{\circ}+75^{\circ} \right )=45^{\circ}\\ BC&=3 \end{cases}\\ & \end{aligned}}\\\hline \textrm{Gambar}&\textrm{Solusi}\\\hline \begin{aligned}&\textrm{perhatikanlah}\\ &\textrm{ilustrasi gambar}\\ &\textrm{di bawah} \end{aligned}&\begin{aligned}\textrm{Dengan}\: \, &\textbf{aturan sinus},\: \textrm{yaitu}:\\ \displaystyle \frac{AB}{\sin \angle C}&=\displaystyle \frac{BC}{\sin \angle A}\\ \displaystyle \frac{AB}{\sin 45^{\circ}}& =\displaystyle \frac{3}{\sin 60^{\circ}}\\ AB&=\displaystyle \frac{3}{\sin 60^{\circ}}\times \sin 45^{\circ}\\ &=\displaystyle \frac{3}{\displaystyle \frac{1}{2}\sqrt{3}}\times \displaystyle \frac{1}{2}\sqrt{2}\\ &=\sqrt{6} \end{aligned}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{19}.&\textrm{Panjang sisi AD sebuah segiempat ABCD berikut adalah}... \end{array}..

\begin{array}{ll}\\ .\, \: \: \quad&\begin{array}{llllll}\\ \textrm{a}.&2\sqrt{7}\: \: \textrm{cm}&&&\textrm{d}.&8\: \: \textrm{cm}\\ \textrm{b}.&4\sqrt{6}\: \: \textrm{cm}&\textrm{c}.&2\sqrt{19}\: \: \textrm{cm}&\textrm{e}.&6\: \: \textrm{cm}\\\\ &&&&&(\textbf{Soal UN 2015}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{a}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Proses}}\\\hline \textrm{Aturan Sinus}&\textrm{Aturan Cosinus}\\\hline \multicolumn{2}{|c|}{\textrm{saat mencari}}\\\hline \textrm{AC}&\textrm{AD}\\\hline \begin{aligned}\displaystyle \frac{AC}{\sin \angle B}&=\displaystyle \frac{BC}{\sin \angle BAC}\\ \displaystyle \frac{AC}{\sin 45^{\circ}}&=\displaystyle \frac{5\sqrt{2}}{\sin 30^{\circ}}\\ AC&=\displaystyle \frac{5\sqrt{2}}{\sin 30^{\circ}}\times \sin 45^{\circ}\\ &=\displaystyle \frac{5\sqrt{2}}{\displaystyle \frac{1}{2}}\times \displaystyle \frac{1}{2}\sqrt{2}\\ &=10 \end{aligned}&\begin{aligned}AD^{2}&=AC^{2}+CD^{2}-2.AC.CD.\cos 30^{\circ}\\ &=10^{2}+\left ( 4\sqrt{3} \right )^{2}-2.10.\left (4\sqrt{3} \right ).\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=100+48-120\\ AD^{2}&=28\\ AD&=\sqrt{48}\\ &=\sqrt{4.7}\\ &=2\sqrt{7} \end{aligned}\\\hline \end{array} \end{array}..

\begin{array}{ll}\\ \fbox{20}.&\textrm{Jika pada segitiga ABC lancip dengan AB}=2\sqrt{2},\: BC=2,\: \textrm{dan}\\ &\angle ABC=\theta ,\: \sin \theta =\displaystyle \frac{1}{3},\: \textrm{maka AC}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\sqrt{2}&&&\textrm{d}.&\displaystyle \frac{3}{2}\sqrt{2}\\ \textrm{b}.&\sqrt{6}&\textrm{c}.&\displaystyle \frac{2}{3}\sqrt{3}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{2}\\\\ &&&&&(\textbf{UM UGM 2010 Mat Das}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Proses penyelesaian}}\\\hline \textrm{Mencari nilai}&\textrm{Aturan Cosinus}\\\cline{2-2} \cos \theta &\textrm{AC}\\\hline \begin{aligned}\cos \theta &=\sqrt{1-\sin ^{2}\theta}\\ &=\sqrt{1-\left ( \displaystyle \frac{1}{3} \right )^{2}}\\ &=\sqrt{\displaystyle \frac{8}{9}}\\ &=\displaystyle \frac{2}{3}\sqrt{2}\\ & \end{aligned}&\begin{aligned}AC^{2}&=AB^{2}+BC^{2}-2.AB.BC.\cos \theta \\ &=\left ( 2\sqrt{2} \right )^{2}+2^{2}-2.\left ( 2\sqrt{2} \right ).2.\left ( \displaystyle \frac{2}{3}\sqrt{2} \right )\\ &=8+4-\displaystyle \frac{32}{3}\\ AC^{2}&=\displaystyle \frac{4}{3}\\ AC&=\displaystyle \frac{2}{3}\sqrt{3} \end{aligned}\\\hline \end{array} \end{array}..

 

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Contoh Soal Bab Trigonometri (K13 Revisi)

\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai dari}\: \: \displaystyle \frac{9}{4}\pi \, rad\: \: \textrm{adalah}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&378^{\circ}&&&\textrm{d}.&405^{\circ}\\ \textrm{b}.&385^{\circ}&\textrm{c}.&395^{\circ}&\textrm{e}.&415^{\circ} \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{aligned}\displaystyle \frac{9}{4}\pi \, rad&=\displaystyle \frac{9}{4}\times 180^{\circ}\\ &=9\times 45^{\circ}\\ &=405^{\circ} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Nilai dari}\: \: \sin 120^{\circ}\cos 240^{\circ}-\sin 480^{\circ}\cos 120^{\circ}\: \: \textrm{adalah}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&\sqrt{3}&&&\textrm{d}.&1\\ \textrm{b}.&\sqrt{2}&\textrm{c}.&\displaystyle \frac{1}{2}\sqrt{3}&\textrm{e}.&0 \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{e}\\ &\begin{aligned}&\sin 120^{\circ}\cos 240^{\circ}-\sin 480^{\circ}\cos 120^{\circ}\\ &=\left ( +\sin \left \langle 180^{\circ}-60^{\circ} \right \rangle \right ).\left ( -\cos \left \langle 180^{\circ}+60^{\circ} \right \rangle \right )-\left ( \sin \left \langle 360^{\circ}+120^{\circ} \right \rangle \right ).\left ( -\cos \left \langle 180^{\circ}-60^{\circ} \right \rangle \right )\\ &=\left (\sin 60^{\circ} \right ).\left ( -\cos 60^{\circ} \right )-\left ( \sin 120^{\circ} \right ).\left ( -\cos 60^{\circ} \right )\\ &=-\left ( \sin 60^{\circ}.\cos 60^{\circ} \right )+\left ( \sin 60^{\circ}.\cos 60^{\circ} \right )\\ &=0 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Nilai untuk}\: \: \sin 2220^{\circ}\: \: \textrm{adalah}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&0&&&\textrm{d}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.&\displaystyle \frac{1}{2}&\textrm{c}.&\displaystyle \frac{1}{2}\sqrt{2}&\textrm{e}.&1 \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{aligned}\sin 2220^{\circ}&=\sin \left \langle 6\times 360^{\circ}+60^{\circ} \right \rangle \\ &=\sin \left \langle 0^{\circ}+60^{\circ} \right \rangle \\ &=\sin 60^{\circ}\\ &=\displaystyle \frac{1}{2}\sqrt{3} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Jika}\: \: M(-2,-2)\: \: \textrm{nilai}\: \cos \angle XOM=....\\ &\begin{array}{llllll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\sqrt{3}&&&\textrm{d}.&-\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{2}&\textrm{c}.&\displaystyle \frac{1}{2}\sqrt{3}&\textrm{e}.&1 \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ &\cos \angle XOM=....\: ?\\ & \end{aligned}}\\\hline \textrm{Titik}&\textrm{Kuadran(Daerah)}&\textrm{Sudut}\\\hline M(-2,-2)&\begin{aligned}&\textrm{III}\\ &\left ( 180^{\circ}+\alpha \right ) \end{aligned}&\tan \angle XOM=\displaystyle \frac{y}{x}\\\hline \multicolumn{1}{c|}{.}&\Large\textbf{PROSES}&\multicolumn{1}{|c}{.}\\\cline{2-2} \multicolumn{1}{r|}{.}&\begin{aligned}\tan \angle XOM&=\displaystyle \frac{y}{x}\\ \tan \alpha &=\displaystyle \frac{-2}{-2}=1\\ &=45^{\circ}\\ \textrm{sudutnya},\: \alpha &=180^{\circ}+45^{\circ}\\ &=225^{\circ}\\ \textrm{Sehingga},\: \cos \alpha &=\cos 225^{\circ}\\ &=-\cos 45^{\circ}\\ &=-\displaystyle \frac{1}{2}\sqrt{2} \end{aligned}&\multicolumn{1}{|l}{.}\\\cline{2-2} \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Jika}\: \: \cos \theta =u\: \: \textrm{dengan}\: \: 0< \theta < \displaystyle \frac{\pi }{2},\: \textrm{maka}\: \: \tan \theta =....\\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&\sqrt{1-u^{2}}\\ \textrm{b}.&\displaystyle \frac{1}{\sqrt{1-u^{2}}}&\textrm{c}.&\displaystyle \frac{u}{\sqrt{1-u^{2}}}&\textrm{e}.&\displaystyle \frac{\sqrt{1-u^{2}}}{u}\\\\ &&&&&(\textbf{SAT Subject Test}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{e}\\ &\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Diketahui}&:\\ &\left\{\begin{matrix} \cos \theta =u=\displaystyle \frac{u}{1}\\\\ 0< \theta < \displaystyle \frac{\pi }{2}\: \: \: \: \: \: \end{matrix}\right.\\ & \end{aligned}}\\\hline \textrm{Posisi}&\textrm{Bantuan proses}&\tan \theta \\\hline \textrm{Di kuadran I}&\begin{aligned}&\textrm{Dengan rumus \textbf{Pythagoras}}\\ &x^{2}+y^{2}=r^{2}\\ &\cos \theta =u=\displaystyle \frac{u}{1}=\frac{x}{r}\\ &\textrm{maka},\\ &y^{2}=r^{2}-x^{2}\\ &y^{2}=1^{2}-u^{2}\\ &y=\sqrt{1-u^{2}} \end{aligned}&\begin{aligned}\tan \theta &=\frac{y}{x}\\ &=\displaystyle \frac{\sqrt{1-u^{2}}}{u}\\ &\\ &\\ &\\ & \end{aligned} \\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Pada setiap}\: \: \alpha \: \: \textrm{berlaku}\\ &\tan \alpha +\cos+\tan (-\alpha ) +\cos (-\alpha )=....\\ &\begin{array}{llllll}\\ \textrm{a}.&0&&&\textrm{d}.&2\left ( \tan \alpha +\cos \alpha \right )\\ \textrm{b}.&2\tan \alpha &\textrm{c}.&2\cos \alpha &\textrm{e}.&2\\\\ &&&&&(\textbf{SAT Subjeck Test}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{c}\\ &\begin{aligned}\tan \alpha &+\cos+\tan (-\alpha ) +\cos (-\alpha )\\ &=\tan \alpha +\cos-\tan \alpha +\cos \alpha \\ &=2\cos \alpha \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Persamaan}\: \: \cot \alpha +\tan \alpha =....\\ &\begin{array}{llllll}\\ \textrm{a}.&\sec \alpha \csc \alpha &&&\textrm{d}.&1-\csc \alpha \\ \textrm{b}.&\tan \alpha \sin \alpha &\textrm{c}.&1-\tan \alpha &\textrm{e}.&\sec ^{2}\alpha \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{a}\\ &\begin{aligned}\cot \alpha +\tan \alpha &=\displaystyle \frac{\cos \alpha }{\sin \alpha }+\displaystyle \frac{\sin \alpha }{\cos \alpha } \\ &=\displaystyle \frac{\cos ^{2}\alpha +\sin ^{2}\alpha }{\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{1}{\cos \alpha .\sin \alpha }\\ &=\sec \alpha \csc \alpha \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Sebuah segitiga ABC dengan panjang AC}=4+2\sqrt{3}\: \: \textrm{dan}\: \: \angle BAC=30^{\circ}.\\ &\textrm{Jika titik D terdapat pada AB sedemikian hingga CD tegak lurus AB}\\ &\textrm{dan panjang BD = 1, maka nilai}\: \: \sin \angle ABC = ....\\ &\begin{array}{llllll}\\ \textrm{a}.&3+2\sqrt{3}&&&\textrm{d}.&\displaystyle \frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right )\\ \textrm{b}.&2+\sqrt{3}&\textrm{c}.&\displaystyle \frac{1}{2}\left ( 1+\sqrt{3} \right )&\textrm{e}.&\left ( \sqrt{2}+\sqrt{6} \right )\\\\ &&&&&(\textbf{SIMAK UI 2010 MatDas}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{d}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\sin \angle BAC&=\displaystyle \frac{DC}{BC}\\ \sin 30^{\circ}&=\displaystyle \frac{DC}{4+2\sqrt{3}}\\ DC&=\left (4+2\sqrt{3} \right ).\sin 30^{\circ}\\ DC&=2+\sqrt{3}\\ AC^{2}&=AC^{2}+CD^{2}\\ &=1+\left ( 2+\sqrt{3} \right )^{2}\\ AC&=\sqrt{8+4\sqrt{3}}=2\sqrt{2+\sqrt{3}} \end{aligned}&\begin{aligned}\sin \angle ABC&=\displaystyle \frac{CD}{BC}\\ &=\displaystyle \frac{2+\sqrt{3}}{2\sqrt{2+\sqrt{3}}}\\ &=\displaystyle \frac{\sqrt{2+\sqrt{3}}}{2},\quad \textrm{ingat}\: \sqrt{2+\sqrt{3}}=\displaystyle \frac{\sqrt{2}+\sqrt{6}}{2}\\ &=\displaystyle \frac{\displaystyle \frac{\sqrt{2}+\sqrt{6}}{2}}{2}\\ &=\displaystyle \frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right ) \end{aligned}\\\hline \end{array} \end{array}.

.\qquad\: \: \begin{aligned}\textbf{Sebagai}&\: \textbf{ilustrasi}\\ \textrm{perhatik}&\textrm{anlah gambar berikut ini}\\\\ &\qquad\qquad\qquad\qquad\begin{array}{ccc|ccc} &&C&&&\\ &&&&&\\ &4+2\sqrt{3}&&&&\\ &&&&&\\ \angle &30^{\circ}&&&&\setminus \\\hline A&&\multicolumn{2}{c}{D}&1&B \end{array} \end{aligned}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Diketahui}\: \: \sin \alpha +\sin \beta =1\: \: \textrm{dan}\: \: \cos \alpha +\cos \beta =\sqrt{\displaystyle \frac{5}{3}}\\ &\textrm{Nilai}\: \: \sin \alpha \sin \beta +\cos \alpha \cos \beta =....\\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{3}&\textrm{c}.&\displaystyle \frac{1}{2}\sqrt{2}&\textrm{e}.&\displaystyle \frac{1}{3} \end{array}\\ &\\ &\textrm{Jawab}:\quad\textbf{e}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\sin \alpha +\sin \beta &=1\\ \left (\sin \alpha +\sin \beta \right )^{2}&=1^{2}\\ \sin ^{2}\alpha +\sin ^{2}\beta +2\sin \alpha \sin \beta &=1\\ 1+2\sin \alpha \sin \beta &=1\\ \sin \alpha \sin \beta &=0\\ &\\ &\\ & \end{aligned}&\begin{aligned}\cos \alpha +\cos \beta &=\sqrt{\displaystyle \frac{5}{3}}\\ \left (\cos \alpha +\cos \beta \right )^{2}&=\left ( \sqrt{\displaystyle \frac{5}{3}} \right )^{2}\\ \cos ^{2}\alpha +\cos ^{2}\beta +2\cos \alpha \cos \beta &=\displaystyle \frac{5}{3}\\ 1+2\cos \alpha \cos \beta &=\displaystyle \frac{5}{3}\\ \cos \alpha \cos \beta &=\displaystyle \frac{1}{3} \end{aligned}\\\hline \multicolumn{2}{|c|}{\sin \alpha \sin \beta +\cos \alpha \cos \beta =0+\displaystyle \frac{1}{3}=\displaystyle \frac{1}{3}}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{10}.&\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\, \qquad\qquad (3)\quad a-3b+c=0\\ &(2) \quad a-b=7\qquad\qquad (4)\quad a+b+c=12\\ &\begin{array}{llllll}\\ \textrm{a}.&(1),\: (2).\: \textrm{dan}\: (3)\: \textrm{benar}&&&\textrm{d}.&\textrm{hanya}\: (4)\: \textrm{yang benar}\\ \textrm{b}.&(1),\: \textrm{dan}\: (3)\: \textrm{benar}&\textrm{c}.&(2),\: \textrm{dan}\: (4)\: \textrm{benar}&\textrm{e}.&\textrm{semuanya benar}\\\\ &&&&&(\textbf{SIMAK UI 2015 Mat IPA}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\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},\qquad\qquad\qquad&\\ \cos ^{3}\theta -\sin ^{3}\theta &=\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}\qquad\qquad\quad&\\ a-b&=8-1=7\\ a+b+c&=8+1+3=12 \end{aligned} \end{array}.

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Trigonometri (K13 Revisi)

A. Satuan Ukuran Sudut

Sebelumnya silahkan lihat arsip lama berikut ini:

Sebuah sudut akan terbentuk dari dua buah sinar yang berpotongan. Sedangkan sinar sendiri di sini adalah sebuah garis yang berpangkal pada sebuah titik dan memanjang ke suatu arah tertentu.

Perhatikanlah ilustrasi berikut

Dari ilustrasi gambar dua sinar di atas, antara sinar  \underset{OA}{\rightarrow}  dan sinar  \underset{OB}{\rightarrow}  bertemu di titik O sehingga terbentuklah sudut  \angle AOB.

Pada materi di tingkat SMP di kenalkan ukuran sudut dalam derajat dan radian. Selanjutnya secara singkat dapat dituliskan pada satu lingkaran penuh akan terdapat  360^{0} atau 2\pi  radian.

Coba perhatikan ilustrasi berikut!

Jika ditunjukkan dengan tabel ukuran sudutnya adalah sebagai berikut:

\begin{array}{|l|c|c|}\hline \quad\textrm{\textbf{Ukuran Sudut untuk}}&\textrm{\textbf{Derajat}}&\textrm{\textbf{Radian}}\\\hline \textrm{Satu Lingkaran Penuh}&360^{0}&\begin{matrix} 2\pi & rad \end{matrix}\\\hline \textrm{Setengah Lingkaran Penuh}&180^{0}&\begin{matrix} \pi & rad \end{matrix}\\\hline \textrm{Seperempat Lingkaran}&90^{0}&\begin{matrix} \displaystyle \frac{1}{2}\pi & rad \end{matrix} \\\hline \textrm{Seperlima lingkaran}&72^{0}&\begin{matrix} \displaystyle \frac{2}{5}\pi & rad \end{matrix}\\\hline \qquad\qquad\quad\quad\vdots &\vdots &\vdots \\\hline \qquad\qquad\quad\: \, dst&dst&dst\\\hline \end{array}.

Sebagai catatan ukuran sudut yang diubah ke menit dan/atau detik dinamakan sebagai sistem seksagesimal, yaitu:

\begin{aligned}1^{0}&=\begin{cases} {60}' & \text{dibaca} \: \: 60 \: \: \textrm{menit}\\\\ {3600}'' & \text{ dibaca } \: \: 3600\: \: detik \end{cases}\\\\ \textrm{in}&\textrm{gat}\: \, \: {1}'={60}''\: \: (\textrm{satu menit = 60 detik})\\ &\qquad 1^{0}={60}'=60\times {1}'=60\times {60}''={3600}'' \end{aligned}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Ubahlah ke dalam sudut-sudut berikut dalam radian}\\ &\begin{array}{llllllll}\\ \textrm{a}.&3^{0}&\textrm{e}.&90^{0}&\textrm{i}.&210^{0}&\textrm{m}.&300^{0}\\ \textrm{b}.&15^{0}&\textrm{f}.&120^{0}&\textrm{j}.&225^{0}&\textrm{n}.&315^{0}\\ \textrm{c}.&30^{0}&\textrm{g}.&135^{0}&\textrm{k}.&240^{0}&\textrm{0}.&12^{0}{24}'\\ \textrm{d}.&60^{0}&\textrm{h}.&150^{0}&\textrm{l}.&270^{0}&\textrm{p}.&30^{0}{12}'{30}'' \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bahwa}:\quad 180^{0}=180\times 1^{0}=\pi \: rad\Rightarrow 1^{0}=\displaystyle \frac{\pi }{180}\: rad \\ &\begin{array}{|l|l|l|}\hline \begin{aligned}\textrm{a}.\quad 3^{0}&=\cdots \: \pi \: rad\\ 3^{0}&=\displaystyle \frac{3}{180}\pi \: rad\\ &=\displaystyle \frac{1}{60}\pi \: rad \end{aligned}&\begin{aligned}\textrm{c}.\quad 30^{0}&=\cdots \: \pi \: rad\\ 30^{0}&=\displaystyle \frac{30}{180}\pi \: rad\\ &=\displaystyle \frac{1}{6}\pi \: rad \end{aligned}&\begin{aligned}\textrm{f}.\quad 120^{0}&=\cdots \: \pi \: rad\\ 120^{0}&=\displaystyle \frac{120}{180}\pi \: rad\\ &=\displaystyle \frac{2}{3}\pi \: rad \end{aligned}\\\hline \begin{aligned}\textrm{m}.\quad 300^{0}&=\cdots \: \pi \: rad\\ 300^{0}&=\displaystyle \frac{300}{180}\pi \: rad\\ &=\displaystyle \frac{5}{3}\pi \: rad\\ &\\ &\\ &\\ & \end{aligned} &\multicolumn{2}{|c|}{\begin{aligned}\textrm{o}.\quad 12^{0}{24}'&=\cdots \: \pi \: rad\\ 12^{0}{24}'&=\displaystyle \frac{12+\left ( \displaystyle \frac{24}{60} \right )}{180}\pi \: rad\\ &=\displaystyle \frac{12+0,4}{180}\pi \: rad\\ &=\displaystyle \frac{12,4}{180}\pi \: rad\\ &=\displaystyle \frac{31}{450}\pi \: rad \end{aligned}}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Ubahlah ke dalam sudut-sudut berikut dalam derajat}\\ &\begin{array}{llllllll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\pi \: rad&\textrm{e}.&\displaystyle \frac{4}{3}\pi \: rad\\\\ \textrm{b}.&\displaystyle \frac{3}{5}\pi \: rad&\textrm{f}.&2 \: rad\\\\ \textrm{c}.&\displaystyle \frac{2}{9}\pi \: rad&\textrm{g}.&12\: rad\\\\ \textrm{d}.&\displaystyle \frac{7}{12}\pi \: rad&\textrm{h}.&15\pi \: rad \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bahwa}:\quad \pi \: rad=180^{0}\Rightarrow 1\: rad=\displaystyle \frac{180^{0}}{\pi } \\ &\begin{array}{|l|l|l|}\hline \begin{aligned}\textrm{a}.\quad \displaystyle \frac{1}{2}\pi \: rad&=\displaystyle \frac{1}{2}\pi \times \displaystyle \frac{180^{0}}{\pi }\\ &=90^{0} \end{aligned}&\begin{aligned}\textrm{b}.\quad \displaystyle \frac{3}{5}\pi \: rad&=\displaystyle \frac{3}{5}\pi \times \displaystyle \frac{180^{0}}{\pi }\\ &=108^{0} \end{aligned}&\begin{aligned}\textrm{d}.\quad \displaystyle \frac{7}{12}\pi \: rad&=\displaystyle \frac{7}{12}\pi \times \displaystyle \frac{180^{0}}{\pi }\\ &=105^{0} \end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}\textrm{f}.\quad 2\: rad&=2\times \displaystyle \frac{180^{0}}{\pi }\\ &=2\times \displaystyle \frac{180^{0}}{\frac{22}{7}}\\ &=\displaystyle \frac{1260^{0}}{11}\\ &=114,\overline{54}\, ^{0} \end{aligned}}\\\hline \end{array} \end{array}.

Yang belum dibahas, silahkan dikerjakan sebagai latihan.

 

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insyaAllah

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Vektor (Kelas X Peminatan Mat&IA K13 Revisi)

Silahkan kunjungi lembaran lama di

Tambahan Materi Vektor

A. Skalar dan Vektor

  • Besaran skalar adalah besaran yang hanya memiliki besar saja tanpa arah
  • Besaran vektor adalah besaran yang memiliki besar sekaligus arah.

Perhatikanlah tiga ilustrasi berikut ini

\begin{array}{|l|c|c|}\hline &\multicolumn{2}{c|}{\textrm{Vektor}}\\\cline{2-3} \raisebox{0.0ex}[0cm][0cm]{\quad\qquad Istilah} &\textrm{Bidang}\: \left ( \textrm{XY} \right )&\textrm{Ruang}\: \left ( \textrm{XYZ} \right )\\\cline{2-3} &\textrm{R}^{2}&\textrm{R}^{3}\\\hline \textrm{Misal titik P}&\begin{cases} \bar{p} &=(a_{1},a_{2}) \\ \bar{p} &=\begin{pmatrix} a_{1}\\ a_{2} \end{pmatrix} \end{cases}&\begin{cases} \bar{p} &=(a_{1},a_{2},a_{3}) \\ \bar{p} &= \begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix} \end{cases}\\\hline \textrm{Vektor satuan}&\bar{i},\: \bar{j}&\bar{i},\: \bar{j},\: \bar{k}\\\hline \textrm{Vektor satuan dari}\: \: \bar{p}&\multicolumn{2}{|c|}{\hat{p}=\displaystyle \frac{\bar{p}}{\left | \bar{p} \right |}}\\\hline \textrm{Vektor posisi P}&\bar{OP}=\bar{p}=a_{1}\bar{i}+a_{2}\bar{j}&\bar{OP}=\bar{p}=a_{1}\bar{i}+a_{2}\bar{j}+a_{3}\bar{k}\\\hline \textrm{Panjang vektor P}&\left |\bar{p} \right |=\sqrt{a_{1}^{2}+a_{2}^{2}}&\left |\bar{p} \right |=\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}\\\hline \textrm{Sifat-sifat operasi}&\multicolumn{2}{|l|}{\begin{array}{ll}\\ &\textrm{Untuk}\: \: \bar{p}=(a,b)=\begin{pmatrix} a\\ b \end{pmatrix},\: \: \bar{q}=(c,d)=\begin{pmatrix} c\\ d \end{pmatrix}\\ 1.&\bar{p}=\bar{q},\: \textrm{jika}\begin{cases} \left | \bar{p} \right |=\left | \bar{q} \right | \\ \textrm{arah}\: \bar{p}=\textrm{arah}\: \bar{q} \end{cases}\\ 2.&\textrm{Jika}\: \: \bar{w}=\bar{p}+\bar{q},\: \textrm{maka}\: \: \bar{w}=\begin{pmatrix} a+c\\ b+q \end{pmatrix}\\ 3.&\textrm{Jika}\: \: \bar{w}=k\bar{p},\: \textrm{maka}\: \: \bar{w}=\begin{pmatrix} ka\\ kb \end{pmatrix}\\ 4.&\textrm{Lawan vektor(invers)}\: \bar{p}\: \: \textrm{adalah}\: \: -\bar{p}\\ 5.&\bar{p}+\bar{q}=\bar{q}+\bar{p}\\ 6.&\bar{p}+\left ( -\bar{p} \right )=\bar{0}\\ 7.&\bar{p}+\bar{0}=\bar{0}+\bar{p}=\bar{p}\\ 8.&(k+l)\bar{p}=k\bar{p}+l\bar{p}\\ 9.&k\left (\bar{p}+\bar{q} \right )=k\bar{p}+k\bar{q} \end{array}}\\\hline \end{array}.

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Contoh Soal Fungsi Komposisi dan Invers

\begin{array}{ll}\\ \fbox{1}.&\textrm{Diketahui fungsi}\: \: f(2x)=8x-9\: \: \textrm{dan}\: \: g(3x+1)=6x+3.\\ &\textrm{Rumus untuk}\: \: \left ( f+g \right )(x)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&6x+8&&&\textrm{d}.&14x-6\\ \textrm{b}.&6x-8&\textrm{c}.&14x+6&\textrm{e}.&6x-6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} f(2x) &=8x-9\quad \Rightarrow \quad f(x)=f\left ( 2\left ( \displaystyle \frac{x}{2} \right ) \right )=8\left ( \displaystyle \frac{x}{2} \right )-9=4x-9 \\ g(3x+1) &=6x+3\quad \Rightarrow \quad g(x)=g\left ( 3\left ( \displaystyle \frac{x-1}{3} \right )+1 \right )=6\left ( \displaystyle \frac{x-1}{3} \right )+3\\ &\qquad\qquad\qquad\quad\qquad\: \: \: =2x+1 \end{cases}\\ \left ( f+ g \right )&=(4x-9)+(2x+1)\\ &=6x-8 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Jika}\: \: f(x)=3-x,\: \: \textrm{maka}\: \: f\left ( x^{2} \right )+\left (f(x) \right )^{2}-2f(x)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&2x^{2}-6x+4&&&\textrm{d}.&6x+4\\ \textrm{b}.&2x^{2}+4x+6&\textrm{c}.&2x^{2}-4x-6&\textrm{e}.&-4x+6 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\textrm{Diketahui bahwa}\: \: f(x)=3&-x,\: \textrm{sehingga}\\ f\left ( x^{2} \right )+\left (f(x) \right )^{2}-2f(x)&=\left ( 3-x^{2} \right )+\left ( 3-x \right )^{2}-2(3-x)\\ &=\left (3-x^{2} \right )+\left ( 9-6x+x^{2} \right )-\left ( 6-2x \right )\\ &=-x^{2}+x^{2}-6x+2x+3+9-6\\ &=-4x+6 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui beberapa fungsi memiliki sifat-sifat sebagaimana berikut ini}:\\ &(i)\quad \Phi (-x)=-\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &(ii)\quad \Phi (-x)=\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &\textrm{Jika diketahui fungsi}\: \: f\: \: \textrm{dan}\: \: g\: \: \textrm{memiliki sifat}\: \: (i)\: \: \textrm{dan fungsi}\: \: h\: \: \textrm{dan}\: \: k\\ &\textrm{memiliki sifat}\: \: (ii),\: \: \textrm{maka pernyataan berikut yang salah adalah}\: ....\\ &(1)\quad (f+g)(-x)=-(f+g)(x)\\ &(2)\quad (f.k)(-x)=-(f.k)(x)\\ &(3)\quad (h-k)(-x)=(h-k)(x)\\ &(4)\quad (h-g)(-x)=(h-g)(x)\\ &\begin{array}{llllll}\\ \textrm{a}.&(1),(2)\: \textrm{dan}\: (3)&&&\textrm{d}.&(4)\: \textrm{saja}\\ \textrm{b}.&(1)\: \textrm{dan}\: (3)&\textrm{c}.&(2)\: \textrm{dan}\: (4)&\textrm{e}.&\textrm{semuanya benar} \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (\textbf{SIMAK UI 2014 Mat Das})\\\\ &\textrm{Jawab}:\quad \textbf{a} \end{array}.

.\qquad \begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} \Phi (-x)=-\Phi (x) & (\textrm{fungsi ganjil})\begin{cases} f & \text{ misal } f(x)=x \\ g & \text{ misal } g(x)=2x \end{cases} \\\\ \Phi (-x)=\Phi (x) & (\textrm{fungsi genap})\begin{cases} h & \text{ misal } h(x)=x^{2} \\ k & \text{ misal } k(x)=2x^{2} \end{cases} \end{cases}\\ & \end{aligned}}\\\hline \begin{aligned}(1)\quad (f+g)(-x)&=-(f+g)(x)\\ &\textrm{benar}\\ (2)\qquad (f.k)(-x)&=-(f.k)(x)\\ &\textrm{benar}\\ \end{aligned}&\begin{aligned} (3)\quad (h-k)(-x)&=(h-k)(x)\\ &\textrm{benar}\\ (4)\quad (h-g)(-x)&=(h-g)(x)\\ &\textrm{salah} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Diketahui fungsi}\: \: f:R\rightarrow R\: \: \textrm{dan}\: \: g:R\rightarrow R\: \: \textrm{dirumuskan dengan}\\ &f(x)=x-1\: \: \textrm{dan}\: \: g(x)=x^{2}+2x-3.\: \textrm{Fungsi komposisi}\: \: g\: \: \textrm{atas}\: \: f\\ &\textrm{dinotasikan dengan}\\ &\begin{array}{ll}\\ \textrm{a}.&\left ( g\circ f \right )(x)=x^{2}-4\\ \textrm{b}.&\left ( g\circ f \right )(x)=x^{2}-5\\ \textrm{c}.&\left ( g\circ f \right )(x)=x^{2}-6\\ \textrm{d}.&\left ( g\circ f \right )(x)=x^{2}-4x-4\\ \textrm{e}.&\left ( g\circ f \right )(x)=x^{2}-4x-5 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{UN 2016})\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}:\\ &\begin{cases} f(x) &=x-1 \\ g(x) &=x^{2}+2x-3 \end{cases}\\ \left ( g\circ f \right )&=g\left ( f(x) \right )\\ &=\left ( f(x) \right )^{2}+2\left ( f(x) \right )-3\\ &=\left ( x-1 \right )^{2}+2\left ( x-1 \right )-3\\ &=\left (x^{2}-2x+1 \right )+\left ( 2x-2 \right )-3\\ &=x^{2}-4 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Diketahui}\: \: f(x)\: \: \textrm{dan}\: \: g(x)\: \: \textrm{adalah dua polinom yang berbeda dengan}\: \: f(10)=m\\ &\textrm{dan}\: \: g(10)=n.\: \textrm{Jika}\: \: f(x).h(x)=\left (\displaystyle \frac{f(x)}{g(x)}-1 \right )\left ( f(x)+g(x) \right )\: \textrm{dengan}\: \: h(10)=-\displaystyle \frac{16}{15},\\ &\textrm{maka nilai maksimum dari}\: \: \left | m+n \right |=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&8&&&\textrm{d}.&2\\ \textrm{b}.&6\qquad&\textrm{c}.&4\qquad&\textrm{e}.&0 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{SIMAK UI 2014 Mat IPA})\\\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}f(x).h(x)&=\left (\displaystyle \frac{f(x)}{g(x)}-1 \right )\left ( f(x)+g(x) \right )\\ f(x).h(x)&=\left (\displaystyle \frac{f(x)-g(x)}{g(x)} \right )\left ( f(x)+g(x) \right )\\ f(x).g(x).h(x)&=f^{2}(x)-g^{2}(x)\\ f(10).g(10).h(10)&=f^{2}(10)-g^{2}(10)\\ m\times n\times -\displaystyle \frac{16}{15}&=m^{2}-n^{2}\\ 0&=15m^{2}+16mn-15n^{2}\\ 0&=\left ( 3m+5n \right )\left ( 5m-3n \right )\\ (3m+5n)&=0\Rightarrow 3m=-5n\: \: \textrm{atau}\: \: (5m-3n)=0\Rightarrow 5m=3n\\ &\qquad\qquad \displaystyle \frac{m}{n}=-\frac{5}{3}\qquad\qquad\qquad\qquad\qquad\quad \displaystyle \frac{m}{n}=\frac{3}{5}\\ \textrm{maka nilai maksi}&\textrm{mum}\: \: \left | m+n \right |=\left | 3+5 \right |=8 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{6}.&\textrm{Jika}\: \: f(x)=\sqrt{2x+3}\: \: \textrm{dan}\: \: g(x)=x^{2}+1,\: \textrm{maka}\: \: \left ( f\circ g \right )(2)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&2,24&&&\textrm{d}.&6\\ \textrm{b}.&3\qquad&\textrm{c}.&3,61\qquad&\textrm{e}.&6,16 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{SAT Subject Test})\\\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left ( f\circ g \right )(x)&=f\left ( g(x) \right )\\ &=\sqrt{2g(x)+3}\\ &=\sqrt{2\left ( x^{2}+1 \right )+3}\\ \left ( f\circ g \right )(2)&=\sqrt{2\left ( 2^{2}+1 \right )+3}\\ &=\sqrt{2(5)+3}\\ &=\sqrt{13}\\ &\approx 3,61 \end{aligned} \end{array}..

\begin{array}{ll}\\ \fbox{7}.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x^{2}-2}}\: \: \textrm{dan}\: \: \left ( f\circ g \right )(x)=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}},\: \textrm{maka}\: \: g(x+2)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{1}{x+3}&&&\textrm{d}.&x+3\\ \textrm{b}.&\displaystyle \frac{1}{x-2}\qquad&\textrm{c}.&x-2\qquad&\textrm{e}.&x+5 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{UM UGM 2010 Mat Das})\\\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\left ( f\circ g \right )(x)&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ f\left ( g(x) \right )&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ \displaystyle \frac{1}{\sqrt{\left (g(x) \right )^{2}-2}}&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ \left (g(x) \right )^{2}-2&=x^{2}+6x+7\\ \left (g(x) \right )^{2}&=x^{2}+6x+9\\ g(x)&=\sqrt{x^{2}+6x+9}=\sqrt{(x+3)^{2}}\\ g(x)&=x+3\\ g(x+2)&=(x+2)+3\\ &=x+5 \end{aligned} \end{array}..

Atau dapat juga dengan cara alternatif berikut

.\qquad \begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Diketahui}&:\\ &\begin{cases} f(x) =& \displaystyle \frac{1}{\sqrt{x^{2}-2}} \\ \left (f\circ g \right )(x) & = \displaystyle \frac{1}{\sqrt{x^{2}+6x+7}} \end{cases}\\ & \end{aligned}}\\\hline \begin{aligned}f(x) =y&= \displaystyle \frac{1}{\sqrt{x^{2}-2}}\\ y^{2}&=\displaystyle \frac{1}{x^{2}-2}\\ x^{2}-2&=\displaystyle \frac{1}{y^{2}}\\ x^{2}&=\displaystyle \frac{1}{y^{2}}+2\\ x&=\sqrt{\displaystyle \frac{1}{y^{2}}+2}\\ f^{-1}(y)&=\sqrt{\displaystyle \frac{1}{y^{2}}+2}\\ f^{-1}(x)&=\sqrt{\displaystyle \frac{1}{x^{2}}+2} \end{aligned}&\begin{aligned}g(x)&=\left (f^{-1}\circ f\circ g \right )(x)\\ &=\sqrt{\displaystyle \frac{1}{\left ( \displaystyle \frac{1}{\sqrt{x^{2}+6x+7}} \right )^{2}}+2}\\ &=\sqrt{\left ( x^{2}+6x+7 \right )+2}\\ &=\sqrt{\left ( x^{2}+6x+9 \right )}\\ &=\sqrt{(x+3)^{2}}\\ &=x+3\\ g(x+2)&=(x+2)+3\\ &=x+5\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Jika}\: \: g(x)=2x+4\: \: \textrm{dan}\: \: \left ( f\circ g \right )(x)=4x^{2}+8x-3,\: \textrm{maka}\: \: f^{-1}(x)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&x+9&&&\textrm{d}.&\sqrt{x+1}+2\\ \textrm{b}.&\sqrt{x}+2\qquad&\textrm{c}.&x^{2}-4x-3\qquad&\textrm{e}.&\sqrt{x+7}+2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{array}{|c|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Proses Pencarian}}&\textrm{Hasil Invers}\\\hline \begin{aligned}g(x)=y&=2x+4\\ y-4&=2x\\ x&=\displaystyle \frac{y-4}{2}\\ f^{-1}(y)&=\displaystyle \frac{y-4}{2}\\ f^{-1}(x)&=\displaystyle \frac{x-4}{2}\\ & \end{aligned}&\begin{aligned}f(x)&=\left (f\circ g\circ g^{-1} \right )(x)\\ &=4\left ( g^{^{-1}}(x) \right )^{2}+8\left ( g^{-1}(x) \right )-3\\ &=4\left ( \displaystyle \frac{x-4}{2} \right )^{2}+8\left ( \displaystyle \frac{x-4}{2} \right )-3\\ &=\left ( \displaystyle x^{2}-8x+16 \right )+4x-16-3\\ &=x^{2}-4x-3\\ &=x^{2}-4x+4-7\\ &=(x-2)^{2}-7 \end{aligned}&\begin{aligned}f(x)=y&=(x-2)^{2}-7\\ y+7&=(x-2)^{2}\\ \sqrt{y+7}&=(x-2)\\ (x-2)&=\sqrt{y+7}\\ x&=\sqrt{y+7}+2\\ f^{-1}(y)&=\sqrt{y+7}+2\\ f^{-1}(x)&=\sqrt{x+7}+2 \end{aligned} \\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Jika}\: \: f\: \: \textrm{dan}\: \: g\: \textrm{adalah fungsi yang mempunyai invers dan memenuhi}\: \: f(2x)=g(x-3),\\ &\textrm{maka}\: \: f^{1}(x)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&g^{-1}\left ( \displaystyle \frac{x}{2}- \frac{2}{3}\right )&&&\textrm{d}.&2g^{-1}(x)-6\\ \textrm{b}.&g^{-1}\left ( \displaystyle \frac{x}{2} \right )-\displaystyle \frac{2}{3}\qquad&\textrm{c}.&g^{-1}(2x+6)\qquad&\textrm{e}.&2g^{-1}(x)+6 \end{array}\\\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{SBMPTN 2016 Mat Das})\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ \textrm{Diketahui}&:\\ f(2x)&=g(x-3)=x,\: \textrm{sebagai misal}\\ &\begin{cases} f^{-1}(x) &=2x \\ g^{-1}(x) &=x-3 \end{cases}\\ & \end{aligned}}\\\hline \textrm{Proses}&\textrm{Hasil}\\\hline \begin{aligned}g^{-1}(x) &=x-3\\ x&=g^{-1}(x)+3\\ & \end{aligned}&\begin{aligned}f^{-1}(x) &=2x\\ &=2\left ( g^{-1}(x)+3 \right )\\ &=2g^{-1}(x)+6 \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Jika}\: \: f^{-1}(x)=\displaystyle \frac{x-1}{5}\: \: \textrm{dan}\: \: g^{-1}(x)=\displaystyle \frac{3-x}{2},\: \textrm{maka}\: \: \left (f\circ g \right )^{-1}(6)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&-1&&&\textrm{d}.&2\\ \textrm{b}.&0\qquad&\textrm{c}.&1\qquad&\textrm{e}.&3 \end{array}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad (\textbf{UMPTN 1995})\\\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\textrm{Diketahui b}&\textrm{ahwa}:\\ &\begin{cases} f^{-1}(x)&=\displaystyle \frac{x-1}{5} \\ g^{-1}(x)&=\displaystyle \frac{3-x}{2} \end{cases}\\ \left (f\circ g \right )^{-1}(x)&=\left (g^{-1}\circ f^{-1} \right )(x)\\ &=\displaystyle \frac{3-\left ( \displaystyle \frac{x-1}{5} \right )}{2}\\ \left (f\circ g \right )^{-1}(6)&=\displaystyle \frac{3-\left ( \displaystyle \frac{6-1}{5} \right )}{2}\\ &=\displaystyle \frac{3-1}{2}=\frac{2}{2}\\ &=1 \end{aligned} \end{array}.

 

Sumber Referensi

  1. Kanginan, M,. 2017. Matematika untuk Siswa SMA-MA/SMA-MAK Kelas X (Cetakan ke-2). Bandung: Srikandi Empat Widya Utama.
  2. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.

 

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