Lanjutan Sifat-Sifat Logaritma (Kelas X MIPA)

Sifat-Sifat Logaritma

\begin{array}{|c|}\hline \begin{matrix}\ \bullet \quad a^{^{^{a}\log b}}=b\\ \bullet \quad ^{a}\log b\times d=\: ^{a}\log b+ \: ^{a}\log d\\ \bullet \quad ^{a}\log \displaystyle \frac{b}{d}=\: ^{a}\log b-\: ^{a}\log d\\ \bullet \quad ^{a}\log b=\displaystyle \frac{^{^{x}}\log b}{^{^{x}}\log a}\\ \bullet \quad ^{a}\log b=\displaystyle \frac{1}{^{^{b}}\log a}\\ \bullet \quad ^{a}\log b=m\Rightarrow \: ^{b}\log a=\displaystyle \frac{1}{m}\\ \bullet \quad ^{^{a^{n}}}\log b^{m}=\: \displaystyle \frac{m}{n}\times \: ^{a}\log b\\ \bullet \quad ^{a}\log b\times \: ^{b}\log c\times \: ^{c}\log d=\: ^{a}\log d\\ \bullet \quad ^{a}\log a=1\\ \bullet \quad ^{a}\log 1=0\\ \bullet \quad ^{a}\log a^{n}=n\\ \bullet \quad \log b=\: ^{10}\log b\ \end{matrix}\\\hline \end{array}.

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

\begin{array}{ll}\\ \fbox{1}.&^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24=...\: .\\\\ &\textrm{Solusi}:\\ &\begin{aligned}^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24&=\: ^{2}\log \left ( \displaystyle \frac{12\times 8}{24} \right )\\ &=\: ^{2}\log 16\\ &=\: ^{2}\log 2^{4}\\ &=4 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{2}.&\textrm{Diketahui}\: \: ^{3}\log 7=a,\: \: ^{5}\log 2=b,\: \: \textrm{dan}\: \: ^{2}\log 3=c\\ &\textrm{Nyatakanlah logaritma berikut dalam bentuk}\: \: a,\: b,\: \textrm{dan}\: \: c,\: \: \textrm{yaitu}:\\ &\textrm{a}.\quad ^{7}\log 3\\ &\textrm{b}.\quad ^{4}\log 5\\ &\textrm{c}.\quad ^{21}\log 5\\ &\textrm{d}.\quad ^{6}\log 7\\\\ &\color{black}\textrm{Solusi}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}&\\ ^{7}\log 3&=\displaystyle \frac{1}{^{3}\log 7}\\ &=\displaystyle \frac{1}{a}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}^{4}\log 5&=\displaystyle \frac{1}{^{5}\log 4}\\ &=\displaystyle \frac{1}{^{5}\log 2^{2}}\\ &=\displaystyle \frac{1}{2\: ^{5}\log 2}\\ &=\displaystyle \frac{1}{2b} \end{aligned}\\\hline \begin{aligned}&\\ ^{21}\log 5&=\displaystyle \frac{^{...}\log 5}{^{...}\log 21}\\ &=\displaystyle \frac{^{2}\log 5}{^{2}\log 21}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{\displaystyle \frac{^{3}\log 21}{^{3}\log 2}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{\displaystyle \frac{^{3}\log 3\times 7}{^{3}\log 2}}\\ &=\displaystyle \frac{1}{^{5}\log 2\times \: ^{2}\log 3\times \left ( ^{3}\log 3+\: ^{3}\log 7 \right )}\\ &=\displaystyle \frac{1}{bc(1+a)} \end{aligned}&\begin{aligned}&\\ ^{6}\log 7&=\displaystyle \frac{^{3}\log 7}{^{3}\log 6}\\ &=\displaystyle \frac{^{3}\log 7}{^{3}\log 2\times 3}\\ &=\displaystyle \frac{^{3}\log 7}{^{3}\log 2+\: ^{3}\log 3}\\ &=\displaystyle \frac{^{3}\log 7}{\displaystyle \frac{1}{^{2}\log 3}+\: ^{3}\log 3}\\ &=\displaystyle \frac{a}{\displaystyle \frac{1}{c}+1}\\ &=\displaystyle \frac{ac}{1+c}\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}.

Untuk cara penyelesaian seperti di atas, tidak harus demikian. Sebagai misal saya memberikan proses yang berbeda misal untuk No. 2 (c), tetapi tetap akan menghasilkan hasil yang sana, yaitu:

\color{magenta}\begin{aligned}^{21}\log 5&=\displaystyle \frac{^{...}\log 5}{^{...}\log 21}\\ &=\displaystyle \frac{^{2}\log 5}{^{2}\log 21}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{^{2}\log 3\times \: ^{3}\log 21}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{2}\log 5}}{^{2}\log 3\times \: ^{3}\log (3\times 7)}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{2}\log 5}}{^{2}\log 3\times \: \left ( ^{3}\log 3+\: ^{3}\log 7 \right )}\\ &=\displaystyle \frac{\displaystyle \frac{1}{b}}{c(1+a)}\\ &=\displaystyle \frac{1}{bc(1+a)} \end{aligned}.

\color{blue}\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui bahwa}\: \: \: ^{4}\log 5=a\\ &\textrm{a}.\quad \textrm{Carilah nilai}\: \: \: ^{4}\log 10\\ &\textrm{b}.\quad \textrm{Tunjukkan bahwa}\: \: \: ^{0,1}\log 1,25=\displaystyle \frac{2-2a}{2a+1}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}&\\ ^{4}\log 10&=\: ^{4}\log (2\times 5)\\ &=\: ^{4}\log 2+\: ^{4}\log 5\\ &=\: ^{2^{2}}\log 2+\: a\\ &=\: \displaystyle \frac{1}{2}.\: ^{2}\log 2+\: a\\ &=\: \displaystyle \frac{1}{2}+a\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ ^{0,1}\log 1,25&=\displaystyle \frac{^{4}\log 1,25}{^{4}\log 0,1}\\ &=\displaystyle \frac{^{4}\log \displaystyle \frac{125}{100}}{^{4}\log \displaystyle \frac{1}{10}}\\ &=\displaystyle \frac{^{4}\log 125-\: ^{4}\log 100}{^{4}\log 10^{-1}}\\ &=\displaystyle \frac{^{4}\log 5^{3}-\: ^{4}\log 10^{2}}{-\: ^{4}\log 10}\\ &=\displaystyle \frac{3.\: ^{4}\log 5-\: 2.\: ^{4}\log 10}{-\: ^{4}\log 10}\\ &=\displaystyle \frac{3a-2\left ( \displaystyle \frac{1}{2}+a \right )}{-\left ( \displaystyle \frac{1}{2}+a \right )}\\ &=\displaystyle \frac{a-1}{-a-\displaystyle \frac{1}{2}}\times \displaystyle \frac{-2}{-2}\\ &=\displaystyle \frac{2-2a}{2a+1}\qquad \color{black}\blacksquare \end{aligned}\\\hline \end{array} \end{array}.

\color{red}\begin{array}{ll}\\ \fbox{4}.&\textrm{Jika}\: \: ^{2019}\log \displaystyle \frac{1}{x}=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2019}\: ,\: \textrm{maka hasil dari}\: \: \left ( 2x-3y \right )\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}^{2019}\log \displaystyle \frac{1}{x}&=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2019}\\ ^{2019}\log \displaystyle \frac{1}{x}&=\: ^{y}\log \displaystyle \frac{1}{2019}\\ ^{2019}\log \displaystyle x^{-1}&=\: ^{y}\log \displaystyle (2019)^{-1}\\ -\:\: ^{2019}\log x&=-\: \: ^{y}\log 2019\\ ^{2019}\log x&=\: ^{y}\log 2019,&\textnormal{dipenuhi saat}\: \: x=y=2019 \end{aligned}\\\\ &(2x-3y)=2x-3x=-x=-2019 \end{array}.

\color{green}\begin{array}{ll}\\ \fbox{5}.&\textrm{Bila diberikan}\: \: ^{2}\log \left ( ^{8}\log x \right )=\: ^{8}\log \left ( ^{2}\log x \right ),\: \textrm{maka hasil dari}\: \: \left ( ^{2}\log x \right )^{2}\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}^{2}\log \left ( ^{8}\log x \right )&=\: ^{8}\log \left ( ^{2}\log x \right )\\ ^{2}\log \left ( ^{8}\log x \right )&=\: ^{2^{3}}\log \left ( ^{2}\log x \right )\\ ^{2}\log \left ( ^{8}\log x \right )&=\: ^{2}\log \left ( ^{2}\log x \right )^{\frac{1}{3}}\\ ^{8}\log x&=\left ( ^{2}\log x \right )^{\frac{1}{3}}\\ \left (^{2^{3}}\log x \right )^{3}&=\: ^{2}\log x\\ \left ( \displaystyle \frac{1}{3}\: .^{2}\log x \right )^{3}&=\: ^{2}\log x\\ \displaystyle \frac{1}{27}\: .\left (^{2}\log x \right )^{3}&=\: ^{2}\log x\\ \left ( ^{2}\log x \right )^{2}.\left ( ^{2}\log x \right )&=27.\: ^{2}\log x\\ \left ( ^{2}\log x \right )^{2}&=27 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ \fbox{6}.&\textrm{Tentukanlah nilai}\: \: a+b\: \: \textrm{dimana}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{adalah bilangan riil positif}.\\ &^{7}\log \left ( 1+a^{2} \right )-\: ^{7}\log 25=\: ^{7}\log \left ( 2ab-15 \right )-\: ^{7}\log \left ( 25+b^{2} \right )\\\\ &\textrm{Jawab}:\\ &\begin{aligned}^{7}\log \displaystyle \frac{\left ( 1+a^{2} \right )}{25}&=\: ^{7}\log \displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ \textrm{diambil}&\: \textrm{persamaannya, maka}\\ \displaystyle \frac{\left ( 1+a^{2} \right )}{25}&=\displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ \displaystyle \left ( 1+a^{2} \right )\left ( 25+b^{2} \right )&=25\left ( 2ab-15 \right )\\ &\begin{cases} \left ( 1+a^{2} \right ) & \text{ faktor dari} \: \: 25,\: \: a> 0,\: a\in \mathbb{R} \\ &\textrm{atau}\\ \left ( 25+b^{2} \right ) & \text{ faktor dari }\: \: 25,\: \: b> 0,\: b\in \mathbb{R}\: \: \: \textrm{juga} \end{cases}\\ \end{aligned}\\\\ &\begin{array}{|c|l|c|l|c|c|c|}\hline \textrm{No}&a&\left ( 1+a^{2} \right )&b&\left ( 25+b^{2} \right )&\textrm{Keterangan}&a+b\\\hline 1&2&5&10&125&\textrm{Memenuhi}&12\\\hline 2&7&50&\cdots &\cdots &\textrm{tidak memenuhi}&\cdots \\\hline 3&\cdots &\cdots &5&50&\textrm{tidak memenuhi}&\cdots \\\hline \end{array} \end{array}.

\color{green}\begin{array}{ll}\\ \fbox{7}.&\textrm{Jika}\: \: 60^{a}=3\: \: \textrm{dan}\: \: 60^{b}=5\: ,\: \textrm{maka hasil dari}\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}\\\\ &\textrm{Jawab}:\\ &\\ &\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ &\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases}\\ & \end{aligned}}\\\hline (1-a-b)&2(1-b)&\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )\\\hline \begin{aligned}1-a-b&=1-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log 60-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log \displaystyle \frac{60}{3\times 5}\\ &=\: ^{60}\log 4\\ &=\: ^{60}\log 2^{2} \end{aligned}&\begin{aligned}2(1-b)&=2\left ( 1-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log 60-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log \displaystyle \frac{60}{5} \right )\\ &=2\left ( ^{60}\log 12 \right )\\ &=\: ^{60}\log 12^{2} \end{aligned}&\begin{aligned}\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )&=\displaystyle \frac{\: ^{60}\log 2^{2}}{\: ^{60}\log 12^{2}}\\ &=\displaystyle \frac{2\: ^{60}\log 2}{2\: ^{60}\log 12}\\ &=\displaystyle \frac{\: ^{60}\log 2}{\: ^{60}\log 12}\\ &=\: ^{12}\log 2 \end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}&=12^{^{\: ^{12}\log 2}}=2\\ & \end{aligned}}\\\hline \end{array} \end{array}.

\color{green}\begin{array}{ll}\\ \fbox{8}.&\textrm{Diberikan bilangan riil positif}\: \: x,\: y,\: \textrm{dan}\: z\: \: \textrm{yang memenuhi persamaan}\\ &2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0.\\ &\textrm{Jika nilai}\: \: xy^{5}z\: \: \textrm{dapat dinyatakan dengan}\: \: \displaystyle \frac{1}{2^{\displaystyle \frac{p}{q}}}\: \: \textrm{dengan}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{bilangan asli yang saling prima},\\ &\textrm{maka nilai dari}\: \: p+q=....\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}2\: ^{x}\log (2y)&=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0\\ \textrm{maka}&\\ ^{x}\log (2y)&=\: ^{2x}\log (4y)\quad\Rightarrow \quad \log (2y)\times \log (2x)=\log x\times \log (4y)............1\\ ^{x}\log (2y)&=\: ^{4x}\log (8yz)\quad\Rightarrow \quad \log (2y)\times \log (4x)=\log x\times \log (8yz)............2\\ ^{2x}\log (4y)&=\: ^{4x}\log (8yz)\quad\Rightarrow \quad \log (4y)\times \log (4x)=\log (2x)\times \log (8yz)............ 3\\ & \end{aligned} \end{array}.
\color{green}\begin{aligned}\textrm{Perhatikan persamaan}\: \: 2&,\: \textrm{yaitu}:\\ \log (2y)\times \log (4x)&=\log x\times \log (8yz)\\ \log (2y)\times \left (\log (2x)+\log 2 \right )&=\log x\times \log (8yz)\\ \log (2y)\times \log (2x)+\log (2y)\times \log 2&=\log x\times \log (8yz)\\ \log x\times \log (4y)+\log (2y)\times \log 2&=\log x\times \log (8yz)&\textnormal{persamaan}\: \: 1\: \: \textrm{disubstitusikan}\\ \log (2y)&=\displaystyle \frac{\log x\times \log (8yz)-\log x\times \log (4y)}{\log 2}\\ \log (2y)&=\displaystyle \frac{\log x\times \left ( \log \displaystyle \frac{8yz}{4y} \right )}{\log 2}\\ \log (2y)&=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\: ..................4 \end{aligned}.
\color{green}\begin{aligned}\textrm{Perhatikan juga persamaan}\: \: 3&,\: \textrm{yaitu}:\\ \log (4y)\times \log (4x)&=\log (2x)\times \log (8yz)\\ \left (\log (2y)+\log 2 \right )\times \log (4x)&=\log (2x)\times \log (8yz)\\ \log (2y)\times \log (4x)+\log 2\times \log (4x)&=\log (2x)\times \log (8yz)\\ \log x\times \log (8yz)+\log 2\times \log (4x)&=\log (2x)\times \log (8yz)&\textnormal{persamaan}\: \: 2\: \: \textrm{disubstitusikan}\\ \log 2\times \log (4x)&=\log (2x)\times \log (8yz)-\log x\times \log (8yz)\\ \log 2\times \log (4x)&=\log (8yz)\times \left ( \log (2x)-\log x \right )\\ \log 2\times \log (4x)&=\log (8yz)\times \left ( \log \displaystyle \frac{2x}{x} \right )\\ \log 2\times \log (4x)&=\log (8yz)\times \log 2\\ \log 4x&=\log (8yz)\\ 4x&=8yz\\ \displaystyle \frac{x}{z}&=2y\: .....................................5 \end{aligned}.
\color{green}\begin{aligned}\textrm{dari persamaan}\: \: 4&\: \: \textrm{dan}\: \: 5\\ \log (2y)&=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ \log \left ( \displaystyle \frac{x}{z} \right )&=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ \log 2\left ( \log x-\log z \right )&=\log x\times \log (2z)\\ \log 2\times \log x-\log 2\times \log z&=\log x\times \left ( \log 2+\log z \right )\\ \log 2\times \log x-\log 2\times \log z&=\log x\times \log 2+\log x\times \log z\\ -\log 2\times \log z&=\log x\times \log z\\ \log 2^{-1}&=\log x\\ \displaystyle \frac{1}{2}&=x\: ..................................................6 \end{aligned}.
\color{blue}\begin{array}{|c|c|c|c|}\hline \textrm{persamaan}\: \: 6&\textrm{persamaan}\: \: 5&\textrm{persamaan}\: \: 2&\textrm{Menentukan nilai}\: \: z\\\hline \begin{aligned}&\\ x=\displaystyle \frac{1}{2}&\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned} \log 2y\times \log (4x)&=\log x\times \log (8yz)\\ \log 2y\times \log (4(2yz))&=\log x\times \log (8yz)\\ \log 2y\times \log (8yz)&=\log x\times \log (8yz)\\ \log (2y)&=\log x\\ 2y&=x\\ y&=\displaystyle \frac{1}{2}x\\ &=\displaystyle \frac{1}{2}\times \frac{1}{2}\\ &=\displaystyle \frac{1}{4}\: ..................7\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ \displaystyle \frac{1}{2}&=2\left ( \displaystyle \frac{1}{4} \right )z\\ 1&=z\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}.
\color{green}\begin{aligned}\textrm{maka nilai untuk}&\: \: xy^{5}z\: \: \textrm{adalah}\\ xy^{5}z&=\left ( \displaystyle \frac{1}{2} \right ).\left ( \displaystyle \frac{1}{4} \right )^{5}.1\\ &=\displaystyle \frac{1}{2\times 4^{5}}\\ &=\displaystyle \frac{1}{2\times \left ( 2^{2} \right )^{5}}=\displaystyle \frac{1}{2^{1+10}}\\ &=\displaystyle \frac{1}{2^{11}}=\displaystyle \frac{1}{2^{^{\frac{11}{1}}}}=\displaystyle \frac{1}{2^{^{\frac{p}{q}}}}\quad \begin{cases} p & =11 \\ q & =1 \end{cases}\quad \textrm{dan jelas bahwa} \: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{saling prima}\\ \textrm{Jadi},&\\ p+q&=11+1=12\\ & \end{aligned}.

Sumber Referensi

  1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
  2. Kurnianingsih, S., Kuntarti, dan Sulistiyono. 2007. Matematika SMA dan MA untuk Kelas X Semester 1 Standar Isi 2006. Jakarta: ESIS.
  3. Sembiring, S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.
  4. Sembiring, S., Suparmin, S. 2015. Pena Emas OSN Matematika SMA. Bandung: YRAMA WIDYA.

Sumber Internet

Lanjutan: Eksponen dan Logaritma

Identitas Trigonometri (Kelas XI MIPA)

Identitas Trigonometri adalah persamaan-persamaan yang berlaku untuk semua nilai pengganti peubah atau variabelnya yang di dalamnya terkandung perbandingan trigonometri.

\color{blue}\begin{array}{|c|c|}\hline Pythagoras&\begin{cases} \sin ^{2}\alpha +\cos ^{2}\alpha =1 \\ \sec ^{2}\alpha =\tan ^{2}\alpha +1 \\ \csc ^{2}\alpha =\cot ^{2}\alpha +1 \end{cases}\\\hline \textrm{Setengah}&\textrm{Satu}\\\hline \begin{cases} \sin \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{2}} \\ \cos \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1+\cos \alpha }{2}} \\ \tan \frac{1}{2}\alpha =\pm \sqrt{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }} \end{cases}&\begin{cases} \sin \alpha =\displaystyle \frac{1}{\csc \alpha } \\ \cos \alpha =\displaystyle \frac{1}{\sec \alpha } \\ \tan \alpha =\displaystyle \frac{1}{\cot \alpha } \\ \tan \alpha =\displaystyle \frac{\sin \alpha }{\cos \alpha } \\ \cot \alpha =\displaystyle \frac{\cos \alpha }{\sin \alpha } \end{cases}\\\hline \textrm{Dua}&\textrm{Tiga}\\\hline \begin{cases} \sin 2\alpha =2\sin \alpha \cos \alpha \\ \cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha \\ \tan 2\alpha =\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \end{cases}&\begin{cases} \sin 3\alpha =3\sin \alpha -4\sin ^{3}\alpha \\ \cos 3\alpha =4\cos ^{3}\alpha -3\cos \alpha \\ \tan 3\alpha =\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \end{cases}\\\hline \end{array}

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

\begin{array}{ll}\\ 1.&\textrm{Dengan identitas}\: \: \sin ^{2}\gamma +\cos ^{2}\gamma =1,\: \textrm{buktikanlah identitas-identitas berikut}!\\ &\begin{array}{lllll}\\ \textrm{a}.&\left ( \sin \gamma -\cos \gamma \right )^{2}=1-\sin 2\gamma &\textrm{f}.&\displaystyle \frac{1+\sin \gamma }{\cos \gamma }+\displaystyle \frac{\cos \gamma }{1+\sin \gamma }=2\sec \gamma \\ \textrm{b}.&\displaystyle \frac{\csc ^{2}\gamma -1}{\csc ^{2}\gamma }=\cos ^{2}\gamma &\textrm{g}.&\displaystyle \frac{1}{1+\sin \gamma }+\displaystyle \frac{1}{1-\sin \gamma }=2\sec ^{2}\gamma \\ \textrm{c}.&\sqrt{\displaystyle \frac{1-\sin ^{2}\gamma }{1-\cos ^{2}\gamma }}=\cot \gamma &\textrm{h}.&\left ( \sec \gamma -\tan \gamma \right )^{2}=\displaystyle \frac{1-\sin \gamma }{1+\sin \gamma }\\ \textrm{d}.&\displaystyle \frac{1+\cos \gamma }{\sin ^{2}\gamma }=\displaystyle \frac{1}{1-\cos \gamma }&\textrm{i}.&\left ( \cot \gamma -\csc \gamma \right )^{2}=\displaystyle \frac{1-\cos \gamma }{1+\cos \gamma }\\ \textrm{e}.&\displaystyle \frac{1+\cos \gamma }{\sin \gamma }+\displaystyle \frac{\sin \gamma }{1+\cos \gamma }=2\csc \gamma &\textrm{j}.&\displaystyle \frac{\sin \gamma -2\sin ^{3}\gamma }{2\cos ^{3}\gamma -\cos \gamma }=\tan \gamma \end{array} \end{array}

Bukti:

\color{blue}\begin{aligned}1.\textrm{a}.\quad\left ( \sin \gamma -\cos \gamma \right )^{2}&=\left ( \sin \gamma -\cos \gamma \right )\times \left ( \sin \gamma -\cos \gamma \right )\\ &=\sin ^{2}\gamma -2\sin \gamma \cos \gamma +\cos ^{2}\gamma \\ &=\sin ^{2}\gamma +\cos ^{2}\gamma -2\sin \gamma \cos \gamma \\ &=1-\sin 2\gamma \qquad \color{black}\blacksquare \end{aligned}.

\begin{aligned}1.\textrm{b}.\quad\displaystyle \frac{\csc ^{2}\gamma -1}{\csc ^{2}\gamma }&=1-\displaystyle \frac{1}{csc^{2}\gamma }\\ &=1-\sin ^{2}\gamma \\ &=\cos ^{2}\gamma \qquad \blacksquare \end{aligned}.

\begin{aligned}1.\textrm{c}.\quad\sqrt{\displaystyle \frac{1-\sin ^{2}\gamma }{1-\cos ^{2}\gamma }}&=\sqrt{\displaystyle \frac{\cos ^{2}\gamma }{\sin ^{2}\gamma }}\\ &=\sqrt{\cot ^{2}\gamma }\\ &=\cot \gamma \qquad \blacksquare \end{aligned}.

\color{blue}\begin{aligned}1.\textrm{j}.\quad\displaystyle \frac{\sin \gamma -2\sin ^{3}\gamma }{2\cos ^{3}\gamma -\cos \gamma }&=\displaystyle \frac{\sin \gamma \left ( 1-2\sin ^{2}\gamma \right )}{\cos \gamma \left ( 2\cos ^{2}\gamma -1 \right )}\\ &=\tan \gamma \times \displaystyle \frac{\cos 2 \gamma }{\cos 2\gamma }\\ &=\tan \gamma \qquad \color{black}\blacksquare \end{aligned}

\begin{array}{ll}\\ 2.&\textrm{Dengan identitas}\: \: 1+\tan ^{2}\beta=\sec ^{2}\beta ,\: \textrm{buktikanlah identitas-identitas berikut}!\\ &\begin{array}{lllll}\\ \textrm{a}.&\displaystyle \frac{1+\tan ^{2}\beta }{\csc ^{2}\beta }=\tan ^{2}\beta &\textrm{f}.&1+\tan ^{2}\beta =\displaystyle \frac{\tan ^{2}\beta }{\sin ^{2}\beta }\\ \textrm{b}.&\displaystyle \frac{1}{\sec \beta +\tan \beta }=\sec \beta -\tan \beta &\textrm{g}.&\displaystyle \frac{\tan ^{2}\beta }{1+\tan ^{2}\beta }+\displaystyle \frac{\cot ^{2}\beta }{1+\cot ^{2}\beta }=1 \\ \textrm{c}.&\displaystyle \frac{1-\tan ^{2}\beta }{1+\tan ^{2}\beta }=2\cos ^{2}\beta -1&\textrm{h}.&\displaystyle \frac{1}{\sec \beta +\tan \beta }+\displaystyle \frac{1}{\sec \beta -\tan \beta }=2\sec \beta \\ \textrm{d}.&\displaystyle \frac{\sec \beta +\tan \beta }{\sec \beta -\tan \beta }=\left ( \sec \beta +\tan \beta \right )^{2}&\textrm{i}.&\left ( 1+\tan ^{2}\beta \right )\left ( 1+\displaystyle \frac{1}{\tan ^{2}\beta } \right )=\sec ^{2}\beta \csc ^{2}\beta\\ \textrm{e}.&\displaystyle \frac{\sec \beta -\tan \beta }{\sec \beta +\tan \beta }=\displaystyle \frac{\cos ^{2}\beta }{\left ( 1+\sin \beta \right )^{2}}&\textrm{j}.&\displaystyle \frac{\sec \beta -\tan \beta }{\sec \beta +\tan \beta }=1-2\sec \beta \tan \beta \end{array} \end{array}.

Bukti:

\color{blue}\begin{aligned}2.\textrm{f}.\quad 1+\tan ^{2}\beta &=\sec ^{2}\beta \\ &=\displaystyle \frac{1}{\cos ^{2}\beta }\\ &=\displaystyle \frac{1}{\cos ^{2}\beta }\times \displaystyle \frac{\sin ^{2}\beta }{\sin ^{2}\beta }\\ &=\displaystyle \frac{\sin ^{2}\beta }{\cos ^{2}\beta }\times \displaystyle \frac{1}{\sin ^{2}\beta }\\ &=\tan ^{2}\beta \times \displaystyle \frac{1}{\sin ^{2}\beta }\\ &=\displaystyle \frac{\tan ^{2}\beta }{\sin ^{2}\beta } \qquad \color{black}\blacksquare \end{aligned}.

\begin{array}{ll}\\ 3.&\textrm{Dengan identitas}\: \: 1+\cot ^{2}\alpha =\csc ^{2}\alpha ,\: \textrm{buktikanlah identitas-identitas berikut}!\\ &\begin{array}{lllll}\\ \textrm{a}.&\left ( \csc \alpha -\cot \alpha \right )^{2}=\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }&\textrm{f}.&\displaystyle \frac{\tan \alpha }{1-\cot \alpha }+\displaystyle \frac{\cot \alpha }{1-\tan \alpha }=1+\tan \alpha +\cot \alpha \\ \textrm{b}.& \sec ^{2}\alpha -\csc ^{2}\alpha =\tan ^{2}\alpha -\cot ^{2}\alpha &\textrm{g}.&1+\displaystyle \frac{\cot ^{2}\alpha }{1+\csc \alpha }=\csc \alpha \\ \textrm{c}.&\sqrt{\sec ^{2}\alpha +\csc ^{2}\alpha }=\tan \alpha +\cot \alpha &\textrm{h}.&\displaystyle \frac{\tan \alpha }{\left ( 1+\tan ^{2}\alpha \right )^{2}}+\displaystyle \frac{\cot \alpha }{\left ( 1+\cot ^{2}\alpha \right )^{2}}=\sin \alpha \cos \alpha \\ \textrm{d}.&\displaystyle \tan ^{2}\alpha +\cot ^{2}\alpha +2=\sec ^{2}\alpha \csc ^{2}\alpha &\textrm{i}.&\left ( \sec ^{2}\alpha -1 \right )\left ( \csc ^{2}\alpha -1 \right )=1\\ \textrm{e}.&\displaystyle \frac{1}{\csc \alpha -\cot \alpha }+\displaystyle \frac{1}{\csc \alpha +\cot \alpha }=2\csc \alpha &\textrm{j}.&\cot ^{2}\alpha \left ( \displaystyle \frac{\sec \alpha -1}{1+\sin \alpha } \right )+\sec ^{2}\alpha \left ( \displaystyle \frac{\sin \alpha -1}{1+\sec \alpha } \right )=0 \end{array} \end{array}.

Bukti:

\color{blue}\begin{aligned}3&.\textrm{j}.\quad \cot ^{2}\alpha \left ( \displaystyle \frac{\sec \alpha -1}{1+\sin \alpha } \right )+\sec ^{2}\alpha \left ( \displaystyle \frac{\sin \alpha -1}{1+\sec \alpha } \right )\\ &=\left ( \csc ^{2}\alpha -1 \right ) \left ( \displaystyle \frac{\sec \alpha -1}{1+\sin \alpha } \right )+\sec ^{2}\alpha \left ( \displaystyle \frac{\sin \alpha -1}{1+\sec \alpha } \right )\\ &=\left ( \displaystyle \frac{1}{\sin ^{2}\alpha } -1 \right ) \left ( \displaystyle \frac{\displaystyle \frac{1}{\cos \alpha } -1}{1+\sin \alpha } \right )+\displaystyle \frac{1}{\cos ^{2}\alpha } \left ( \displaystyle \frac{\sin \alpha -1}{1+\displaystyle \frac{1}{\cos \alpha } } \right )\\ &=\left ( \displaystyle \frac{1-\sin ^{2}\alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{\displaystyle \frac{1-\cos \alpha }{\cos \alpha }}{1+\sin \alpha } \right )+\displaystyle \frac{1}{\cos ^{2}\alpha } \left ( \displaystyle \frac{\sin \alpha -1}{\displaystyle \frac{\cos \alpha +1}{\cos \alpha } } \right )\\ \end{aligned}.
\color{blue}\begin{aligned} .\: &=\left ( \displaystyle \frac{\cos ^{2}\alpha }{\sin ^{2}\alpha .\cos \alpha } \right ) \left ( \displaystyle \frac{1-\cos \alpha }{1+\sin \alpha } \right )+\displaystyle \frac{\cos \alpha }{\cos ^{2}\alpha } \left ( \displaystyle \frac{\sin \alpha -1}{\cos \alpha +1} \right )\\ .\: &=\left ( \displaystyle \frac{\cos \alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{1-\cos \alpha }{1+\sin \alpha } \right )\times \left ( \displaystyle \frac{1-\sin \alpha }{1-\sin \alpha } \right )+\displaystyle \frac{1 }{\cos \alpha } \left ( \displaystyle \frac{\sin \alpha -1}{\cos \alpha +1} \right )\times \left ( \displaystyle \frac{\cos \alpha -1}{\cos \alpha -1} \right )\\ .\: &=\left ( \displaystyle \frac{\cos \alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{1-\sin \alpha -\cos \alpha +\cos \alpha \sin \alpha }{1-\sin^{2} \alpha } \right )+\displaystyle \frac{1 }{\cos \alpha } \left ( \displaystyle \frac{\sin \alpha \cos \alpha -\sin \alpha -\cos \alpha +1}{\cos ^{2}\alpha -1} \right )\\ \end{aligned}.
\color{blue}\begin{aligned} .\: &=\left ( \displaystyle \frac{\cos \alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{1-\sin \alpha -\cos \alpha +\cos \alpha \sin \alpha }{\cos^{2} \alpha } \right )+\displaystyle \frac{1 }{\cos \alpha } \left ( \displaystyle \frac{\sin \alpha \cos \alpha -\sin \alpha -\cos \alpha +1}{-\sin ^{2}\alpha } \right )\\ &=\left ( \displaystyle \frac{1}{\sin ^{2}\alpha \cos \alpha }-\displaystyle \frac{1}{\sin ^{2}\alpha \cos \alpha } \right ) \left ( 1-\sin \alpha -\cos \alpha +\cos \alpha \sin \alpha \right )\\ .\: &=0\qquad \color{black}\blacksquare \end{aligned}.

\colorbox{magenta}{\LARGE{\fbox{\LARGE{\fbox{LATIHAN SOAL}}}}}.

Untuk soal-soal yang belum ditunjukkan buktinya, silahkan gunakan sebagai latihan mandiri

Sumber Referensi

  1. Kurnia, N.,  dkk. 2017. Jelajah Matematika 2 SMA Kelas XI Peminatan MIPA(Edisi Revisi 2016). Jakarta: Yudistira.
  2. Tampomas, H. 1999. Seribu Pena Matematika SMU Jilid 1 Kelas 1. Jakarta: ERLANGGA.

Rumus Jumlah dan Selisih Dua Sudut Sinus dan Kosinus (Kelas XI MIPA)

A. Rumus Jumlah dan Selisih Dua Sudut

\color{blue}\begin{cases} \sin \left ( \alpha +\gamma \right ) & =\sin \alpha \cos \gamma +\cos \alpha \sin \gamma \\ \color{green}\sin \left ( \alpha -\gamma \right ) & =\sin \alpha \cos \gamma -\cos \alpha \sin \gamma \\ \cos \left ( \alpha +\gamma \right ) & =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \\ \color{green}\cos \left ( \alpha -\gamma \right ) & =\cos \alpha \cos \gamma +\sin \alpha \sin \gamma \\ \tan \left ( \alpha +\gamma \right ) & =\displaystyle \frac{\tan \alpha +\tan \gamma }{1-\tan \alpha \tan \gamma } \\ \color{green}\tan \left ( \alpha -\gamma \right ) & =\displaystyle \frac{\tan \alpha -\tan \gamma }{1+\tan \alpha \tan \gamma } \end{cases}.

Untuk bentuk \color{blue}\sin \left ( \alpha +\beta \right )=\sin \alpha \cos \beta +\cos \alpha \sin \beta akan diberikan buktinya sebagaimana berikut ini
Perhatikanlah ilustrasi segitiga ABC berikut

Sebagai buktinya adalah, untuk ilustrasi segitiga kita tambah keterangan seabagimana berikut, yaitu

Perhatikanlah ΔAA’C dan ΔAA’B

\color{magenta}\begin{aligned}\displaystyle \frac{AC}{\sin 90^{0}}&=\displaystyle \frac{CA'}{\sin \alpha }=\displaystyle \frac{AA'}{\sin \angle C}\\ AA'&=AC.\sin \angle C\\ &=AC.\sin \left ( 90^{0}-\alpha \right )\\ &=AC.\cos \alpha\\ &\color{blue}\textnormal{dengan cara yang kurang lebih sama akan diperoleh juga}\\ AA'&=AB.\cos \beta\\ &\textnormal{selanjutnya kita tentukan luas seperti perinyah soal}\\ \left [ ABC \right ]&=\left [ AA'C \right ]+\left [ AA'B \right ]\\ \displaystyle \frac{1}{2}.AB.AC.\sin \left ( \alpha +\beta \right )&=\displaystyle \frac{1}{2}.AC.AA'.\sin \alpha +\displaystyle \frac{1}{2}.AB.AA'.\sin \beta \\ \sin \left ( \alpha +\beta \right )&=\displaystyle \frac{AC.AA'.\sin \alpha }{AB.AC}+\displaystyle \frac{AB.AA'.\sin \beta }{AB.AC}\\ &=\displaystyle \frac{AA'}{AB}.\sin \alpha +\displaystyle \frac{AA'}{AC}.\sin \beta \\ &=\displaystyle \frac{\left ( AB.\cos \beta \right )}{AB}.\sin \alpha +\displaystyle \frac{\left ( AC.\cos \alpha \right )}{AC}.\sin \beta \\ \color{black}\sin \left ( \alpha +\beta \right )&=\sin \alpha .\cos \beta +\cos \alpha .\sin \beta \quad \color{black}\blacksquare \end{aligned}.

B. Rumus Jumlah dan Selisih Sinus dan Kosinus

\color{blue}\begin{cases} \sin X+\sin Y & =2\sin \frac{1}{2}\left (X+Y \right )\cos \frac{1}{2}\left ( X-Y \right ) \\ \sin X-\sin Y & =2\cos \frac{1}{2}\left (X+Y \right )\sin \frac{1}{2}\left ( X-Y \right )\\ \cos X+\cos Y & =2\cos \frac{1}{2}\left (X+Y \right )\cos \frac{1}{2}\left ( X-Y \right ) \\ \cos X-\cos Y & =-2\sin \frac{1}{2}\left (X+Y \right )\sin \frac{1}{2}\left ( X-Y \right ) \end{cases}.

Untuk mendapatkan rumus di atas, ambil contoh untuk persamaan rumus no,3, yaitu:

\color{blue}\begin{array}{|cll|c|}\hline &\begin{aligned}\cos \left ( \alpha +\gamma \right )&=\cos \alpha \cos \beta -\sin \alpha \sin \gamma \\ \cos \left ( \alpha -\gamma \right )&=\cos \alpha \cos \beta +\sin \alpha \sin \gamma \\ \end{aligned}&&\\\hline &&+&\textrm{Misalkan}\\ &\begin{aligned}\cos \left ( \alpha +\gamma \right )+\cos \left ( \alpha -\gamma \right )&=2\cos \alpha \cos \gamma \end{aligned}&&X=\alpha +\gamma \\ &&&Y=\alpha -\gamma\\ &\color{magenta}\textbf{Proses perubahannya adalah sebagai berikut}&&\\\hline &\underset{\textrm{perhatikan}}{\underbrace{\begin{matrix} \begin{array}{ccc}\\ &X=\alpha +\gamma &\\ &Y=\alpha -\gamma &+\\\hline{2-2}\\ &X+Y=2\alpha &\\ &2\alpha =X+Y&\\ &\alpha =\displaystyle \frac{X+Y}{2}& \end{array} & \textrm{dan} & \begin{array}{ccc}\\ &X=\alpha +\gamma &\\ &Y=\alpha -\gamma &-\\\hline\\ &X-Y=2\gamma &\\ &2\gamma =X-Y&\\ &\gamma =\displaystyle \frac{X-Y}{2}& \end{array} \end{matrix}}} &&\\\hline &\color{magenta}\textbf{Sehingga rumus akan menjadi}&&\\ &\cos X+\cos Y =2\cos \displaystyle \frac{\left ( X+Y \right )}{2}\cos \frac{\left ( X- Y\right )}{2}&&\\\hline \end{array}.

. C. Rumus bentuk Perkalian Trigonometri

\color{blue}\begin{array}{|c|c|}\hline \textrm{Sejenis}&\textrm{Tidak Sejenis}\\\hline \begin{cases} 2\cos \alpha \cos \gamma &=\cos \left ( \alpha +\gamma \right )+\cos \left ( \alpha -\gamma \right )\\ - 2\sin \alpha \sin \gamma &=\cos \left ( \alpha +\gamma \right )-\cos \left ( \alpha -\gamma \right ) \end{cases}&\begin{cases} 2\sin \alpha \cos \gamma &=\sin \left ( \alpha +\gamma \right )+\sin \left ( \alpha -\gamma \right )\\ 2\cos \alpha \sin \gamma &=\sin \left ( \alpha +\gamma \right )-\sin \left ( \alpha -\gamma \right ) \end{cases}\\\hline \end{array}.

Fungsi (Matematika Wajib Kelas X)

A. Pendahuluan fungsi

Pada awal bab ini kita akan ingatkan kembali tentang notasi, domain(daerah asal), kodomain(daerah kawan), serta range(daerah hasil) fungsi.

Perhatikan tabel berikut

\color{blue}\begin{array}{|l|l|}\hline \textrm{Fungsi}&\textrm{Fungsi atau Pemetaan dari A ke B adalah}\\ &\textrm{sebuah relasi khusus yang memasangkan setiap}\: \: x\in A\\ &\textrm{ke tepat satu}\: \: y\in B\\\hline \textrm{Notasi}&f:x \rightarrow y\: \: \: \textrm{atau}\: \: \: f:x \rightarrow f(x) \\\hline \textrm{Dibaca}&\textrm{fungsi}\: \: f\: \: \textrm{memetakan}\: \: x\in A\: \: \textrm{ke}\: \: y\in B\\\hline \: \: \: \: \textrm{A}&\textrm{Domain atau daerah asal fungsi atau}\quad D_{f}\\\hline \: \: \: \: \, x&\textrm{prapeta(sebelum dipetakan)}\\\hline \: \: \: \: \textrm{B}&\textrm{Kodomain atau daerah kawan fungsi atau}\quad K_{f}\\\hline \: \: \: \: \, y&\textrm{peta(bayangan dari prapeta) adalah Range atau}\quad R_{f}\\\hline \end{array}.

B. Sifat-Sifat Fungsi

\color{blue}\begin{array}{|c|c|l|}\hline \textrm{Injektif(satu-satu)}&\textrm{Surjektif(pada)}&\textrm{Bijektif(korespondensi satu-satu)}\\\hline \begin{aligned}&\textrm{Jika setiap anggota}\\ &\textrm{himpunan A memiliki}\\ &\textrm{bayangan berbeda di}\\ &\textrm{himpunan B} \end{aligned}&\begin{aligned}&\textrm{Jika setiap anggota}\\ &\textrm{himpunan di B}\\ &\textrm{mempunyai prapeta}\\ &\textrm{di himpunan A} \end{aligned}&\begin{aligned}&\textrm{Jika fungsi yang injektif dan}\\ &\textrm{sekaligus juga surjektif}\\ &\\ & \end{aligned}\\\hline \end{array}.

C. Operasi Aljabar Fungsi

\color{blue}\begin{array}{|c|c|}\hline \textrm{Aljabar Fungsi}&\textrm{Daerah Asal}\\\hline (f+g)(x)=f(x)+g(x)&D_{_{(f+g)}}=D_{_{f}}\cap D_{_{g}}\\\hline (f-g)(x)=f(x)-g(x)&D_{_{(f-g)}}=D_{_{f}}\cap D_{_{g}}\\\hline (f.g)(x)=f(x).g(x)&D_{_{(f.g)}}=D_{_{f}}\cap D_{_{g}}\\\hline \left ( \displaystyle \frac{f}{g} \right )(x)=\displaystyle \frac{f(x)}{g(x)}&D_{_{\left ( \frac{f}{g} \right )}}=D_{_{f}}\cap D_{_{g}}\: , \\ &\textrm{dengan}\: \: g(x)\neq 0 \\\hline \end{array}.

D. Macam-Macam Fungsi

\color{blue}\begin{array}{|l|l|l|}\hline \textrm{Fungsi Linear}&\textrm{Fungsi Konstan}&f(x)=c\\\hline &\textrm{Fungsi Identitas}&f(x)=x\\\hline &\textrm{Fungsi linear/garis lurus}&f(x)=ax+b\\\hline \textrm{Fungsi Kuadrat}&\textrm{Fungsi Kuadrat/parabola}&f(x)=ax^{2}+bx+c,\quad a\neq 0\\\hline \textrm{Fungsi Rasional}&\textrm{Fungsi Pecahan}&f(x)=\displaystyle \frac{p(x)}{q(x)}\\\hline &\textrm{Fungsi Modulus(nilai mutlak)}&f(x)=\left | x \right |\\\hline \textrm{Fungsi Khusus} &\textrm{Fungsi tangga}&f(x)=\left \lfloor x \right \rfloor\\\hline &\textrm{Fungsi genap dan ganjil}&\begin{cases} \textrm{Fungsi ganjil} & f(-x)=-f(x) \\ \textrm{Fungsi genap} & f(-x)=f(x) \end{cases}\\\hline \end{array}.

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

\color{fuchsia}\begin{array}{ll}\\ 1.&\textrm{Diketahui 2 humpuan sebagai berikut}:\\ &\begin{cases} \textrm{P} & =\left \{ -2,-1,0,1,2 \right \} \\ \textrm{Q} & =\left \{ 0,1,2,5,7 \right \} \end{cases}\\ &\textrm{Di antara relasi dari P ke Q berikut manakah yang merupakan fungsi}\\ &\textrm{a}.\quad \textrm{A}=\left \{ (-2,0),(-1,0),(0,0),(1,0),(2,0) \right \}\\ &\textrm{b}.\quad \textrm{B}=\left \{ (-2,1),(-1,2),(0,5),(1,7),(-2,2) \right \}\\ &\textrm{c}.\quad \textrm{C}=\left \{ (-2,0),(-1,1),(0,2),(1,5),(2,7) \right \}\\\\ &\textrm{Jawab}:\\ &\textrm{Semuanya Fungsi kecuali}\textbf{ poin b)} \end{array}

\color{fuchsia}\begin{array}{ll}\\ 2.&\textrm{Tentukanlah daerah asal dari fungsi beberapa berikut}:\\ &\begin{array}{lllllllll}\\ \textrm{a}.&f(x)=x-3&\textrm{g}.&f(x)=\displaystyle \frac{\left | x \right |}{x}\\ \textrm{b}.&f(x)=\displaystyle \frac{6}{x^{2}-2x-8}&\textrm{h}.&f(x)=\left \lfloor x \right \rfloor\qquad \textrm{catatan}\: \: \left \lfloor x \right \rfloor\: \: \textrm{adalah bulat terbesar atau sama dengan}\: \: x\\ \textrm{c}.&f(x)=\displaystyle \frac{x^{2}-3x}{x^{2}-2x-15}&\textrm{i}.&f(x)=\left | x \right |+\left \lfloor x \right \rfloor\\ \textrm{d}.&y+2=x^{2}-5x+5&\textrm{j}.&f(x)=\sqrt{x^{2}-16}\\ \textrm{e}.&f(x)=\left | x-3 \right |&\textrm{k}.&f(x)=\sqrt{2x^{2}-50}\\ \textrm{f}.&f(x)=3-\left | 2x-1 \right |&\textrm{l}.&f(x)=\displaystyle \frac{2\sqrt{x}}{x-3} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|c|}\hline (\textrm{a})&(\textrm{b})&(\textrm{c})\\\hline \begin{aligned}f(x)&=x-3\\ \textrm{selu}&\textrm{ruh bilangan real}\\ &x\: \: \textrm{akan terdefinisi}\\ &\textrm{atau tetap bernilai}\\ &\textrm{real}\\ \textrm{sehi}&\textrm{ngga},\\ D_{f}&=\left \{ x|x\in \mathbb{R} \right \}\\ & \end{aligned}&\begin{aligned}f(x)&=\displaystyle \frac{6}{x^{2}-2x-8}\\ \textrm{terd}&\textrm{efinisi ketika}\\ &\textrm{penyebut \textit{tidak} sama}\\ &\textrm{dengan}\: \: 0, \: \: \textrm{yaitu}:\\ &\begin{aligned}x^{2}-2x-8&\neq 0\\ (x-4)(x+2)&\neq 0\\ x\neq 4\: \: \textrm{dan}\: \: x&\neq -2 \end{aligned}\\ D_{f}&=\left \{ x|x\in \mathbb{R},\: x\neq 4\: \: \textrm{dan}\: \: x\neq -2 \right \} \end{aligned}&\begin{aligned}f(x)&=\left | x-3 \right |\\ D_{f}&=\left \{ x|x\in \mathbb{R} \right \}\\ \textrm{teta}&\textrm{pi pada \textit{range} fungsinya}\\ &\textrm{hanya akan berupa}\\ &\textrm{bilangan positif saja}.\\ \textrm{yait}&\textrm{u}:\\ R_{f}&=\left \{ y|y\in \mathbb{R},\: y\geq 0 \right \}\\ & \end{aligned}\\\hline \end{array} \end{array}.

\color{fuchsia}\begin{array}{ll}\\ 3.&\textrm{Jika}\: \left | x \right |\: \textrm{menyatakan nilai mutlak}\\ &\textrm{dan}\: \left \lfloor x \right \rfloor\: \textrm{menyatakan bilangan bulat terbesarnatau sama dengan}\: x\\ &\textrm{misalkan}\: \: \left \lfloor 1,6 \right \rfloor=1,\: \left \lfloor \pi \right \rfloor=3\\ &\textrm{Jika diberikan}\: \: f(x)=\left | x \right |+\left \lfloor x \right \rfloor,\: \textrm{maka tentukanlah nilai untuk}\\ &\textrm{a}.\quad f\left ( -3,5 \right )+f\left ( 2,5 \right )\\ &\textrm{b}.\quad f\left ( -1,5 \right )+f\left ( 3,5 \right )\\\\ &\textrm{Jawab}:\\ &\begin{array}{l}\\ \begin{aligned}\textrm{a}.f\left ( -3,5 \right )+f\left ( 2,5 \right )&=\left | -3,5 \right |+\left \lfloor -3,5 \right \rfloor+\left | 2,5 \right |+\left \lfloor 2,5 \right \rfloor\\ &=3,5+\left ( -4 \right )+2,5+2\\ &=4 \end{aligned}\\ \begin{aligned}\textrm{b}.f\left ( -1,5 \right )+f\left ( 3,5 \right )&=\left | -1,5 \right |+\left \lfloor -1,5 \right \rfloor+\left | 3,5 \right |+\left \lfloor 3,5 \right \rfloor\\ &=1,5+(-2)+3,5+3\\ &=6 \end{aligned} \end{array} \end{array}.

\color{green}\begin{array}{ll}\\ 4.&\textrm{Jika diketahui relasi}\: \: f\: \: \textrm{dengan kondisi}\\ &\begin{array}{l} (\textrm{a}).\quad f(1)=1\\ (\textrm{b}).\quad f(2x)=4f(x)+6\\ (\textrm{c}).\quad f(x+2)=f(x)+12x+12 \end{array}\\ &\textrm{maka nilai}\: \: f(14)\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(1)&=1\\ f(2.1)=f(2)&=4f(1)+6=4.1+6=10\\ f(1+2)=f(3)&=f(1)+12.1+12\\ f(3)&=1+12+12=25\\ f(3+2)=f(5)&=f(3)+12.3+12\\ f(5)&=25+36+12=73\\ f(5+2)=f(7)&= f(5)+12.5+12\\ f(7)&=73+60+12=145\\ f(7.2)=f(14)&=4.f(7)+6\\ f(14)&=4.145+6=580+6\\ &=586 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ 5.&\textbf{(OSK 2013)}\textrm{Fungsi}\: \: f\: \: \textrm{didefinisikan oleh}\: \: f(x)=\displaystyle \frac{kx}{2x+3},\quad x=-\displaystyle \frac{3}{2}.\\ & \textrm{Tentukanlah nilai}\: \: k\: \: \textrm{agar}\: \: f(f(x))=x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{kx}{2x+3},\quad x\neq -\frac{2}{3}\\ f\left ( f(x) \right )&=x\\ x&=f\left ( f(x) \right )\\ x&=\displaystyle \frac{k\left ( \displaystyle \frac{kx}{2x+3} \right )}{2\left ( \displaystyle \frac{kx}{2x+3} \right )+3}\\ x&=\displaystyle \frac{\displaystyle \frac{k^{2}x}{2x+3}}{\displaystyle \frac{2kx+3(2x+3)}{2x+3}}\\ x&=\displaystyle \frac{k^{2}x}{2kx+6x+9}\\ 2kx+6x+9&=k^{2}\\ 0&=k^{2}-2xk-6x-9\\ 0&=(k+3)(k-2x-3)\\ &\quad k=-3\: \: \textrm{atau}\: \: k=2x+3 \end{aligned} \end{array}.

Logaritma (Kelas X Pemintan)

A. Mengingat Kembali Bilangan Bentuk Pangkat

\color{blue}\textrm{Perhatikanlah Soal-Jawab berikut}:\\ \begin{array}{ll}\\ 1.&\displaystyle 2^{5}=\underset{5}{\underbrace{2\times 2\times 2\times 2\times 2}}=32\\ 2.&\displaystyle 2^{4}=\underset{4}{\underbrace{2\times 2\times 2\times 2}}=16\\ 3.&\displaystyle 2^{3}=\underset{3}{\underbrace{2\times 2\times 2}}=8\\ 4.&\displaystyle 2^{2}=\underset{2}{\underbrace{2\times 2}}=4\\ 5.&\displaystyle 2^{1}=\underset{1}{\underbrace{2}}=2\\ 6.&\displaystyle 2^{0}=1\\ 7.&\displaystyle 2^{-1}=\displaystyle \frac{1}{2}\\ 8.&\displaystyle 2^{-2}=\displaystyle \frac{1}{4}\\ \vdots &\vdots \end{array}.

B. Logaritma

Logaritma adalah invers dari perpangkatan.

Kita ambil contoh pada soal-jawab di atas 2^{x}=32 , maka untuk mencari nilai x itu kita mungkin langsung dapat menjawab 5, tetapi bagai mana jika pertanyaannya \colorbox{Magenta}{2}^{y}=\colorbox{lime}{27}, bagaimana kita menentukan nilai y tersebut?

Oleh karnanya kita membutuhkan balikan(invers) dari bentuk perpangkatan bilangan di atas, yaitu Logaritma yang selanjutnya dalam penyebutannya cukup disebutkan dengan “log” saja.

Selanjutnya coba perhatikanlah untuk \colorbox{Magenta}{a} bilangan basis(pokok) dengan \colorbox{magenta}a,b>0 dan \colorbox{magenta}a\neq 1, maka berlaku:

^{\colorbox{magenta}a}\log\colorbox{lime}b=\color{blue}c.

C. Sifat-Sifat Operasi Aljabar Pada Logaritma

\color{magenta}\textrm{Perhatikanlah tabel berikut}\\ \begin{array}{|c|c|}\hline \LARGE\colorbox{blue}{Logaritma}&\LARGE\colorbox{green}{Eksponen}\\\hline \LARGE ^{a}\log b=c&\LARGE a^{c}=b\\\hline \multicolumn{2}{|c|}{\begin{matrix}\\ \bullet \quad a^{^{^{a}\log b}}=b\\\\ \bullet \quad ^{a}\log b\times d=\: ^{a}\log b+ \: ^{a}\log d\\\\ \bullet \quad ^{a}\log \displaystyle \frac{b}{d}=\: ^{a}\log b-\: ^{a}\log d\\\\ \bullet \quad ^{a}\log b=\displaystyle \frac{^{^{x}}\log b}{^{^{x}}\log a}\\\\ \bullet \quad ^{a}\log b=\displaystyle \frac{1}{^{^{b}}\log a}\\\\ \bullet \quad ^{a}\log b=m\Rightarrow \: ^{b}\log a=\displaystyle \frac{1}{m}\\\\ \bullet \quad ^{^{a^{n}}}\log b^{m}=\: \displaystyle \frac{m}{n}\times \: ^{a}\log b\\\\ \bullet \quad ^{a}\log b\times \: ^{b}\log c\times \: ^{c}\log d=\: ^{a}\log d\\\\ \bullet \quad ^{a}\log a=1\\\\ \bullet \quad ^{a}\log 1=0\\\\ \bullet \quad ^{a}\log a^{n}=n\\\\ \bullet \quad \log b=\: ^{10}\log b\\ \end{matrix}}\\\hline \end{array}.

\LARGE\colorbox{yellow}{CONTOH SOAL}.

\color{blue}\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai logaritma berikut}\\ &\textrm{a}.\quad ^{10}\log 10000\qquad \textrm{d}.\quad ^{2}\log 0,125\\ &\textrm{b}.\quad ^{10}\log 0,001\qquad \textrm{e}.\quad ^{3}\log 81\\ &\textrm{c}.\quad ^{2}\log 8\qquad\qquad\: \: \textrm{f}.\quad ^{3}\log \displaystyle \frac{1}{81}\\\\ &\textbf{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\: \: \textrm{Karena}&\: 10^{4}=10000\\ \textrm{maka}\: &^{10}\log 10000=4 \end{aligned}&\begin{aligned}\textrm{d}.\: \: \textrm{Karena}&\: 2^{-3}=\displaystyle \frac{1}{8}=0,125\\ \textrm{maka}\: &^{2}\log 0,125=-3 \end{aligned}\\\hline \begin{aligned}\textrm{b}.\: \: \textrm{Karena}&\: 10^{-3}=\displaystyle \frac{1}{10^{3}}=0,001\\ \textrm{maka}\: &^{10}\log 0,001=-3 \end{aligned}&\begin{aligned}\textrm{e}.\: \: \textrm{Karena}&\: 3^{4}=81\\ \textrm{maka}\: &^{3}\log 81=4 \end{aligned}\\\hline \begin{aligned}\textrm{c}.\: \: \textrm{Karena}&\: 2^{3}=8\\ \textrm{maka}\: &^{2}\log 8=3 \end{aligned}&\begin{aligned}\textrm{f}.\: \: \textrm{Karena}&\: 3^{-4}=\displaystyle \frac{1}{3^{4}}=\frac{1}{81}\\ \textrm{maka}\: &^{3}\log \displaystyle \frac{1}{81}=-4 \end{aligned}\\\hline \end{array} \end{array}.

\color{blue}\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai}\: \: a\: \: \textrm{yang memenuhi}\: \: ^{a}\log 81=4\\\\ &\textrm{Jawab}:\\ &\begin{aligned}^{a}\log 81=4\Leftrightarrow a^{4}&=81\\ a^{4}&=(\pm 3)^{4}\\ \textrm{karena}\: a> 0&,\: \: \textrm{maka}\: \: a=3 \end{aligned} \end{array}.

\color{blue}\begin{array}{ll}\\ 3.&\textrm{Tuliskan kebalikan dari bentuk kesamaan berikut}\\ &\textrm{a}.\quad 4^{^{\frac{3}{2}}}=8\qquad\qquad\qquad \textrm{c}.\quad ^{2}\log 32=5\\ &\textrm{b}.\quad \left ( 2\sqrt{2} \right )^{^{-\frac{2}{3}}}=\displaystyle \frac{1}{2}\: \: \: \qquad \textrm{d}.\quad ^{100}\log 0,1=-\displaystyle \frac{1}{2}\\\\ &\textrm{Jawab}:\\ &\textrm{a}.\quad \begin{aligned}4^{^{\frac{3}{2}}}=8\: \: \Leftrightarrow \: \: ^{4}\log 8=\displaystyle \frac{3}{2} \end{aligned}\\ &\textrm{b}.\quad \begin{aligned}\left ( 2\sqrt{2} \right )^{^{-\frac{2}{3}}}=\displaystyle \frac{1}{2}\: \: \Leftrightarrow \: \: ^{2\sqrt{2}}\log \displaystyle \frac{1}{2}=-\frac{2}{3} \end{aligned}\\ &\textrm{c}.\quad \begin{aligned}^{2}\log 32=5\: \: \Leftrightarrow \: \: 2^{5}=32 \end{aligned}\\ &\textrm{d}.\quad \begin{aligned}^{100}\log 0,1=-\displaystyle \frac{1}{2}\: \: \Leftrightarrow \: \: 100^{^{-\frac{1}{2}}}=0,1 \end{aligned} \end{array}.

\color{green}\begin{array}{ll}\\ 4.&\color{black}\textrm{Tentukanlah nilai logaritma berikut}\\ &\textrm{a}.\quad ^{2}\log 64\qquad\qquad\qquad \textrm{d}.\quad ^{27}\log \displaystyle \frac{1}{9}\\ &\textrm{b}.\quad ^{25}\log 125\, \quad\qquad\qquad \textrm{e}.\quad ^{\frac{1}{25}}\log 5\\ &\textrm{c}.\quad ^{81}\log 2187\: \: \: \: \quad\quad\qquad \textrm{f}.\quad ^{5}\log \displaystyle \frac{1}{25}\\\\ &\color{black}\textrm{Jawab}:\\ &\textrm{a}.\quad ^{2}\log 64=6\\ &\textrm{b}.\quad ^{25}\log 125=\displaystyle \frac{3}{2}\\ &\textrm{c}.\quad ^{81}\log 2187=\displaystyle \frac{7}{4}\\ &\textrm{d}.\quad ^{27}\log \displaystyle \frac{1}{9}=-\frac{2}{3}\\ &\textrm{e}.\quad ^{\frac{1}{25}}\log 5=-\displaystyle \frac{1}{2}\\ &\textrm{f}.\quad ^{5}\log \displaystyle \frac{1}{25}=-2 \end{array}.

Sebagai catatan adalah cara di atas dengan menggunakan pengertian dasar logaritma tanpa diselesakan dengan menggunkan sifat-sifat logaritma

PUSTAKA

  • Budhi, W.S. Widodo, U. 2017. Matematika untuk SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Turunan Fungsi Trigonometri

A. Pengenalan

Mengingat kembali pendukung konsep turunan untuk fungsi trigonometri, dengan mengerjakan soal-soal berikut

\begin{array}{ll}\ 1.&\textrm{Jika}\: \: f(x)=x^{2}\: \: \textrm{Tentukanlah nilai}\: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ 2.&\textrm{Gradien suatu kurva}\: \: f(x)\: \: \textrm{di}\: \: (a,b)\: \: \textrm{dinyatakan dengan}\\ &m=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}.\: \textrm{Jika diketahui}\: \: f(2x+1)=4x+7,\\ &\textrm{tentukanlah gradien garis singgung}\: \: f(x)\: \: \textrm{di}\: \: (a,b) \end{array}.

Pembahasan untuk soal no.1 sebagai berikut
\begin{aligned}\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}&=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{(x+h)^{2}-x^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+2xh+h^{2}-x^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{2xh+h^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: 2x+h\\ &=2x+0\\ &=2x \end{aligned}.

Untuk no. 2-nya adalah

\begin{aligned}f(2x+1)&=4x+7\\ f(2x+1)&=2(2x+1)+5\\ \textrm{atau}&\: \textrm{dapat kita menjadi lebih sederhana dengan} \\ f(x)&=2x+5 \end{aligned}.
maka,
\begin{aligned}m&=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{2(x+h)+5-(2x+5)}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{2x+2h+5-2x-5}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{2h}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: 2\\ &=2 \end{aligned}.

Statistika (Kelas XII Wajib)

A. Pengenalan

\color{blue}\begin{array}{|c|l|l|}\hline \textrm{No}&\: \: \: \textrm{Istilah}&\textrm{Pengertian}\\\hline 1.&\textrm{Statistika}&\textrm{Cabang ilmu tentang cara mengumpulkan, menyusun} \\ &&\textrm{penyajian, dan penganalisaan dari suatu data}\\\hline 2.&\textrm{Statistik}&\textrm{Data yang telah tersusun ke dalam daftar atau diagram}\\\hline 3.&\textrm{Populasi}&\textrm{Keseluruhan objek dari hasil penelitian yang memenuhi}\\ &&\textrm{syarat tertentu}\\\hline 4.&\textrm{Sampel}&\textrm{Bagian dari populasi yang dapat mewakili}\\ && \textrm{seluruh populasi}\\\hline \end{array}.

Sebagai tambahan penjelasan
\color{blue}\begin{array}{|l|l|}\hline \qquad\qquad\textrm{Istilah}&\qquad\qquad\textrm{Pengertian dan atau Penjelasan}\\\hline \textrm{Statistika}&\textrm{Lihat pengertian di atas}\\ Statistik&\textrm{Hasil pengolahan data}.\\ \textrm{Statistika deskriptif}&\textrm{Statistika baik yang berkenaan dengan kegiatan}\\ & \textrm{pengumpulan,penyajian, penyederhanaan atau}\\ & \textrm{penganalisaan, sertapenentuan khusus dari suatu}\\ &\textrm{data tanpa penarikan suatu kesimpulan}.\\ populasi&\textrm{Keseluruhan objek yang akan diteliti}.\\ \textrm{Sampel (Contoh)}&\textrm{Bagian dari populasi yang diamati}.\\ \textrm{Data}&\textrm{Kumpulan dari datum}.\\ \textrm{Datum}&\textrm{Informasi atau catatan keterangan dari penelitian}.\\ \textrm{Data kualitatif}&\textrm{Data yang menunjukkan sifat atau kondisi objek}.\\ \textrm{Data kuantitatif}&\textrm{Data yang menunjukkan jumlah objek}.\\ \textrm{Data ukuran (Data kontinu)}&\textrm{Data yang diperoleh dengan cara mengukur}\\ & \textrm{besaran objek}.\\ \textrm{Data cacahan (Data diskrit)}&\textrm{Data yang diperoleh dengan cara mencacah,}\\ & \textrm{membilang atau menghitung banyak objek}.\\\hline \end{array}.

B. Pengumpulan Data
Pengumpulan data dapat dilakukan dengan metode wawancara, pengamatan langsung(observasi) dan bisa juga dengan menggunakan angket.
Setiap keterangan yang diperoleh dalam pengamatan dinamakan datum, dan dari sekumpulan datum inilah nantinya yang disebut data.

C. Penyajian Data
Dalam statika data statistik dapat disajikan dalam berbagai bentuk, mmenyesuaikan jenisnya data. Data statistik dapat berupa daftar bilangan yang memiliki kondisi tertentu sebagai misal berupa data tunggal. Selain data statistik dapat dinyatakan dalam daftar bilangan, data juga bisa dinyatakan dalam bentuk tabel/daftar distribusi frekuensi, atupun diagram.

\begin{array}{|l|l|}\hline \textrm{Bentuk Diagram}&\textrm{BentukDaftar Distribusi Frekuensi}\\\hline \begin{aligned}\blacklozenge &\: \textrm{Diagram garis}\\ \blacklozenge &\: \textrm{Diagram batang daun}\\ \blacklozenge &\: \textrm{Diagram kotak garis}\\ &\\ & \end{aligned}&\begin{aligned}\blacklozenge &\: \textrm{Daftar distribusi data tunggal}\\ \blacklozenge &\: \textrm{Daftar distribusi frekuensi data berkelompok}\\ \blacklozenge &\: \textrm{Daftar distribusi frekuensi relatif}\\ \blacklozenge &\: \textrm{Daftar distribusi kumulatif}\\ \blacklozenge &\: \textrm{Histogram, poligon frekuensi, dan ogif }\end{aligned}\\\hline \end{array}.

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

\begin{array}{|c|c|}\hline \color{blue}\textrm{Daftar distribusi data tunggal}&\color{magenta}\textrm{Daftar distribusi data berkelompok}\\\hline n\geq 30&n\geq 30\\\hline \begin{aligned}\textrm{Sebagai}&\: \textrm{misal data jumlah anak}\\ \textrm{dari}\: &\: \textrm{30 karyawan sebuah}\\ \textrm{peru}&\textrm{sahaan}\\ 3&\, 2\, 0\, 1\, 4\, 2\, 2\, 2\, 1\, 2\\ 3&\, 0\, 3\, 2\, 1\, 1\, 2\, 1\, 2\, 2\\ 2&\, 1\, 2\, 2\, 0\, 3\, 1\, 1\, \, 2\, 3 \end{aligned}&\begin{aligned}\textrm{Langkah}&-\textrm{langkah membuat daftar}\\ \textrm{dengan}\: &\textrm{mentukan hal-hal berikut}\\ 1.&Jangkauan(J)=x_{maks}-x_{min}\\ 2.&\textit{Banyak kelas}(k)=1+3,3\times \log n\\ 3.&\textit{Panjang kelas}(c)=\displaystyle \frac{J}{k}\\ 4.&\textit{Batas kelas pertama}=\textit{datum terkecil} \end{aligned}\\\hline \color{blue}\begin{array}{|c|c|}\hline \textrm{Juml anak}&\textrm{Frekuensi}\\\hline 0&3\\ 1&8\\ 2&13\\ 3&5\\ 4&1\\\hline \textrm{Jumlah}&30\\\hline \end{array}&\begin{aligned}\textrm{Buat}&\textrm{lah Daftar distribusi frekuensi}\\ &\textrm{untuk data berikut}\\ &79\, 68\, 60\, 73\, 62\, 78\, 56\, 64\, 68\, 53\, 72\, 67\, 64\, 62\\ &58\, 53\, 52\, 72\, 59\, 74\, 52\, 71\, 62\, 51\, 70\, 60\, 72\, 68\\ &74\, 67\, 70\, 70\, 57\, 55\, 62\, 52\, 61\, 77\, 63\, 63\\ & \end{aligned}\\\hline \end{array}.
\begin{aligned}\textrm{Dari}&\: \textrm{data di atas kita mendapatkan}\\ 1.\quad&J=x_{maks}-x_{min}=79-51=28\\ 2.\quad&k=1+3,3\times \log 40=1+3,3\times (1,602)=1+(5,2866)=6,2866\\ &\: \: \,\approx 6\qquad (\textrm{ada 6 kelas dari hasil pembulatan})\\ 3.\quad&c=\displaystyle \frac{J}{k}=\frac{28}{6}=4,\bar{66}\approx 5\quad (\textrm{pembulatan})\\ 4.\quad&x_{min}=51,\: \textrm{maka kelas interval pertama adalah}=51-55\\ 5.\quad&\textrm{Hasil yang diperoleh}\\ &\color{blue}\begin{array}{|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}\\\hline 51-55&7\\ 56-60&6\\ 61-65&9\\ 66-70&8\\ 71-75&7\\ 76-80&3\\\hline \textrm{Jumlah}&40\\\hline \end{array} \end{aligned}.

D. Daftar Bilangan
Dafar bilangan di sini yang dimaksud adalah pada data tunggal. Sebagai contoh data berat badan 5 orang dalam kg yaitu: 40. 42, 45, 41, 43.

E. Tabel Distribusi Frekuensi
Lihat contoh ilustrasi di atas pada poin C, kita sajikan lagi
\color{blue}\begin{array}{|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}\\\hline 51-55&7\\ 56-60&6\\ 61-65&9\\ 66-70&8\\ 71-75&7\\ 76-80&3\\\hline \textrm{Jumlah}&40\\\hline \end{array}.

F. Daftar Distribusi Frekuensi Relatif

frekuensi relatif suatu kelas interval dapat dihitung dengan rumus

\begin{aligned}f_{r}&=\displaystyle \frac{f_{i}}{n}\times 100\% \end{aligned}.
Sehingga pada tabel poin E akan didapatkan tabel distribusi frekuensin relatifnya sebagaimana berikut:

\color{blue}\begin{array}{|c|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}&\textrm{Frekuensi Relatif}\\\hline 51-55&7&0,175\\ 56-60&6&0,15\\ 61-65&9&0,225\\ 66-70&8&0,2\\ 71-75&7&0,175\\ 76-80&3&0,075\\\hline \textrm{Jumlah}&40&\\\hline \end{array}.

G. Daftar Distribusi Frekuensi Kumulatif

a. kurang dari

Jumlah seluruh frekuensi yang nilainya kurang dari atau sama dengan nilai tepi atas pada tiap kelas dari data yang diberikan.

Sebagai misal data berikut yang diambilkan dari dari sebelumnya, yaitu:

\color{blue}\begin{array}{|c|c|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}&\textrm{Nilai}&f_{k}\leq \\\hline 51-55&7&55,5&7\\ 56-60&6&60,5&13\\ 61-65&9&65,5&22\\ 66-70&8&70,5&30\\ 71-75&7&75,5&37\\ 76-80&3&80,5&40\\\hline \textrm{Jumlah}&40&&\\\hline \end{array}.

b. lebih dari

Kebalikan dari yang kurang dari di atas dan hasilnya adalah sebagai berikut

\color{red}\begin{array}{|c|c|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}&\textrm{Nilai}&f_{k}\geq \\\hline 51-55&7&50,5&40\\ 56-60&6&55,5&33\\ 61-65&9&60,5&27\\ 66-70&8&65,5&18\\ 71-75&7&70,5&10\\ 76-80&3&75,5&3\\\hline \textrm{Jumlah}&40&&\\\hline \end{array}.

H. Histogram, Poligon Frekuensi, dan Ogif

\color{blue}\begin{array}{|c|l|l|}\hline 1.&\textrm{Histogram}&\textrm{Diagram batang yang alasnya adalah menunjukkan}\\ &&\textrm{panjang kelas interval sedangkan tingginya merupa}-\\ &&\textrm{kan frekuensi kelas interval}\\\hline 2.&\textrm{Poligon}&\textrm{Titik-titik tengah pada tiap diagram batang pada his}-\\ &frekuensi&\textrm{togram dihubungkan dengan garis lurus}\\\hline 3.&Ogif&\textrm{Kurva mulus dari daftar distribusi frekuensi kumulatif}\\\hline &positif&\textrm{jika disusun dari}\: \: f_{k}\leq \\ &negatif&\textrm{jika disusun dari}\: \: f_{k}\geq \\\hline \end{array}.

\colorbox{yellow}{\LARGE{\fbox{\LARGE{\fbox{CONTOH SOAL}}}}}\begin{array}{ll}\\ \fbox{1}.&\textrm{Dari tabel sebagaimana berikut(dari tabel di atas)}\\ &\begin{array}{|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}\\\hline 51-55&7\\ 56-60&6\\ 61-65&9\\ 66-70&8\\ 71-75&7\\ 76-80&3\\\hline \textrm{Jumlah}&40\\\hline \end{array}\\ &\textrm{Buatlah menjadi daftar distribusi frekuensi relatif}?\\ & \end{array}

\color{blue}\textbf{Pembahasan}\\ \begin{array}{|c|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}&\textrm{Frekuensi Relatif}\\\hline 51-55&7&0,175=17,5\%\\ 56-60&6&0,15=15\%\\ 61-65&9&0,225=22,5\%\\ 66-70&8&0,2=20\%\\ 71-75&7&0,175=17,5\%\\ 76-80&3&0,075=7,5\%\\\hline \textrm{Jumlah}&40&100\%\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{2}&\textrm{Pada tabel di atas buatkanlah ke daftar distribusi frekuensi kumulatif}\\ &\textrm{kurang dari}\\ \\ &\textrm{Pembahasan}:\\ &\begin{array}{|c|c|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}&\textrm{Nilai}&f_{k}\leq \\\hline 51-55&7&\leq 55,5&7\\ 56-60&6&\leq 60,5&13\\ 61-65&9&\leq 65,5&22\\ 66-70&8&\leq 70,5&30\\ 71-75&7&\leq 75,5&37\\ 76-80&3&\leq 80,5&40\\\hline \textrm{Jumlah}&40&&\\\hline \end{array}\Rightarrow \begin{array}{l}\\ \textbf{sebagai penjelasan}\\ \\ 13=7+6\\ 22=7+6+9\\ 30=7+6+9+8\\ 37=7+6+9+8+7\\ 40=7+6+9+8+7+3\\ \\ \\ \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{3}&\textrm{Pada tabel di atas buatkanlah ke daftar distribusi frekuensi kumulatif}\\ &\textrm{lebih dari}\\ \\ &\textrm{Pembahasan}:\\ &\begin{array}{|c|c|c|c|}\hline \textrm{Kelas Interval}&\textrm{Frekuensi}&\textrm{Nilai}&f_{k}\leq \\\hline 51-55&7&\geq 50,5&40\\ 56-60&6&\geq 55,5&33\\ 61-65&9&\geq 60,5&27\\ 66-70&8&\geq 65,5&18\\ 71-75&7&\geq 70,5&10\\ 76-80&3&\geq 75,5&3\\\hline \textrm{Jumlah}&40&&\\\hline \end{array}\Rightarrow \begin{array}{l}\\ \textbf{sebagai penjelasan}\\ 40\\ 33=40-7\\ 27=40-7-6\\ 18=40-7-6-9\\ 10=40-7-6-9-8\\ 3=40-7-6-9-8-7\\ \\ \\ \end{array} \end{array}.