Pertidaksamaan Logaritma

 Perhatikanlah grafik fungsi logaritma  

$f:x\to {}^a\log x$.

Ada 2 macam pilihan untuk nilai basisnya. Sesuai sifat-sifat logaritma, basis atau bilangan pokok akan mempengaruhi nilai suatu logaritma. Karena basis logaritma harus positif dan tidak boleh sama dengan 1, maka basis ini dapat dipecah menjadi 2 macam, yaitu: 

Saat basisnya lebih besar dari 1 atau a>1, maka grafiknya adalah sebagai berikut:

dan saat basisnya berada pada saat  0<a<1, maka gambar grafiknya adalah sebagai berikut:

Dari dua ilustrasi di atas dapat ditemukan dua hal, yaitu:

• pada fungsi moton naik (saat a>1), jika  $x_1<x_2$, maka  $f(x_1)<f(x_2)$.
• pada fungsi moton turun (saat 0<a<1), jika  $x_1<x_2$, maka  $f(x_1)>f(x_2)$.

Dari uraian di atas dapat disimpulkan bahwa:

$\begin{array}{|c|}\hline \begin{array}{ll}1.& \text{Untuk basis}a>1\\ & \text{dengan}f(x)>0,g(x)>0:\\ & \bullet {}^a\log f(x)\ge {}^a\log g(x),\text{maka}f(x)\ge g(x)\\ & \bullet {}^a\log f(x)>{}^a\log g(x),\text{maka}f(x)>g(x)\\ & \bullet {}^a\log f(x)\le {}^a\log g(x),\text{maka}f(x)\le g(x)\\ & \bullet {}^a\log f(x)<{}^a\log g(x),\text{maka}f(x)<g(x)\\ \\ 2.& \text{Untuk basis}0<a<1\\ & \text{dengan}f(x)>0,g(x)>0:\\ & \bullet {}^a\log f(x)\ge {}^a\log g(x),\text{maka}f(x)\le g(x)\\ & \bullet {}^a\log f(x)>{}^a\log g(x),\text{maka}f(x)<g(x)\\ & \bullet {}^a\log f(x)\le {}^a\log g(x),\text{maka}f(x)\ge g(x)\\ & \bullet {}^a\log f(x)<{}^a\log g(x),\text{maka}f(x)>g(x)\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{ll}1.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.{}^3\log (x+1)>2\\ & \text{b}.{}^{{.}^{\frac{1}{3}}}\log (x+1)>2\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Diketahui}:{}^3\log (x+1)>2\\ & \text{Syarat numerusnya}\\ & f(x)>0\iff x+1>0\iff x>-1\\ & \text{Proses lanjutan penyelesaian}\\ & {}^3\log (x+1)>2\iff {}^3\log (x+1)>{}^3\log 3^2\\ & \text{Karena}a=3,\text{maka}f(x)>p\\ & (x+1)>3^2\\ & \iff x+1>9\iff x>8\\ & \text{Karena},x>8\text{berada di daerah}x>-1,\\ & \text{maka}x>8\mathbf{\text{memenuhi}}\\ & \text{Jadi},\text{HP}=\{x|x>8\}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Diketahui}:{}^{{.}^{\frac{1}{3}}}\log (x+1)>2\\ & \text{Syarat numerusnya}\\ & f(x)>0\iff x+1>0\iff x>-1\\ & \text{Proses lanjutan penyelesaian}\\ & {}^{{.}^{\frac{1}{3}}}\log (x+1)>2\iff {}^{{.}^{\frac{1}{3}}}\log (x+1)>{}^{{.}^{\frac{1}{3}}}\log {\left({\displaystyle \frac{1}{3}}\right)}^2\\ & \text{Karena}a={\displaystyle \frac{1}{3},\text{maka}f(x)<p}\\ & (x+1)<{\left({\displaystyle \frac{1}{3}}\right)}^2\\ & \iff x+1<{\displaystyle \frac{1}{9}\iff x<{\displaystyle \frac{1}{9}-1\iff x<-{\displaystyle \frac{8}{9}}}}\\ & \text{maka yang}\mathbf{\text{memenuhi}}-1<x<-{\displaystyle \frac{8}{9}}\\ & \text{Jadi},\text{HP}=\{x|-1<x<-{\displaystyle \frac{8}{9}}\}\end{array}\end{array}$.

$\begin{array}{ll}2.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.-1<\log (x-5)<2\\ & \text{b}.{}^2\log x+{}^2\log (x-3)>2\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Diketahui}\\ & -1<\log (x-5)<2\\ & \iff -1<{}^{10}\log (x-5)<2\text{mohon ingat}\\ & \iff (-1){}^{10}\log 10<{}^{10}\log (x-5)<(2){}^{10}\log 10\\ & \iff {}^{10}\log {10}^{-1}<{}^{10}\log (x-5)<{}^{10}\log {10}^2\\ & \iff {10}^{-1}<x-5<{10}^2\\ & \iff {\displaystyle \frac{1}{10}<x-5<100}\\ & \iff {\displaystyle \frac{1}{10}+5<x-5+5<100+5}\\ & \iff 5{\displaystyle \frac{1}{10}<x<105}\\ & \text{Jadi},\text{HP}=\left\{x|5{\displaystyle \frac{1}{10}<x<105}\right\}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Diketahui}\\ & {}^2\log x+{}^2\log (x-3)>2\\ & \iff {}^2\log x+{}^2\log (x-3)>{}^2\log 2^2\\ & \iff {}^2\log x(x-3)>{}^2\log 2^2\\ & \text{Syarat numerusnya}\\ & f(x)>0\iff x>0\text{dan}(x-3)>0\\ & \text{atau}x>0\text{atau}x>3\\ & \text{Proses lanjutan penyelesaian}\\ & {}^2\log x(x-3)>{}^2\log 2^2\\ & \text{Karena}a=2,\text{maka}f(x)>p\\ & x(x-3)>2^4\iff x^2-3x>4\\ & x^2-3x-4>0\iff (x+1)(x-4)>0\\ & \iff x<-1\text{atau}x>4\\ & \text{Karena},x>4\text{berada di daerah}x>3,\\ & \text{maka}x>4\mathbf{\text{memenuhi dan yang}}\\ & \mathbf{\text{lainnya tidak memenuhi}}\\ & \text{Jadi},\text{HP}=\{x|x>4\}\end{array}\end{array}$.

$\begin{array}{ll}3.& \text{Tentukan himpunan penyelesaian dari}\\ & {({}^b\log x)}^2+10<7.{}^b\log x,\text{dengan}b>1\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {({}^b\log x)}^2+10<7.{}^b\log x\\ & \iff {({}^b\log x)}^2-7.({}^b\log x)+10<0\\ & \iff ({}^b\log x-2)({}^b\log x-5)<0\\ & \text{Penyelesaiannya}:2<{}^b\log x<5\\ & \iff (2){.}^b\log b<{}^b\log x<(5){.}^b\log b\\ & \iff {}^b\log b^2<{}^b\log x<{}^b\log b^5\\ & \iff b^2<x<b^5\\ & \text{Jadi},\text{ HP}=\{x|b^2<x<b^5\}\end{array}\end{array}$.

$\begin{array}{ll}4.& \text{Tentukan penyelesaian dari pertidaksamaan}\\ & {}^3\log (5x-2)<{}^3\log (3x+8)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \bullet \text{Syarat basis},a=3.\text{Jelas}3>0,\ne 1\\ & \bullet \text{Syarat numerusnya},\text{dari bentuk}{}^a\log f(x)>{}^a\log g(x)\\ & \begin{array}{lll}\begin{array}{|cc|}\hline f(x)& g(x)\\ \begin{array}{rl}& 5x-2>0\\ & \iff 5x>2\\ & \iff x>{\displaystyle \frac{2}{5}}\end{array}& \begin{array}{rl}& 3x+8>0\\ & \iff 3x>-8\\ & \iff x>-{\displaystyle \frac{8}{3}}\end{array}\\ \hline\end{array}& \Rightarrow & \begin{array}{rl}& \text{Dari keduanya}\\ & \text{kita pilih yang}\\ & x>{\displaystyle \frac{2}{5}}\end{array}\end{array}\\ & \bullet \text{Proses penyelesaian}\\ & .\begin{array}{rl}& f(x)<g(x)\\ & \iff 5x-2<3x+8\\ & \iff 5x-3x<8+2\\ & \iff 2x<10\\ & \iff x<{\displaystyle \frac{10}{2}}\\ & \iff x<5.\\ & \text{Jadi, solusinya adalah}:\{\begin{array}{ll}1& :x>{\displaystyle \frac{2}{5},\text{dan}}\\ 2& :x<5\end{array}\\ & \text{maka solusinya adalah}:{\displaystyle \frac{2}{5}<x<5}\end{array}\end{array}\end{array}$

$\text{LATIHAN SOAL}$.

$\begin{array}{ll}& \text{Tentukan himpunan penyelesaian dari}\\ & \text{pertidaksamaan berikut}!\\ \\ & 1.-1<\log (2x-3)<2\\ & 2.{}^3\log (2x-3)+{}^3\log x>3\\ & 3.{}^6\log (x^2+x-6)-1\ge 0\\ & 4.{}^4\log (4^x.4)\le 2-x\\ & 5.{}^{{.}^{\frac{1}{3}}}\log (x-2)<2\\ & 6.{}^{{.}^{\frac{1}{2}}}\log x^2-{}^{{.}^{\frac{1}{2}}}\log (x+3)>-4\\ & 7.4({}^{{.}^{\frac{1}{2}}}\log m)<{}^{{.}^{\frac{1}{3}}}\log 81\\ & 8.{}^2\log (x+2)<{}^{(x+2)}\log (8x^2+32x+32)\\ & 9.{}^2\log (2x+2)>6.{}^{(x+1)}\log 2\\ & 10.{}^2\log (1-{}^2\log x)<2\\ & 11.{}^6\log (x^2-x)<1\\ & 12.{}^x\log (x+12)-3.{}^x\log 4+1\ge 0\\ & 13.{}^5\log x^2\le {}^5\log 5x\\ & 14.{\displaystyle \frac{2^x-4}{{}^2\log x-2}\le 0}\\ & 15.{\displaystyle \frac{{}^2\log x+{}^2\log (x-1)}{{}^2\log 4}\le 1}\\ & 16.\log |x+1|\ge \log 3+\log |2x-1|\\ \end{array}$.

DAFTAR PUSTAKA

1. Sembiring,S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: Srikandi Empat Widya Utama.
2. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: Tiga Serangkai Pustaka Mandiri.

Persamaan Logaritma 5

E. Persamaan Logaritma Bentuk  $A{({}^a\log f(x))}^2+B({}^a\log f(x))+C=0$.

Himpunan penyelesaian dari bentuk ini adalah kurang lebih sama dengan persamaan kuadrat, baik dengan cara dimisalkan terlebih dahulu ataupun tidak, 

Jika dimisalkan, maka bentuknya akan semakin sederhana dan dan lebih efektif.

Adapun solusi dari persamaan kuadrat sendiri adalah:

1. memfaktorkan
2. melengkapkan kuadrat sempurna
3. rumus abc

Catatan : Syarat numerus dan basisnya mengikuti, yaitu untuk numerus harus positif dan basisnya selain harus positif juga tidak boleh sama dengan 1.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \text{a}.{({}^2\log x)}^2-6({}^2\log x)+8=0\\ & \text{b}.2{}^3{\log }^{2}x+2{}^3\log x-12=0\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Dengan tanpa pemisalan}\\ & {({}^2\log x)}^2-6({}^2\log x)+8=0\\ & \iff ({}^2\log x-2)({}^2\log x-4)=0\\ & \iff {}^2\log x=2\text{atau}{}^2\log x=4\\ & \iff x=2^2=4\text{atau}x=2^4=16\\ & \text{Jadi},\text{HP}=\{4,16\}\\ \text{b}.& \text{Dengan tanpa pemisalan juga}\\ & 2{}^3{\log }^{2}x+2{}^3\log x-12=0\\ & {}^3{\log }^{2}x{+}^3\log x-6=0\\ & \iff ({}^3\log x+3)({}^3\log x-2)=0\\ & \iff {}^3\log x=-3\text{atau}{}^3\log x=2\\ & \iff x=3^{-3}={\displaystyle \frac{1}{27}\text{atau}x=3^2=9}\\ & \text{Jadi},\text{HP}=\left\{{\displaystyle \frac{1}{27},9}\right\}\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah himpunan penyelesaian dari}\\ & \text{a}.{(\log x)}^2-6(\log x)+8=0\\ & \text{b}.{(\log x)}^2-\log x^3-10=0\\ & \text{c}.{({}^3\log x)}^2+2({}^3\log x)-3=0\\ & \text{d}.{}^5{\log }^{2}x-{}^5\log x^4+{}^5\log 125=0\\ \end{array}$.

DAFTAR PUSTAKA

1. Sembiring,S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: Srikandi Empat Widya Utama.

Persamaan Logaritma 4

 D. Persamaan Logaritma Bentuk  ${}^{h(x)}\log f(x)={}^{h(x)}\log g(x)$.

Syarat penyelesaian dari bentuk:

$\begin{array}{rl}& \text{Jika}{}^{h(x)}\log f(x)={}^{h(x)}\log g(x)\\ & \text{dengan}f(x)\text{dan}g(x)\text{keduanya positif}\\ & \text{serta}h(x)>0,\text{dan}h(x)\ne 1,\\ & \text{maka}f(x)=g(x)\end{array}$.

$\begin{array}{rl}& \mathbf{\text{atau}}\\ & \text{Pernyataan}{}^{h(x)}\log f(x)={}^{h(x)}\log g(x)\\ & \text{akan bernilai benar jika}\\ & (1)h(x)>0,h(x)\ne 1\\ & (2)f(x)>0,g(x)>0\\ & (3)f(x)=g(x)\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.{}^x\log (2x-3)={}^x\log (x-1)\\ & \text{b}.{}^x\log (2x^2+11x-6)={}^x\log (x^2+10x)\\ & \text{c}.{}^x\log (x-1)+{\displaystyle \frac{1}{{}^{x+6}\log x}=2+{\displaystyle \frac{1}{{}^2\log x}}}\\ & \text{d}.{}^{2x-1}\log (x^3+3x^2-4x-1)={}^{2x-1}\log (2x^2+3)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\mathbf{\text{a}}.\text{Dik}& \text{etahui}\\ & {}^x\log (2x-3)={}^x\log (x-1)\\ (\ast )& \text{Syarat numerus dan bilangan pokok}\\ & \begin{array}{|ll|}\hline \text{Syarat}& \text{hasil}\\ h(x)>0,h(x)\ne 1& x>0\\ f(x)>0& \begin{array}{rl}2x& -3>0\\ \iff & 2x>3\\ \iff & x>{\displaystyle \frac{3}{2}}\end{array}\\ g(x)>0& \begin{array}{rl}x-& 1>0\\ \iff & x>1\end{array}\\ \hline\end{array}\\ & \text{Syarat numerusnya},x>{\displaystyle \frac{3}{2}}\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff 2x-3=x-1\\ & \iff 2x-x=3-1\\ & \iff x=2\\ & \text{Karena}x>{\displaystyle \frac{3}{2},}\\ & \text{maka nilai}x=2\text{memenuhi}\\ (\ast )& \text{Jadi},\text{HP}=\left\{2\right\}\end{array}\\ & \begin{array}{rl}\mathbf{\text{b}}.\text{Dik}& \text{etahui}\\ & {}^x\log (2x^2+11x-6)={}^x\log (x^2+10)\\ (\ast )& \text{Syarat numerus dan bilangan pokok}\\ & \begin{array}{|ll|}\hline \text{Syarat}& \text{hasil}\\ h(x)>0,h(x)\ne 1& x>0\\ f(x)>0& \begin{array}{rl}2x^2& +11x-6>0\\ \iff & (x+6)(2x-1)>0\\ \iff & x<-6\text{atau}x>{\displaystyle \frac{1}{2}}\end{array}\\ g(x)>0& \begin{array}{rl}{x}^2& +10x>0\\ \iff & x(x+10)>0\\ \iff & x<-10\text{atau}x>0\end{array}\\ \hline\end{array}\\ & \text{Syarat numerusnya},x>{\displaystyle \frac{1}{2}}\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff 2x^2+11x-6=x^2+10x\\ & \iff x^2+x-6=0\\ & \iff (x+3)(x-2)=0\\ & \iff x=-3\text{atau}x=2\\ & \text{Karena}x>{\displaystyle \frac{1}{2},}\\ & \text{maka nilai}x=2\text{memenuhi}\\ (\ast )& \text{Jadi},\text{HP}=\left\{2\right\}\end{array}\\ & \begin{array}{rl}\mathbf{\text{c}}.\text{Dik}& \text{etahui}\\ & {}^x\log (x-1)+{\displaystyle \frac{1}{{}^{x+6}\log x}=2+{\displaystyle \frac{1}{{}^2\log x}}}\\ & \iff {}^x\log (x-1)+{}^x\log (x+6)={}^x\log x^2+{}^x\log 2\\ & \iff {}^x\log (x-1)(x+6)={}^x\log 2x^2\\ & \iff {}^x\log x^2+5x-6={}^x\log 2x^2\\ (\ast )& \text{Syarat numerus dan bilangan pokok}\\ & \begin{array}{|ll|}\hline \text{Syarat}& \text{hasil}\\ h(x)>0,h(x)\ne 1& x>0\\ f(x)>0& \begin{array}{rl}{x}^2& +5x-6>0\\ \iff & (x+6)(x-1)>0\\ \iff & x<-6\text{atau}x>1\end{array}\\ g(x)>0& \begin{array}{rl}2x^2& >0\\ \iff & x>0\end{array}\\ \hline\end{array}\\ & \text{Syarat numerusnya},x>{\displaystyle 1}\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff x^2+5x-6=2x^2\\ & \iff x^2-5x+6=0\\ & \iff (x-2)(x-3)=0\\ & \iff x=2\text{atau}x=3\\ & \text{Karena}x>{\displaystyle 1,}\\ & \text{maka nilai}x=2\text{dan}x=3\text{memenuhi}\\ (\ast )& \text{Jadi},\text{HP}=\{2,3\}\end{array}\\ & \begin{array}{rl}\mathbf{\text{d}}.\text{Dik}& \text{etahui}\\ & {}^{2x-1}\log (x^3+3x^2-4x-1)={}^{2x-1}\log (2x^2+3)\\ (\ast )& \text{Syarat numerus dan bilangan pokok}\\ & \begin{array}{|ll|}\hline \text{Syarat}& \text{hasil}\\ h(x)>0,h(x)\ne 1& 2x-1>0\iff x>{\displaystyle \frac{1}{2}}\\ f(x)>0& \begin{array}{rl}{x}^3& +3x^2-4x-1>0\\ \bullet & \text{Susah difaktorkan}\\ \bullet & \text{gunakan uji nilai}\end{array}\\ g(x)>0& \begin{array}{rl}2x^2& +3>0\\ \bullet & a>0,D<0\\ \bullet & \mathbf{\text{Definit positif}}\end{array}\\ \hline\end{array}\\ & \text{Syarat basis/bilangan pokoknya},x>{\displaystyle \frac{1}{2}}\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff x^3+3x^2-4x-1=2x^2+3\\ & \iff x^3+x^2-4x-4=0\\ & \iff x^2(x+1)-4(x+1)=0\\ & \iff (x+1)(x^2-4)=0\\ & \iff (x+1)(x+2)(x-2)=0\\ & \iff x=-1\text{atau}x=-2\text{atau}x=2\\ & \text{Karena basisnya}x>{\displaystyle \frac{1}{2},}\\ & \text{maka nilai yang memenuhi hanya}x=2\text{saja}\\ & \text{dan nilai untuk numerusnya juga memenuhi}\\ & \text{yaitu}:(2{)}^3+3(2{)}^2-4(2)-1=11>0\\ & \text{demikian pula untuk}:2(2{)}^3+3=19>0\\ (\ast )& \text{Jadi},\text{HP}=\left\{2\right\}\end{array}\end{array}$.

Penjelasan untuk jawaban 1. d  tentang definit positif  di sini dan di sini

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 2.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.{}^x\log (2x+3)={}^x\log (x+7)\\ & \text{b}.{}^x\log (x+12)-{}^x\log (4x+1)=0\\ & \text{c}.{}^{x-2}\log (x^2-3)={}^{x-2}\log x\\ & \text{d}.{}^{3x-2}\log (x^2-2x+4)={}^{3x-2}\log (5-4x)\\ \end{array}$

Persamaan Logaritma 3

C. Persamaan Logaritma Bentuk  ${}^a\log f(x)={}^a\log g(x)$.

Syarat penyelesaian dari bentuk ini adalah numerusnya harus positif serta basisnya juga harus positif dan tidak berupa angka 1.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.2\log x=\log (x+6)\\ & \text{b}.\log (2x-3)=\log (x^2-3x+1)\\ & \text{c}.{}^3\log (x^2+3x+2)={}^3\log (5x+5)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\mathbf{\text{a}}.\text{Dik}& \text{etahui}\\ & 2\log x=\log (x+6)\\ & \iff \log x^2=\log (x+6)\\ (\ast )& \text{Syarat numerus}\\ & \begin{array}{ll}\begin{array}{rl}& f(x)>0\\ \iff & x^2>0\\ \iff & x>0\end{array}& \begin{array}{rl}& g(x)>0\\ \iff & x+6>0\\ \iff & x>-6\end{array}\end{array}\\ & \text{Sehingga syarat numerusnya},x>0\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff x^2=x+6\\ & \iff x^2-x-6=0\\ & \iff (x+2)(x-3)=0\\ & \iff x=-2\text{atau}x=3\\ & \text{Karena}x>0,\text{yang memenuhi}x=3\\ (\ast )& \text{Jadi},\text{HP}=\left\{3\right\}\end{array}\\ & \begin{array}{rl}\mathbf{\text{b}}.\text{Dik}& \text{etahui}\\ & \log (2x-3)=\log (x^2-3x+1)\\ (\ast )& \text{Syarat numerus}\\ & \begin{array}{ll}\begin{array}{rl}& f(x)>0\\ \iff & 2x-3>0\\ \iff & 2x>3\\ \iff & x>{\displaystyle \frac{3}{2}}\end{array}& \begin{array}{rl}& g(x)>0\\ \iff & x^2-3x+1>0\\ \iff & x<{\displaystyle \frac{3-\sqrt{5}}{2}}\\ & \text{atau}x>{\displaystyle \frac{3+\sqrt{5}}{2}}\end{array}\end{array}\\ & \text{Syarat numerusnya},x>{\displaystyle \frac{3+\sqrt{5}}{2}}\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff (2x-3)=(x^2-3x+1)\\ & \iff -x^2+5x-4=0\\ & \iff x^2-5x+4=0\\ & \iff (x-1)(x-4)=0\\ & \iff x=1\text{atau}x=4\\ & \text{Karena}x>{\displaystyle \frac{3+\sqrt{5}}{2},}\\ & \text{yang memenuhi adalah}x=4\\ (\ast )& \text{Jadi},\text{HP}=\left\{4\right\}\end{array}\\ & \begin{array}{rl}\mathbf{\text{c}}.\text{Dik}& \text{etahui}\\ & {}^3\log (x^2+3x+2)={}^3\log (5x+5)\\ (\ast )& \text{Syarat numerus}\\ & \begin{array}{ll}\begin{array}{rl}& f(x)>0\\ \iff & x^2+3x+2>0\\ \iff & (x+1)(x+2)>0\\ \iff & x<-2\text{atau}x>-1\end{array}& \begin{array}{rl}& g(x)>0\\ \iff & 5x+5>0\\ \iff & x+1>0\\ \iff & x>-1\end{array}\end{array}\\ & \text{Syarat numerusnya},x>-1\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff x^2+3x+2=5x+5\\ & \iff x^2-2x-3=0\\ & \iff (x+1)(x-3)=0\\ & \iff x=-1\text{atau}x=3\\ & \text{Karena}x>-1,\\ & \text{yang memenuhi adalah}x=3\\ (\ast )& \text{Jadi},\text{HP}=\left\{3\right\}\end{array}\end{array}$.

Catatan:

Penjelasan untuk soal no.1 b ada berkaitan dengan penentuan akar ${\displaystyle \frac{3\pm \sqrt{5}}{2}}$ , silahlkan Anda klik di sini

Berikut soal yang berbasis seolah-olah berbeda, tetapi setelah Anda cermati, maka Anda akan dengan mudah menentukan penyelesaiannya.

$\begin{array}{ll}2.& \text{Tentukan himpunan penyelesaian dari}\\ & {}^{0,25}\log (x-4)+{}^{16}\log (x+2)=0\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Dik}& \text{etahui}\\ & {}^{0,25}\log (x-4)+{}^{16}\log (x+2)=0\\ & {\iff }^{{.}^{\frac{1}{4}}}\log (x-4)+{}^{16}\log (x+2)=0\\ & {\iff }^{{.}^{4^{-1}}}\log (x-4)+{}^{{.}^{4^2}}\log (x+2)=0\\ & \iff -{}^4\log (x-4)+{\displaystyle \frac{1}{2}{.}^4\log (x+2)=0}\\ & \iff {\displaystyle \frac{1}{2}{.}^4\log (x+2)={}^4\log (x-4)}\\ & \iff {}^4\log (x+2)=2.{}^4\log (x-4)\\ & \iff {}^4\log (x+2)={}^4\log (x-4{)}^2\\ & \iff {}^4\log (x+2)={}^4\log (x^2-8x+16)\\ (\ast )& \text{Syarat numerus}\\ & \begin{array}{ll}\begin{array}{rl}& f(x)>0\\ \iff & x-4>0\\ \iff & x>4\end{array}& \begin{array}{rl}& g(x)>0\\ \iff & x+2>0\\ \iff & x>-2\end{array}\end{array}\\ & \text{Sehingga syarat numerusnya},x>4\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff x^2-8x+16=x+2\\ & \iff x^2-8x-x+16-2=0\\ & \iff x^2-9x+14=0\\ & \iff (x-2)(x-7)=0\\ & \iff x=2\text{atau}x=7\\ & \text{Karena}x>4,\text{yang memenuhi}x=7\\ (\ast )& \text{Jadi},\text{HP}=\left\{7\right\}\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 3.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.{}^2\log x+{}^2\log (x-1)={}^2\log (x+3)\\ & \text{b}.\log x+\log 2=\log (x+2)\\ & \text{c}.\log (x^2-4x-5)=\log (x-5)\\ & \text{d}.\log (x^2-2x-8)=\log (3x-4)\\ \end{array}$.

DAFTAR PUSTAKA

1. Sembiring,S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: Srikandi Empat Widya Utama.
2. Kurnia, N., dkk. 2016. Jelajah Matematika SMA Kelas X Peminatan MIPA. Jakarta: YUDHISTIRA.

Persamaan Logaritma 2

 B. Persamaan Logaritma Bentuk  ${}^a\log f(x)={}^b\log f(x)$.

Syarat penyelesaian dari bentuk ini adalah numerusnya harus positif serta basisnya juga harus positif. Ketika basisnya berbeda, maka numerusnya cukup sama dengan 1. Hal ini dikarenakan nilai logaritma akan sama dengan 0 jika numerusnya berupa angka 1 dan basisnya bilangan positif. Sebagaimana ilustrasi contoh berikut ini

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.\log (x+6)={}^2\log (x+6)\\ & \text{b}.\log (2x-3)={}^3\log (2x-3)\\ & \text{c}.{}^4\log (x^2-x+1)=\log (x^2-x+1)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \text{Karena basisnya berbeda, maka cukup}\\ & \text{numerusnya}=1,\text{yaitu}:x+6=1.\\ & \text{Sehingga}x=-5\\ & \text{Jadi},\text{HP}=\left\{7\right\}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Karena basisnya berbeda, maka cukup}\\ & \text{numerusnya}=1,\text{yaitu}:2x-3=1.\\ & \text{Sehingga}x=2\\ & \text{Jadi},\text{HP}=\left\{2\right\}\end{array}\\ & \begin{array}{rl}\text{c}.& \text{Karena basisnya berbeda, maka cukup}\\ & \text{numerusnya}=1,\text{yaitu}:x^2-x+1=1.\\ & \text{Sehingga}\\ & x^2-x=0\\ & \iff x(x-1)=0\\ & \iff x=0\text{atau}x=1\\ & \text{Jadi},\text{HP}=\{0,1\}\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 2.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.{}^3\log (5x-4)=\log (5x-4)\\ & \text{b}.\log (2x-1)={}^5\log (2x-1)\\ & \text{c}.\log (2x^2+6x-5)={}^8\log (2x^2+6x-5)\\ & \text{d}.{}^2\log (x^2-4x+6)=\log (x^2-4x+6)\end{array}$.

DAFTAR PUSTAKA

1. Sembiring,S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: Srikandi Empat Widya Utama.
2. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: Tiga Serangkai Pustaka Mandiri.

Persamaan Logaritma 1

 A. Persamaan Logaritma Bentuk  ${}^a\log f(x)={}^a\log p$.

Syarat yang harus dipenuhi numerus harus berupa bilangan positif demikian juga bilangan basisnya dan khus bilangan basisnya ketambahan syarat yang harus terpenuhi yaitu tidak boleh sama dengan 1.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.\log (4x-5)=\log 3\\ & \text{b}.\log (2x^2-x)=1\\ & \text{c}.{}^3\log (x^2-3x+5)=1\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \text{Diketahui numerus}:4x-5\\ & \text{1. Syarat numerus}:f(x)>0\\ & 4x-5>0\iff x>{\displaystyle \frac{5}{4}}\\ & \text{2. Persamaan}\\ & \log (4x-5)=\log 3\\ & \iff 4x-5=3\\ & \iff 4x=8\\ & \iff x=2\\ & \text{3. Simpulan}\\ & \text{Karena}x>{\displaystyle \frac{5}{4},}\\ & \text{maka}x=2\text{memenuhi}\\ & \text{Jadi},\text{HP}=\left\{2\right\}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Diketahui numerus}:2x^2-x\\ & \text{1. Syarat numerus}:f(x)>0\\ & 2x^2-x>0\iff x(2x-1)>0\\ & \iff x<0\text{atau}x>{\displaystyle \frac{1}{2}}\\ & \text{2. Persamaan}\\ & \log (2x^2-x)=1\\ & \log (2x^2-x)=\log 10\\ & \iff 2x^2-x=10\\ & \iff 2x^2-x-10=0\\ & \iff (2x-5)(x+2)=0\\ & \iff x=-2\text{atau}x={\displaystyle \frac{5}{2}}\\ & \text{3. Simpulan}\\ & \text{Karena nilai}x\text{memenuhi}\\ & \text{syarat numerus}2x^2-x>0\\ & \text{maka}x=-2\text{dan}x={\displaystyle \frac{5}{2}\text{memenuhi}}\\ & \text{Jadi},\text{HP}=\{-2,{\displaystyle \frac{5}{2}}\}\end{array}\\ & \begin{array}{rl}\text{c}.& \text{Diketahui numerus}:x^2-3x+5\\ & \text{1. Syarat numerus}:f(x)>0\text{memenuhi}.\\ & x^2-3x+5>0\iff \text{Nilai}D=b^2-4ac>0\\ & \text{artinya numerus definit positif}\\ & \text{2. Persamaan}\\ & {}^3\log (x^2-3x+5)=1\\ & {}^3\log (x^2-3x+5)={}^3\log 3\\ & \iff x^2-3x+5=3\\ & \iff x^2-3x+2=0\\ & \iff (x-1)(x-2)=0\\ & \iff x=1\text{atau}x=2\\ & \text{3. Simpulan}\\ & \text{Karena nilai}x\text{memenuhi}\\ & \text{syarat numerus}{x}^2-3x+5>0\\ & \text{maka}x=1\text{dan}x=2\text{memenuhi}\\ & \text{Jadi},\text{HP}=\{1,2\}\end{array}\end{array}$.

Untuk materi difinit positif silahkan klik di sini

$\begin{array}{l}\\ 2.& \text{Tentukan himpunan penyelesaian dari}\\ & \log x+\log (2x-1)=1\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Persamaan di atas adalah persamaan}\\ & \text{logaritma model}:{}^a\log f(x)={}^a\log p.\\ & \text{dengan bentuknya}:\\ & {}^a\log f_1(x)+{}^a\log f_2(x)={}^a\log p\\ & \text{Diketahui numerus}:\\ & f_1(x)=x\text{dan}{f}_2(x)=2x-1\\ & \text{1. Syarat numerus}:f(x)>0\\ & \begin{array}{cc}{f}_1(x)& f_2(x)\\ \begin{array}{rl}& x>0\\ & \\ & \end{array}& \begin{array}{rl}& 2x-1>0\\ & 2x>1\\ & x>{\displaystyle \frac{1}{2}}\end{array}\end{array}\\ & \text{Sehingga syarat numerusnya}:x>{\displaystyle \frac{1}{2}}\\ & \text{2. Persamaan}\\ & \log (2x^2-x)=1\\ & \log (2x^2-x)=\log 10\\ & \iff 2x^2-x=10\\ & \iff 2x^2-x-10=0\\ & \iff (2x-5)(x+2)=0\\ & \iff x=-2\text{atau}x={\displaystyle \frac{5}{2}}\\ & \text{3. Simpulan}\\ & \text{Nilai}x\text{yang memenuhi}\\ & \text{syarat numerus}2x^2-x>0\\ & \text{hanya ada satu, yaitu}:x={\displaystyle \frac{5}{2}}\\ & \text{Jadi},\text{HP}=\left\{{\displaystyle \frac{5}{2}}\right\}\end{array}\end{array}$.

Catatan:

Coba bandingkan penyelesaian no. 1.b dan no. 2, secara sifat operasi logaritma soal sama, tetapi karena spesifikasi dari numerus tiap tipe soal, maka perlakuannya berbeda.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 3.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.{}^3\log (5x-4)=2\\ & \text{b}.\log x+\log (2x-1)=1\\ & \text{c}.\log (2x^2+6x-5)=1\\ & \text{d}.{}^2\log (x^2-4x+6)=1\\ & \text{e}.{}^2\log (x-4)+{}^2\log (x-6)=3\\ \\ & \end{array}$

Logaritma

A. Pendahuluan

Silahkan kunjungi alamat ini di sini

B. Sifat-Sifat

$\begin{array}{|l|}\hline \text{Logaritma}\\ {}^a\log b=c\Rightarrow a^c=b\\ \bullet {}^a\log x+{}^a\log y={}^a\log xy\\ \bullet {}^a\log x-{}^a\log y={}^a\log {\displaystyle \frac{x}{y}}\\ \bullet {}^a\log x={\displaystyle \frac{{}^m\log x}{{}^m\log a}}\\ \bullet {}^a\log b\times {}^b\log c={}^a\log c\\ \bullet {}^{a^m}\log b^n={\displaystyle \frac{n}{m}\times {}^a\log b}\\ \bullet {\displaystyle a^{{}^a\log b}=b}\\ \bullet {}^a\log b={\displaystyle \frac{1}{{}^b\log a}}\\ \bullet {}^a\log 1=0\\ \bullet {}^a\log a=1\\ \{\begin{array}{ll}a\ne 0& \\ a>0& (\text{bilangan pokok})\\ x,y>0& (\text{numerus})\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Hitunglah}\\ & \text{a}.{}^{36}\log 6\\ & \text{b}.{}^8\log {\displaystyle \frac{1}{4}}\\ & \text{c}.{}^2\log 3+{}^2\log 12-{}^2\log 9\\ & \text{d}.{}^{16}\log \sqrt[3]{25}\times {}^5\log {\displaystyle \frac{1}{4}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& {}^{36}\log 6={}^{6^2}\log 6^1={\displaystyle \frac{1}{2}\times {}^6\log 6}\\ & ={\displaystyle \frac{1}{2}\times \underset{1}{\underbrace{{}^6\log 6}}={\displaystyle \frac{1}{2}\times 1={\displaystyle \frac{1}{2}}}}\\ & \text{atau}\\ & {}^{36}\log 6={\displaystyle \frac{1}{{}^6\log 36}={\displaystyle \frac{1}{{}^{6^{{.}^1}}\log 6^2}}}\\ & ={\displaystyle \frac{1}{\frac{2}{1}\times {}^6\log 6}={\displaystyle \frac{1}{2\times 1}={\displaystyle \frac{1}{2}}}}\\ \text{b}.& {}^8\log 4={}^{2^3}\log 2^2={\displaystyle \frac{2}{3}\times \underset{1}{\underbrace{{}^2\log 2}}}\\ & ={\displaystyle \frac{2}{3}\times 1={\displaystyle \frac{2}{3}}}\end{array}\\ & \begin{array}{rl}\text{c}.& {}^2\log 3+{}^2\log 12-{}^2\log 9\\ & ={}^2\log \left({\displaystyle \frac{3\times 12}{9}}\right)={}^2\log {\displaystyle \frac{36}{9}}\\ & ={}^2\log 4={}^2\log 2^2=2\times \underset{1}{\underbrace{{}^2\log 2}}\\ & =2\times 1=2\end{array}\\ & \begin{array}{rl}\text{d}.& {}^{16}\log \sqrt[3]{25}\times {}^5\log {\displaystyle \frac{1}{4}}\\ & ={}^{4^{{.}^2}}\log 5^{{.}^{\frac{2}{3}}}\times {}^{{.5}^{{.}^1}}\log 4^{-1}\\ & =\left({\displaystyle \frac{\left({\displaystyle \frac{2}{3}}\right)}{2}\times {}^4\log 5}\right)\times \left({\displaystyle \frac{-1}{1}\times {}^5\log 4}\right)\\ & ={\displaystyle \frac{1}{3}\times -1\times {}^4\log 5\times {}^5\log 4}\\ & =-{\displaystyle \frac{1}{3}\times {}^4\log 4,(\text{ingat})}\\ & =-{\displaystyle \frac{1}{3}\times \underset{1}{\underbrace{{}^4\log 4}}}\\ & =-{\displaystyle \frac{1}{3}\times 1}\\ & =-{\displaystyle \frac{1}{3}}\end{array}\end{array}$.

C. Persamaan Logaritma

Bentuk-bentuk persamaan logaritma secara umum adalah persamaan dengan numerus ataupun bilangan basis/pokok yang memuat variabel x.

$\begin{array}{rl}1.& {}^a\log f(x)={}^a\log p\\ 2.& {}^a\log f(x)={}^b\log f(x)\\ 3.& {}^a\log f(x)={}^a\log g(x)\\ 4.& {}^{h(x)}\log f(x)={}^{h(x)}\log g(x)\\ 5.& A{({}^a\log f(x))}^2+B({}^a\log f(x))+C=0\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 2.& \text{Hitunglah}\\ & \text{a}.\log (x-5)=\log 3\\ & \text{b}.\log (2x^2-x)=1\\ & \text{c}.{}^3\log (x^2-3x+5)=1\\ \\ & \text{Jawab}:\\ & \text{Yang dibahas hanya no.2a, yaitu}:\\ & \begin{array}{rl}\text{a}.& \text{Diketahui numerus}:x-5\\ & \text{1. Syarat numerus}:f(x)>0\\ & x-5>0\iff x>5\\ & \text{2. Persamaan}\\ & \log (x-5)=\log 3\\ & \iff x-5=3\\ & \iff x=8\\ & \text{3. Simpulan}\\ & \text{Karena}x>5,\\ & \text{maka}x=8\text{memenuhi}\\ & \text{Jadi},\text{HP}=\left\{8\right\}\end{array}\end{array}$.

Contoh Soal 5 Transformasi Geometri

$\begin{array}{l}\\ 21.& \text{Jika setiap titik pada grafik dengan}\\ & \text{dengan persamaan}y=\sqrt{x}\text{dicerminkan}\\ & \text{terhadap garis}y=x,\text{maka persamaan}\\ & \text{grafik yang dihasilkan adalah}...\\ & \begin{array}{l}\\ \text{a}.y=x^2,x\ge 0& & \\ \text{b}.y=-\sqrt{x},x\ge 0& \\ \text{c}.y=-x^2,x\le 0& \\ \text{d}.y=\sqrt{-x},x\le 0\\ \text{e}.y=-\sqrt{-x},x\le 0\end{array}\\ \\ & \mathbf{\text{UMB Tahun 2011 Kode 152}}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ y& =\sqrt{x},\text{atau}{y}^2=x\\ \mathbf{\text{Alt}}& \mathbf{\text{ernatif 1}}\\ \text{mak}& \text{a}\text{saat dicerminkan terhadap}\\ \text{gari}& \text{s}y=x,\text{adalah}{x}^2=y\\ \text{atau}& y=x^2.\\ \mathbf{\text{Alt}}& \mathbf{\text{ernatif 2}}\\ \text{Jika}& \text{ingin dikerjakan dengan rumus}\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =M_{x=y}\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}y\\ x\end{array}\right)\\ \text{Sela}& \text{njutnya hasilnya disubstitusikan}\\ \text{ke p}& \text{ersamaan}y=\sqrt{x}\Rightarrow x^{\prime }=\sqrt{y^{\prime }}\\ \sqrt{y^{\prime }}& =x^{\prime }\text{maka}\\ y^{\prime }& ={\left(x^{\prime }\right)}^2\text{selanjutnya}\\ y& =x^2\end{array}\end{array}$.

Sebelum dicerminkan terhadap garis y=x

Gambar kurva/grafik setelah cerminkan terhadap garis y=x

$\begin{array}{l}\\ 22.& \text{Transformasi}T\text{adalah pencerminan}\\ & \text{terhadap garis}y={\displaystyle \frac{x}{3}\text{dilanjutkan oleh}}\\ & \text{pencerminan terhadap garis}y=-3x.\\ & \text{Matriks yang bersesuian dengan}\\ & \text{transformasi}T\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)& & \\ \text{b}.\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)& \\ \text{c}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)& \\ \text{d}.\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ \text{e}.\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\end{array}\\ \\ & \mathbf{\text{SBMPTN Tahun 2013 Kode 433}}\\ \\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \text{sebu}& \text{ah persamaan garis lurus}\\ \text{dapa}& \text{t dituliskan dengan}:y=mx\\ \text{Dike}& \text{tahui pula bahwa ada 2 garis}:\\ y_1& ={\displaystyle \frac{1}{3}x\text{dan}{y}_2=-3x}\\ \text{seba}& \text{gai representasi transformasi}T.\\ \text{Kare}& \text{na}{m}_1\times m_2=\left({\displaystyle \frac{1}{3}}\right)(-3)=-1\\ \text{bera}& \text{rti 2 garis di atas saling tegak}\\ \text{luru}& \text{s dan hal ini seperti rotasi 2}\\ \text{kali}& {90}^{\circ }\text{atau}{180}^{\circ }\\ \text{Jadi},& T=\left(\begin{array}{cc}\cos {180}^{\circ }& -\sin {180}^{\circ }\\ \sin {180}^{\circ }& \cos {180}^{\circ }\end{array}\right)\\ \iff & T=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.
3. Sharma,S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.

Contoh Soal 4 Transformasi Geometri

$\begin{array}{l}\\ 16.& \text{Bayangan titik A(2,4) dicerminkan }\\ & \text{terhadap garis}y-x=0\text{dilanjutkan}\\ & \text{ke garis}x\sqrt{3}-3y=0\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.A^{\prime }(2+\sqrt{3},-1+2\sqrt{3})& & \\ \text{b}.A^{\prime }(2+\sqrt{3},1-2\sqrt{3})& \\ \text{c}.A^{\prime }(1-\sqrt{3},-2+\sqrt{3})& \\ \text{d}.A^{\prime }(-2+\sqrt{3},1+2\sqrt{3})\\ \text{e}.A^{\prime }(2-\sqrt{3},1-2\sqrt{3})\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ & \{\begin{array}{ll}x\sqrt{3}-3y=0& \iff y={\displaystyle \frac{1}{3}\sqrt{3}x}\\ & \iff y=\tan {30}^{\circ }.x\\ \\ x-y=0& \iff y=x\end{array}\\ \\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {2.30}^{\circ }& \sin {2.30}^{\circ }\\ \sin {2.30}^{\circ }& -\cos {2.30}^{\circ }\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}\sqrt{3}}\\ {\displaystyle \frac{1}{2}\sqrt{3}}& -{\displaystyle \frac{1}{2}}\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}2\\ 4\end{array}\right)\\ & =\left(\begin{array}{c}\sqrt{3}+2\\ -1+2\sqrt{3}\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Jika}{T}_1=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\text{dan}{T}_2=\left(\begin{array}{cc}-2& 5\\ -1& 3\end{array}\right)\\ & \text{maka bayangan garis}x+y+1=0\\ & \text{oleh}{T}_2\circ T_1\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.x-2y-1=0& & \\ \text{b}.x+2y-1=0& \\ \text{c}.x+2y+1=0& \\ \text{d}.x-2y+1=0\\ \text{e}.x+y-1=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T_2\circ T_1\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}-2& 5\\ -1& 3\end{array}\right)\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}-2+5& -4+5\\ -1+3& -2+3\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}3& 1\\ 2& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}3x+y\\ 2x+y\end{array}\right)\\ \text{Dipe}& \text{roleh}\\ & \begin{array}{l}\\ x^{\prime }& =3x+y\\ y^{\prime }& =2x+y-\\ x^{\prime }-y^{\prime }& =x\\ \iff x& =x^{\prime }-y^{\prime }....(1)\\ \text{maka}\\ y& =x^{\prime }-3x\\ & =x^{\prime }-3(x^{\prime }-y^{\prime })\\ & =3y^{\prime }-2x^{\prime }....(2)\end{array}\\ \text{Sehin}& \text{gga}\\ x+y& +1=0\\ x^{\prime }-& y^{\prime }+3y^{\prime }-2x^{\prime }+1=0\\ -x^{\prime }+& 2y^{\prime }+1=0\\ x^{\prime }-& 2y^{\prime }-1=0\\ \text{maka}& \text{bayangan garisnya}\\ x-2& y-1=0\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Garis}2x+y+4=0\text{ditranslasikan}\\ & \text{oleh}\left(\begin{array}{c}-2\\ 5\end{array}\right)\text{dilanjutkan transformasi}\\ & \text{oleh}\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\text{persamaan bayangannya}\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.2x+y+3=0& & \\ \text{b}.2x-3y+3=0& \\ \text{c}.2x+3y+3=0& \\ \text{d}.3x+2y+3=0\\ \text{e}.3x-2y+3=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}-2\\ 5\end{array}\right)=\left(\begin{array}{c}x-2\\ y+5\end{array}\right)\\ \left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)& =\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & =\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{c}x-2\\ y+5\end{array}\right)\\ & =\left(\begin{array}{c}x-2+2y+10\\ y+5\end{array}\right)\\ & =\left(\begin{array}{c}x+2y+8\\ y+5\end{array}\right)\\ \text{Diper}& \text{oleh}\\ & \begin{array}{l}\\ x^{″}& =x+2y+8\\ 2y^{″}& =2y+10-\\ x^{″}-2y^{″}& =x-2\\ \iff x& =x^{″}-2y^{″}+2....(1)\\ \text{maka}\\ y& =y^{″}-5....(2)\end{array}\\ \text{sehin}& \text{gga}\\ 2x+& y+4=0\\ 2(x^{″}& -2y^{″}+2)+(y^{″}-5)+4=0\\ 2x^{″}-& 3y^{″}+3=0\\ \text{maka}& \text{bayangan garisnya}\\ 2x-& 3y+3=0\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Diketahui}M\text{adalah pencerminan terhadap}\\ & \text{garis}y=-x\text{dan}T\text{adalah transformasi}\\ & \text{yang dinyatakan oleh matriks}\left(\begin{array}{cc}2& 3\\ 0& -1\end{array}\right)\\ & \text{Koordinat bayangan titik}A(2,-8)\text{oleh}\\ & \text{transformasi}M\text{dilanjutkan oleh}T\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(-10,2)& & \\ \text{b}.(-2,-10)& \\ \text{c}.(10,2)& \\ \text{d}.(-10,-2)\\ \text{e}.(2,10)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui bahwa}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T\circ M\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}2& 3\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{cc}0-3& -2+0\\ 0+1& 0+0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{cc}-3& -2\\ 1& 0\end{array}\right)\left(\begin{array}{c}2\\ -8\end{array}\right)\\ & =\left(\begin{array}{c}-6+16\\ 2+0\end{array}\right)\\ & =\left(\begin{array}{c}10\\ 2\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Jika}W\text{adalah transformasi oleh}\\ & \text{matriks}\left(\begin{array}{cc}1& 0\\ 3& 1\end{array}\right),\text{maka titik mula}\\ & \text{dari}{W}^{\prime }(-2,5)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(-11,-2)& & \\ \text{b}.(11,-2)& \\ \text{c}.(-2,11)& \\ \text{d}.(2,11)\\ \text{e}.(12,11)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dimi}& \text{salkan}:\\ A& =\left(\begin{array}{c}-2\\ 5\end{array}\right),\text{dan}\\ W& =\left(\begin{array}{cc}1& 0\\ 3& 1\end{array}\right),\text{serta}X=\left(\begin{array}{c}x\\ y\end{array}\right)\\ \text{mak}& \text{a}\\ & \begin{array}{|c|}\hline \begin{array}{rl}A& =BX\\ B^{-1}A& =B^{-1}BX\\ B^{-1}A& =I.X\\ B^{-1}A& =X\\ X& =B^{-1}A\end{array}\\ \hline\end{array}\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\displaystyle \frac{1}{\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|}\left(\begin{array}{cc}1& 0\\ -3& 1\end{array}\right)\left(\begin{array}{c}-2\\ 5\end{array}\right)}\\ & =1.\left(\begin{array}{c}-2+0\\ 6+5\end{array}\right)\\ & =\left(\begin{array}{c}-2\\ 11\end{array}\right)\end{array}\end{array}$

Contoh Soal 3 Transformasi Geometri

$\begin{array}{l}\\ 11.& \text{Titik A(1,-2) dirotasikan sejauh}{15}^{\circ }\\ & \text{kemudian dilanjutkan}{75}^{\circ }\text{dengan pusat }\\ & O(0,0)\text{maka bayangan akhir titik A adalah}...\\ & \begin{array}{l}\\ \text{a}.(-2,1)& & \text{d}.(2,1)\\ \text{b}.(-1,2)& \text{c}.(1,2)& \text{e}.(-2,-1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos ({\theta }_1+{\theta }_2)& -\sin ({\theta }_1+{\theta }_2)\\ \sin ({\theta }_1+{\theta }_2)& \cos ({\theta }_1+{\theta }_2)\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos ({75}^{\circ }+{15}^{\circ })& -\sin ({75}^{\circ }+{15}^{\circ })\\ \sin ({75}^{\circ }+{15}^{\circ })& \cos ({75}^{\circ }+{15}^{\circ })\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2\\ 1\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Jika garis}3x-2y+5=0\text{dicerminkan }\\ & \text{terhadap garis}y=-x\text{kemudian}\\ & \text{didilatasikan dengan pusat (1,-2) }\\ & \text{dengan faktor skala 2, maka persamaan}\\ & \text{bayangannya adalah}....\\ & \begin{array}{l}\\ \text{a}.x-2y-10=0& & \\ \text{b}.x+2y-10=0& \\ \text{c}.x-6y+5=0& \\ \text{d}.x+2y-12=0\\ \text{e}.2x-3y+18=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}\text{Proses}& \text{untuk refleksinya}\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-y\\ -x\end{array}\right)\\ \text{proses}& \text{dilatasinya}\\ \left(\begin{array}{c}{x}^{″}\\ y^{″}\end{array}\right)& =\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\left(\begin{array}{c}{x}^{\prime }-1\\ y^{\prime }+2\end{array}\right)+\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2x^{\prime }-2\\ 2y^{\prime }+4\end{array}\right)+\left(\begin{array}{c}1\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2x^{\prime }-1\\ 2y^{\prime }+2\end{array}\right)\\ & =\left(\begin{array}{c}2(-y)-1\\ 2(-x)+2\end{array}\right)\\ & \{\begin{array}{ll}x& =-{\displaystyle \frac{1}{2}(y^{″}-2)}\\ y& =-{\displaystyle \frac{1}{2}(x^{″}+1)}\end{array}\end{array}\\ & \begin{array}{rl}\text{Sehingga persam}& \text{aan bayangan}\\ \text{garisnya adalah}:& \\ 3x& -2y+5=0\\ 3(-{\displaystyle \frac{1}{2}(y^{″}-2)})& -2(-{\displaystyle \frac{1}{2}(x^{″}+1)})+5=0\\ -{\displaystyle \frac{3}{2}{y}^{″}+3}& +(x^{″}+1)+5=0\\ 2x& -3y+6+2+10=0\\ 2x& -3y+18=0\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Titik A(4,-4) dicerminkan terhadap}\\ & \text{garis}y=x\tan {15}^{\circ }\text{menghasilkan}\\ & \text{bayangan}{A}^{\prime }(a,b)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.\sqrt{3}& & \text{d}.4\sqrt{3}\\ \text{b}.2\sqrt{3}& \text{c}.3\sqrt{3}& \text{e}.6\sqrt{3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\left(\begin{array}{c}a\\ b\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {2.15}^{\circ }& \sin {2.15}^{\circ }\\ \sin {2.15}^{\circ }& -\cos {2.15}^{\circ }\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {30}^{\circ }& \sin {30}^{\circ }\\ \sin {30}^{\circ }& -\cos {30}^{\circ }\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\right)\left(\begin{array}{c}4\\ -4\end{array}\right)\\ & =\left(\begin{array}{c}2\sqrt{3}-2\\ 2+2\sqrt{3}\end{array}\right)\\ & \{\begin{array}{ll}a& =2\sqrt{3}-2\\ b& =2+2\sqrt{3}\end{array}\\ \text{mak}& \text{a nilai dari}\\ a+b& =(2\sqrt{3}-2+2+2\sqrt{3})\\ & =4\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Lingkaran}{x}^2+y^2-5x+8y+7=0\\ & \text{ditranslasikan oleh}T=\left(\begin{array}{c}m\\ n\end{array}\right)\text{menghasilkan}\\ & \text{bayangan}{x}^2+y^2-9x+2y+6=0.\\ & \text{Nilai}m+n=...\\ & \begin{array}{l}\\ \text{a}.2& & \text{d}.5\\ \text{b}.3& \text{c}.4& \text{e}.6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Dik}& \text{etahui sebuah lingkaran dengan persamaan}:\\ & x^2+y^2-5x+8y+7=0\\ \text{kar}& \text{ena akibat translasi, maka}\\ & \{\begin{array}{ll}x& =x^{\prime }-m\\ y& =y^{\prime }-n\end{array}\\ & x^2+y^2-5x+8y+7=0\\ \text{seh}& \text{ingga}\\ & \iff (x^{\prime }-m{)}^2+(y^{\prime }-n{)}^2-5(x^{\prime }-m)+8(y^{\prime }-n)+7=0\\ & \iff x^{\prime 2}+y^{\prime 2}-2m{x}^{\prime }-2n{y}^{\prime }+m^2+n^2-5x^{\prime }+5m+8y^{\prime }-8n+7=0\\ & \iff x^{\prime 2}+y^{\prime 2}-(2m+5)x^{\prime }+(8-2n)y^{\prime }+m^2+n^2+5m-8n+7=0\\ & \equiv x^{\prime 2}+y^{\prime 2}-9x^{\prime }+2y^{\prime }+6=0(\mathbf{\text{akhir bayangan}})\\ & \{\begin{array}{ll}9& =2m+5\Rightarrow m=2\\ 2& =8-2n\Rightarrow n=3\end{array}\\ \text{Jad}& \text{i , nilai}m+n=2+3=5\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Jika titik A(-2,1) dicerminkan terhadap garis}\\ & y=-{\displaystyle \frac{1}{3}x\sqrt{3},\text{maka bayangan dari}}\\ & \text{titik textit{A} tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.A^{\prime }(1-{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})& & \\ \text{b}.A^{\prime }(-1-{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})& \\ \text{c}.A^{\prime }(-1-{\displaystyle \frac{1}{2}\sqrt{3},{\displaystyle \frac{1}{2}-\sqrt{3}}})& \\ \text{d}.A^{\prime }(1-{\displaystyle \frac{1}{2}\sqrt{3},{\displaystyle \frac{1}{2}-\sqrt{3}}})\\ \text{e}.A^{\prime }(-1+{\displaystyle \frac{1}{2}\sqrt{3},-{\displaystyle \frac{1}{2}+\sqrt{3}}})\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ y& =-{\displaystyle \frac{1}{3}x\sqrt{3}=(-{\displaystyle \frac{1}{3}\sqrt{3}})x}\\ & =(-\tan {30}^{\circ })x=\tan ({180}^{\circ }-{30}^{\circ })x\\ & =\tan {150}^{\circ }.x\\ \text{maka}\theta & ={150}^{\circ }\Rightarrow 2\theta ={300}^{\circ }\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {300}^{\circ }& \sin {300}^{\circ }\\ \sin {300}^{\circ }& -\cos {300}^{\circ }\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)\\ & =\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& -{\displaystyle \frac{1}{2}\sqrt{3}}\\ -{\displaystyle \frac{1}{2}\sqrt{3}}& -{\displaystyle \frac{1}{2}}\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)\\ & =\left(\begin{array}{c}-1-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \sqrt{3}-{\displaystyle \frac{1}{2}}\end{array}\right)\end{array}\end{array}$

Contoh Soal 2 Transformasi Geometri

$\begin{array}{l}\\ 6.& \text{Bayangan untuk titik P(2,5) oleh rotasi }\\ & \text{dengan pusat}\mathit{\text{A}}(1,3)\text{sejauh}{180}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.(1,0)& & \text{d}.(2,0)\\ \text{b}.(0,1)& \text{c}.(0,2)& \text{e}.(1,2)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Karena rotasi de}& \text{ngan pusat A sebesar}{180}^{\circ },\\ \text{maka}& \\ R(A(1,3),{180}^{\circ })& =\left(\begin{array}{cc}\cos {180}^{\circ }& -\sin {180}^{\circ }\\ \sin {180}^{\circ }& \cos {180}^{\circ }\end{array}\right)\\ & =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ \text{sehingga bayang}& \text{an titik P(2,5)-nya adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}2-1\\ 5-3\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-1\\ -2\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Bayangan kurva}xy=6\text{oleh rotasi sebesar}\\ & {\displaystyle \frac{\pi }{2}\text{dengan pusat}O(0,0)\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.xy=-6& & \text{d}.x(y-x)=6\\ \text{b}.xy=6& & \text{e}.x(x+y)=-6\\ \text{c}.x(x-y)=6& \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Karena rotasi d}& \text{engan pusat O sebesar}\\ {\displaystyle \frac{\pi }{2}={90}^{\circ },\text{maka}}& \\ R(O(0,0),{90}^{\circ })& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\\ \text{sehingga bayan}& \text{gan semua titik yang }\\ \text{terletak pada k}& \text{urva adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}-y\\ x\end{array}\right)\\ & \{\begin{array}{ll}x& =y^{\prime }\\ y& =-x^{\prime }\end{array}\\ \text{Selanjunya}\text{unt}& \text{uk bayangan kurvanya }\\ \text{adalah}:& \\ xy& =6\\ y^{\prime }.(-x^{\prime })& =6\\ x^{\prime }{y}^{\prime }& =-6\\ \text{Jadi , persamaa}& \text{n kurva bayangannya}\\ \text{adalah}& xy=-6\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Sebuah lingkaran yang berpusat di (3,4) }\\ & \text{dan menyinggung sumbu-X dicerminkan}\\ & \text{terhadap garis}y=x\text{, maka persamaan }\\ & \text{akhir lingkaran yang terjadi adalah}....\\ & \begin{array}{l}\\ \text{a}.x^2+y^2-8x-6y+9=0& & \\ \text{b}.x^2+y^2+8x+6y+9=0& \\ \text{c}.x^2+y^2+6x+8y+9=0& \\ \text{d}.x^2+y^2-8x-6y+16=0\\ \text{e}.x^2+y^2+8x+6y+16=0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Refleksi l}& \text{ingkaran yang berpusat di (3,4) }\\ \text{dan men}& \text{yinggung sumbu-X, }\\ \text{dengan}r& =(y)=4,\\ \text{maka}\text{per}& \text{samaan lingkarannya adalah}:\\ (x-3{)}^2+& (y-4{)}^2=4^2.\text{Karena}\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}y\\ x\end{array}\right)\\ & \{\begin{array}{ll}x& =y^{\prime }\\ y& =x^{\prime }\end{array}\\ \text{selanjutn}& \text{ya untuk persamaan bayangan }\\ \text{lingkaran}& \text{nya adalah}:\\ & (y^{\prime }-3{)}^2+(x^{\prime }-4{)}^2=4^2,\\ & \mathbf{\text{menjadi}}\\ & (y-3{)}^2+(x-4{)}^2=4^2,\text{atau}:\\ & x^2+y^2-8x-6y+9=0\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}{M}_x\text{adalah pencerminan terhadap sumbu-X }\\ & \text{dan}{M}_{y=x}\text{adalah pencerminan terhadap garis}\\ & y=x,\text{maka matriks transformasi tunggal }\\ & \text{yang mewakili}{M}_x\circ M_{y=x}=....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)& & \text{d}.\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ \text{b}.\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)& & \text{e}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\\ \text{c}.\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ & \{\begin{array}{ll}{M}_x& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ M_{y=x}& =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\end{array}\\ M_x\circ M_{y=x}& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ & =\left(\begin{array}{cc}0+0& 1+0\\ 0-1& 0+0\end{array}\right)\\ & =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Diketahui vektor}\overrightarrow{x}\text{dirotasikan terhadap titik asal}\\ & O\text{sebesar}\theta >0\text{searah jarum jam}.\\ & \text{Kemudian hasilnya dicerminkan terhadap garis}y=0\\ & \text{menghasilkan vektor}\overrightarrow{y}.\\ & \text{Jika}\overrightarrow{y}=A.\overrightarrow{x},\text{maka matriks}A-\text{nya adalah}....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)& & \\ \text{b}.\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)& & \\ \text{c}.\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)& & \\ \text{d}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\\ \text{e}.\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Diketahui bah}\text{wa}:\\ & \{\begin{array}{ll}{M}_x& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ R_{-\theta }& =\left(\begin{array}{cc}\cos (-\theta )& -\sin (-\theta )\\ \sin (-\theta )& \cos (-\theta )\end{array}\right)=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\end{array}\\ & A=M_x\circ R_{-\theta }=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\end{array}\end{array}$.

Contoh Soal 1 Transformasi Geometri

$\begin{array}{l}\\ 1.& \text{Suatu translasi yang memetakan titik P(9,8) }\\ & \text{ke titik}{\text{P}}^{\prime }(14,-2)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.\left(\begin{array}{c}5\\ -10\end{array}\right)& & \text{d}.\left(\begin{array}{c}6\\ 6\end{array}\right)\\ \text{b}.\left(\begin{array}{c}5\\ 6\end{array}\right)& \text{c}.\left(\begin{array}{c}23\\ -10\end{array}\right)& \text{e}.\left(\begin{array}{c}5\\ 2\end{array}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\text{T}+\left(\begin{array}{c}x\\ y\end{array}\right)\\ \text{T}& =\left(\begin{array}{c}{x}^{\prime }-x\\ y^{\prime }-y\end{array}\right)\\ & =\left(\begin{array}{c}14-9\\ -2-8\end{array}\right)\\ & =\left(\begin{array}{c}5\\ -10\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Sebuah transformasi yang didefiniskan oleh}\\ & \{\begin{array}{ll}{x}^{\prime }& =2x+3y\\ y^{\prime }& =3x+2y\end{array}\\ & \text{Maka bayangan titik M}(2,-1)\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(7,10)& & \text{d}.(1,10)\\ \text{b}.(10,7)& \text{c}.(1,4)& \text{e}.(4,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ & \{\begin{array}{ll}{x}^{\prime }& =2x+3y\\ y^{\prime }& =3x+2y\end{array}\\ \begin{array}{c}x=2\\ y=-1\end{array}\}& \Rightarrow \{\begin{array}{ll}{x}^{\prime }& =2(2)+3(-1)=4-3=1\\ y^{\prime }& =3(2)+2(-1)=6-2=4\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Bayangan untuk titik A(1,3) oleh rotasi }\\ & \text{dengan pusat}\mathit{\text{O}}(0,0)\text{sejauh}{90}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.(-1,3)& & \text{d}.(1,-3)\\ \text{b}.(-1,-3)& \text{c}.(-3,1)& \text{e}.(3,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Karena rotasi d}& \text{engan pusat O sebesar}{90}^{\circ },\\ \text{maka}& \\ R(O(0,0),{90}^{\circ })& =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\\ \text{sehingga}& \\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}-3\\ 1\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Suatu lingkaran dengan jari-jari 4 }\\ & \text{dengan pusat di O(0,0) dtranslasikan}\\ & \text{oleh}\text{T}=\left(\begin{array}{c}2\\ -3\end{array}\right),\text{maka luas }\\ & \text{bayangan lingkaran tersebut adalah}\\ & ....\text{satuan luas}\\ & \begin{array}{l}\\ \text{a}.\pi & & \text{d}.8\pi \\ \text{b}.2\pi & \text{c}.4\pi & \text{e}.16\pi \end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \text{Diketahui persamaan lingkaran berpusat}\\ & \text{di O dengan}r=4.\text{Karena translasi adalah}\\ & \text{termasuk transformasi isometri(kongruen)}\\ & \text{maka jari-jari lingkaran bayangannya }\\ & \text{akan sama dengan bendanya. Sehingga}\\ & \text{ luas bayangan lingkarannya}\\ & =\pi r^2=\pi \times 4^2=16\pi \end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Sebuah transformasi yang didefiniskan oleh}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \text{Yang merupakan titik invarian (tidak berubah) }\\ & \text{adalah}...\\ & \begin{array}{l}\\ \text{a}.(0,0)& & \text{d}.(0,-1)\\ \text{b}.(1,-1)& \text{c}.(1,0)& \text{e}.(1,1)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{D}& \text{iketahui bahwa}:\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\\ & \begin{array}{|cccc|}\hline \text{NO}& \text{Titik}& \begin{array}{rl}& \text{Disubstitusikan ke}\\ & \{\begin{array}{ll}{x}^{\prime }& =4-3x\\ y^{\prime }& =2x-y-4\end{array}\end{array}& \begin{array}{rl}& \text{Keterangan}\\ & \text{Titik}\end{array}\\ \text{a}.& (0,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(0)-4=-4\end{array}& \text{Varian}\\ \text{b}& (1,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(-1)-4=-1\end{array}& \mathbf{\text{Invarian}}\\ \text{c}& (1,0)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(0)-4=-2\end{array}& \text{Varian}\\ \text{d}& (0,-1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(0)=4\\ y^{\prime }& =2(0)-(-1)-4=-3\end{array}& \text{Varian}\\ \text{e}& (1,1)& \{\begin{array}{ll}{x}^{\prime }& =4-3(1)=1\\ y^{\prime }& =2(1)-(1)-4=-3\end{array}& \text{Varian}\\ \hline\end{array}\end{array}\end{array}$

Transformasi Geometri (XI Matematika Wajib)

 A. Pengertian 

Transformasi Geometri adalah suatu perubahan objek geometri atau suatu pemetaan dari suatu titik-titik ke himpunan titik-titik yang lain pada bidang kartesius.

Dari pengertian di atas jelas bahwa aturan transformasi sebagaimana fungsi atau pemetaan dan transformasi ini selanjutnya dapat disimbolkan dengan sebuah huruf kapital, misal M, T, R, dan lain sebagainya. Sebagai misal titik P(x,y) oleh transformasi T menghasilkan titik baru yaitu P'(x',y') dan operasi ini dapat dituliskan dengan:

$\begin{array}{rl}& P(x,y)\stackrel{T}{\to }{P}^{\prime }(x^{\prime },y^{\prime })\end{array}$.

B. Matriks Transformasi

Misalkan suatu transfomasi T memetakan sebuah titik A(x,y) ke A'(x',y') 

selanjutnya perhatikan ilustrasi berikut:

$\boxed{{\displaystyle \begin{array}{rl}A(x,y)& \underset{.}{\overset{Transformasi=T}{\to }}{A}^{\prime }(x^{\prime },y^{\prime })=A^{\prime }(ax+by,cx+dy)\\ \\ \Rightarrow & \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\underset{\underset{transformasi}{Matriks}}{\underbrace{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)}}\left(\begin{array}{c}x\\ y\end{array}\right)\end{array}}}$.

C. Jenis-Jenis Transformasi dengan matriks yang sesuaian

1. Translasi (Geseran)

$\begin{array}{|lcc|}\hline \begin{array}{rl}& \text{Jenis}\\ & \text{Transformasi}\end{array}& \text{Rumus}& \text{Matriks}\\ \text{Translasi}& (x,y)\underset{.}{\overset{\left(\begin{array}{c}a\\ b\end{array}\right)}{\to }}(x+a,y+b)& \left(\begin{array}{c}a\\ b\end{array}\right)\\ \hline\end{array}$.

2. Rotasi (Perputaran)

$\begin{array}{rl}& \begin{array}{|lcc|}\hline \begin{array}{rl}& \text{Jenis}\\ & \text{Transformasi}\end{array}& \text{Rumus}& \text{Matriks}\\ \text{Rotasi}& & \\ \begin{array}{rl}& \text{Pusat rotasi}\\ & [O,\alpha ]\end{array}& \begin{array}{rl}& \{\begin{array}{l}{x}^{\prime }=...\\ y^{\prime }=...\end{array}\\ & \begin{array}{rl}& \text{Lihat}\\ & \text{di bawah}\\ & \text{tulisan}\\ & \text{warna}\\ & \text{biru}\end{array}\end{array}& \left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right)\\ \begin{array}{rl}& \text{Pusat}\\ & (a,b)\text{sudut}\alpha \end{array}& \left(\begin{array}{c}{x}^{\prime }-a\\ y^{\prime }-b\end{array}\right)=& \begin{array}{rl}& \text{Lihat}\\ & \text{di bawah}\\ & \text{tulisan}\\ & \text{warna}\\ & \text{merah}\end{array}\\ \hline\end{array}\\ & \{\begin{array}{l}{x}^{\prime }=x\cos \alpha -y\sin \alpha \\ y^{\prime }=x\sin \alpha +y\cos \alpha \end{array}\\ & ▹▹▹▹\left(\begin{array}{c}{x}^{\prime }-a\\ y^{\prime }-b\end{array}\right)=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right).\left(\begin{array}{c}x-a\\ y-b\end{array}\right)\end{array}$.

3. Refleksi (Pencerminan)

$\begin{array}{|lcc|}\hline \text{Refleksi}& & \\ \text{terhadap sumbu}-\text{X}& (x,y)\to (x,-y)& \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\\ \text{terhadap sumbu}-\text{Y}& (x,y)\to (-x,y)& \left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\\ \text{terhadap garis y = x}& (x,y)\to (y,x)& \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ \text{terhadap garis y = -x}& (x,y)\to (-y,-x)& \left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\\ \text{terhadap garis x = h}& (x,y)\to (2h-x,y)& \left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}2h\\ 0\end{array}\right)\\ \text{terhadap garis y = x}& (x,y)\to (y,x)& \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ \text{terhadap garis y = -x}& (x,y)\to (-y,-x)& \left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\\ \text{terhadap garis x = h}& (x,y)\to (2h-x,y)& \left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}2h\\ 0\end{array}\right)\\ \text{terhadap garis y = k}& (x,y)\to (x,2k-y)& \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}0\\ 2k\end{array}\right)\\ \text{pusat}(0,0)\{\begin{array}{l}y=mx\\ m=\tan \alpha \end{array}& & \left(\begin{array}{cc}\cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{array}\right)\\ \hline\end{array}$.

4. Dilatasi (Perkalian)

$\begin{array}{rl}& \begin{array}{|lcc|}\hline \begin{array}{rl}& \text{Jenis}\\ & \text{Transformasi}\end{array}& \text{Rumus}& \text{Matriks}\\ \text{Dilatasi}& & \\ \text{Pusat}[O,k]& (x,y)\to (kx,ky)& \left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right)\\ \begin{array}{rl}& \text{Pusat}(a,b)\\ & \text{faktor skala}k\end{array}& \left(\begin{array}{c}{x}^{\prime }-a\\ y^{\prime }-b\end{array}\right)& \begin{array}{rl}& \text{Lihat}\\ & \text{di bawah}\\ & \text{tulisan}\\ & \text{warna}\\ & \text{merah}\end{array}\\ \begin{array}{rl}& \text{Luas bangun}\\ & \text{ datar}\end{array}& \text{Misal bangun A}& \text{T}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ & {\mathbf{\text{Bangun A}}}^{\prime }& =\text{det T}\times \text{A}\\ \hline\end{array}\\ & ▹▹▹▹▹▹\left(\begin{array}{c}{x}^{\prime }-a\\ y^{\prime }-b\end{array}\right)=\left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right).\left(\begin{array}{c}x-a\\ y-b\end{array}\right)\end{array}$.

Catatan:

Translasi, refleksi, dan rotasi suatu objek adalah bagian dari transformasi yang hanya mengubah posisi objek saja, sehingga jenis transformasi-transformasi ini juga disebut dengan transformasi isometri

D. Bayangan Kurva dan Komposisi Transformasi

$\begin{array}{|ll|}\hline \text{Bayangan Kurva}y=f(x)& \text{Komposisi Transformasi}\\ \begin{array}{rl}\text{Lan}& \text{gkah-langkah}:\\ 1.& \text{Tentukan bayangan titiknya}\\ & (x,y)\to (x^{\prime },y^{\prime })\\ 2.& \text{Salanjutnya tentukan}x\text{dan}y\\ & \text{dalam}{x}^{\prime }\text{dan}{y}^{\prime }\\ 3.& \text{Substitusikan}x\text{dan}y\\ & \text{ke}y=f(x)\end{array}& \begin{array}{rl}\text{Lan}& \text{gkah-langkah}:\\ 1.& \text{Selesaikan sesuai urutan transformasi}\\ & (x,y)\underset{.}{\overset{T_1}{\to }}(x^{\prime },y^{\prime })\underset{.}{\overset{T_2}{\to }}(x^{″},y^{″})\\ 2.& \text{Jika dapat disederhanakan kedua transformasi}\\ & \text{tersebut di atas, maka cukup dengan}\\ & (x,y)\underset{.}{\overset{T_2\circ T_1}{\to }}(x^{″},y^{″})\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah bayangan dari segitiga PQR dengan}\\ & P(0,4),Q(-1,1),\text{dan}R(3,6).\\ & \text{oleh translasi}T=\left(\begin{array}{c}5\\ -2\end{array}\right)\\ \\ & \mathbf{\text{Jawab}}\\ & \{\begin{array}{ll}\left(\begin{array}{c}{x}_P^{{}^{\prime }}\\ y_P^{{}^{\prime }}\end{array}\right)& =T+\left(\begin{array}{c}{x}_P\\ y_P\end{array}\right)=\left(\begin{array}{c}5\\ -2\end{array}\right)+\left(\begin{array}{c}0\\ 4\end{array}\right)=\left(\begin{array}{c}5+0\\ -2+4\end{array}\right)=\left(\begin{array}{c}5\\ 2\end{array}\right)\\ \left(\begin{array}{c}{x}_Q^{{}^{\prime }}\\ y_Q^{{}^{\prime }}\end{array}\right)& =\cdots \text{isilah sendiri}\\ \left(\begin{array}{c}{x}_R^{{}^{\prime }}\\ y_R^{{}^{\prime }}\end{array}\right)& =\cdots \text{isilah sendiri}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah bayangan dari garis}y=2x+4\\ & \text{oleh translasi}T=\left(\begin{array}{c}-1\\ 2\end{array}\right).\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{|cc|}\hline \mathbf{\text{Bayangan Titik-titik}}& \mathbf{\text{Bayangan Garis}}\\ \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =T+\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}-1\\ 2\end{array}\right)+\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}-1+x\\ 2+y\end{array}\right)\\ & \{\begin{array}{ll}{x}^{\prime }& =-1+x\iff x=x^{\prime }+1\\ y^{\prime }& =2+y\iff y=y^{\prime }-2\end{array}\end{array}& \begin{array}{rl}y& =2x+4\\ y^{\prime }-2& =2(x^{\prime }+1)+4\\ y^{\prime }& =2x+2+4+2\\ & =2x+8\\ \text{Jadi},& \mathbf{\text{bayangan garisnya}}\\ \text{adala}& \text{h}:\\ y& =2x+8\\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah bayangan titik A(4,6) oleh rotasi yang berpusat }\\ & \text{di titik P(3,-2) dengan sudut putar sebesar}{90}^{\circ }\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Untuk Ro}& \text{tasi yang berpusat di}(a,b)\text{dengan sudut}\alpha \text{adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right)\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{cc}\cos {90}^{\circ }& -\sin {90}^{\circ }\\ \sin {90}^{\circ }& \cos {90}^{\circ }\end{array}\right)\left(\begin{array}{c}4-3\\ 6-(-2)\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ 8\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}-8\\ 1\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}-5\\ -1\end{array}\right)\\ \text{Jadi},& \text{bayangan titik A adalah}{\text{A}}^{\prime }(-5,-1)\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah bayangan titik A(4,6) }\\ & \text{oleh dilatasi yang berpusat di titik P(3,-2)}\\ & \text{dengan faktor skala}k=2\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Bayangan}& \text{titik A-nya adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}k& 0\\ 0& k\end{array}\right)\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)\\ & =\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\left(\begin{array}{c}4-3\\ 6-(-2)\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\left(\begin{array}{c}1\\ 8\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}2\\ 16\end{array}\right)+\left(\begin{array}{c}3\\ -2\end{array}\right)\\ & =\left(\begin{array}{c}5\\ 14\end{array}\right)\\ \text{Jadi},& \text{bayangan titik A-nya adalah}{\text{A}}^{\prime }(5,14)\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukanlah bayangan titik A(4,6) }\\ & \text{oleh translasi}t\text{dilanjutkan}s\text{dengan}\\ & \text{matriks transformasi berturut-turut }\\ & \text{adalah}T=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)\text{dan}S=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Bayangan}& \text{titik A-nya adalah}:\\ \left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =S\times T\times \left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)\left(\begin{array}{c}4\\ 6\end{array}\right)\\ & =\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right)\left(\begin{array}{c}4\\ 6\end{array}\right)\\ & =\left(\begin{array}{c}26\\ 16\end{array}\right)\\ \text{Jadi},& \text{bayangan titik A-nya adalah}{\text{A}}^{\prime }(26,16)\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Suatu kurva}y={}^3\log (2x-2)\text{memiliki bayangan}\\ & y={}^3\log \left({\displaystyle \frac{2x+3}{3}}\right)\text{oleh translasi}\\ & T=\left(\begin{array}{c}a\\ b\end{array}\right).\text{Tentukanlah nilai}a+b\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}\\ y& ={}^3\log (2x-2)\iff 3^y=2x-2(\text{benda})\\ y& ={}^3\log \left({\displaystyle \frac{2x+3}{3}}\right)\\ & \iff 3^y=\left({\displaystyle \frac{2x+3}{3}}\right)(\mathbf{\text{bayangan}})\\ \text{sehingga}& \text{untuk bayangan}\\ 3^{y^{\prime }-b}& =2(x^{\prime }-a)-2\iff 3^{y^{\prime }}{.3}^{-b}=2(x^{\prime }-a)-2\\ & \iff 3^{y^{\prime }}={\displaystyle \frac{2(x^{\prime }-a)-2}{3^{-b}}={\displaystyle \frac{2x^{\prime }+3}{3}}}\\ \text{Jadi},& \{\begin{array}{ll}a& ={\displaystyle \frac{5}{2}}\\ b& =-1\end{array}\\ \text{Sehingga}& a+b={\displaystyle \frac{5}{2}+(-1)={\displaystyle \frac{3}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tentukanlah bayangan garis}ax+by+c=0\text{oleh transformasi}\\ & \text{yang bersesuaian dengan matriks}\left(\begin{array}{cc}1& -2\\ 3& -4\end{array}\right)\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{|cc|}\hline \mathbf{\text{Proses Awal}}& \mathbf{\text{Penentuan Bayangan}}\\ \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{cc}1& -2\\ 3& -4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)& ={\left(\begin{array}{cc}1& -2\\ 3& -4\end{array}\right)}^{-1}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\\ & ={\displaystyle \frac{1}{\left|\begin{array}{cc}1& -2\\ 3& -4\end{array}\right|}\left(\begin{array}{cc}-4& 2\\ -3& 1\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)}\\ & ={\displaystyle \frac{1}{-4+6}\left(\begin{array}{c}-4x^{\prime }+2y^{\prime }\\ -3x^{\prime }+y^{\prime }\end{array}\right)}\\ & ={\displaystyle \frac{1}{2}\left(\begin{array}{c}-4x^{\prime }+2y^{\prime }\\ -3x^{\prime }+y^{\prime }\end{array}\right)}\\ & \{\begin{array}{ll}x& =-2x^{\prime }+y^{\prime }\\ y& =-{\displaystyle \frac{3}{2}{x}^{\prime }+{\displaystyle \frac{1}{2}{y}^{\prime }}}\end{array}\end{array}& \begin{array}{rl}ax+by+c& =0\\ a(-2x^{\prime }+y^{\prime })+b(-{\displaystyle \frac{3}{2}{x}^{\prime }+\frac{1}{2}{y}^{\prime }})+c& =0\\ -2a{x}^{\prime }-{\displaystyle \frac{3}{2}b{x}^{\prime }+a{y}^{\prime }+{\displaystyle \frac{1}{2}b{y}^{\prime }+c}}& =0\\ (-4a-3b)x^{\prime }+(2a+b)y^{\prime }+2c& =0\\ & \\ \mathbf{\text{Jadi, bayangan garisnya adalah}}:& \\ & \\ (-4a-3b)x+(2a+b)y+2c& =0\\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Diketahui kurva}y=4x^2-9\text{dicerminkan terhadap sumbu-X kemudian}\\ & \text{ditranslasikan dengan}\left(\begin{array}{c}-1\\ 2\end{array}\right).\text{Ordinat titik potong terhadap sumbu-Y adalah}....\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)& =\left(\begin{array}{c}-1\\ 2\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =\left(\begin{array}{c}-1\\ 2\end{array}\right)+\left(\begin{array}{c}x\\ -y\end{array}\right)\\ & =\left(\begin{array}{c}-1+x\\ 2-y\end{array}\right)\\ & \{\begin{array}{ll}x& =x^{\prime }-1\\ y& =2-y^{\prime }\end{array}\end{array}& \begin{array}{rl}y& =4x^2-9\\ (2-y^{\prime })& =4(x^{\prime }-1{)}^2-9\\ -y^{\prime }& =4(x^{\prime 2}-2x^{\prime }+1)-9-2\\ -y^{\prime }& =4x^{\prime 2}-8x^{\prime }+4-11\\ y^{\prime }& =-4x^{\prime 2}+8x^{\prime }+7\\ & \\ \mathbf{\text{Maka}},& \mathbf{\text{persamaan kurva bayangannya}}:\\ y& =-4x^2+8x+7\end{array}\\ \hline\end{array}\\ & \begin{array}{rl}\text{Sehingga}& \text{ordinat dari titik potong terhadap sumbu-Y-nya adalah}:\\ y& =-4x^2+8x+7,\mathbf{\text{atau}}\\ f(x)& =-4x^2+8x+7\\ f(0)& =-4(0{)}^2+8(0)+7\text{saat}x=0(\text{karena memotong sumbu-Y})\\ & =7\\ \text{Jadi}& \text{ordinatnya adalah}y=f(0)=7\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim, 2006. Kompetensi Matematika 3A SMA Kelas XII Program IPA Semester Pertama. Jakarta: YUDHISTIRA.
2. Nugroho, P. A. Gunarto, D. 2013. Big Bank Soal-Bahas MAtematika SMA/MA. Jakarta: WAHYUMEDIA.

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

$\begin{array}{l}\\ 31.& \text{Nilai dari}\\ & \cos {\displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{1}{8}}}& & & \text{d}.& {\displaystyle \frac{1}{2}}\\ \\ \text{b}.& {\displaystyle -\frac{1}{4}}& \text{c}.& 0& \text{e}.& {\displaystyle \frac{1}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}& \cos {\displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\times {\displaystyle \frac{2\sin {\displaystyle \frac{2\pi }{7}}}{2\sin {\displaystyle \frac{2\pi }{7}}}}}\\ & =\left(\sin {\displaystyle \frac{4\pi }{7}-\sin 0}\right)\frac{{\displaystyle \cos \frac{\pi }{7}\cos {\displaystyle \frac{4\pi }{7}}}}{2\sin {\displaystyle \frac{2\pi }{7}}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{4\pi }{7}{\displaystyle \cos \frac{\pi }{7}\cos {\displaystyle \frac{4\pi }{7}}}}}{2\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{\left(\sin {\displaystyle \frac{5\pi }{7}+\sin {\displaystyle \frac{3\pi }{7}}}\right)\cos {\displaystyle \frac{4\pi }{7}}}{4\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{5\pi }{7}\cos {\displaystyle \frac{4\pi }{7}+\sin {\displaystyle \frac{3\pi }{7}\cos {\displaystyle \frac{4\pi }{7}}}}}}{4\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{9\pi }{7}+\sin {\displaystyle \frac{\pi }{7}+\sin {\displaystyle \frac{7\pi }{7}+\sin (-{\displaystyle \frac{\pi }{7}})}}}}{8\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{-\sin {\displaystyle \frac{2\pi }{7}+\sin {\displaystyle \frac{\pi }{7}+0-\sin {\displaystyle \frac{\pi }{7}}}}}{8\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{-\sin {\displaystyle \frac{2\pi }{7}}}{8\sin {\displaystyle \frac{2\pi }{7}}}}\\ & =-{\displaystyle \frac{1}{8}}\end{array}\\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& \cos {\displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}}\\ & =\cos {\displaystyle \frac{4\pi }{7}\cos \frac{2\pi }{7}\cos \frac{\pi }{7}}\\ & ={\displaystyle \frac{1}{2}\left(\cos {\displaystyle \frac{6\pi }{7}+\cos {\displaystyle \frac{2\pi }{7}}}\right)\cos {\displaystyle \frac{\pi }{7}}}\\ & ={\displaystyle \frac{1}{2}(\cos (\pi -{\displaystyle \frac{\pi }{7}})+\cos {\displaystyle \frac{2\pi }{7}})\cos {\displaystyle \frac{\pi }{7}}}\\ & ={\displaystyle \frac{1}{2}(-\cos {\displaystyle \frac{\pi }{7}+\cos {\displaystyle \frac{2\pi }{7}}})\cos {\displaystyle \frac{\pi }{7}}}\\ & ={\displaystyle \frac{1}{2}(-{\cos }^2{\displaystyle \frac{\pi }{7}+\cos {\displaystyle \frac{2\pi }{7}\cos {\displaystyle \frac{\pi }{7}}}})}\\ & ={\displaystyle \frac{1}{4}(-\cos {\displaystyle \frac{2\pi }{7}-\cos 0+\cos {\displaystyle \frac{3\pi }{7}+\cos {\displaystyle \frac{\pi }{7}}}})}\\ & ={\displaystyle \frac{1}{4}(-\cos 0+\cos {\displaystyle \frac{\pi }{7}-\cos {\displaystyle \frac{2\pi }{7}+\cos {\displaystyle \frac{3\pi }{7}}}})}\\ & ={\displaystyle \frac{1}{4}(-1+{\displaystyle \frac{1}{2}})}\\ & ={\displaystyle \frac{1}{4}\times (-\frac{1}{2})}\\ & =-{\displaystyle \frac{1}{8}}\end{array}\end{array}$.

Berikut penjelasan untuk  $\cos {\displaystyle \frac{\pi }{7}-\cos {\displaystyle \frac{2\pi }{7}+\cos {\displaystyle \frac{3\pi }{7}={\displaystyle \frac{1}{2}}}}}$.

$\begin{array}{rl}& \cos {\displaystyle \frac{\pi }{7}-\cos {\displaystyle \frac{2\pi }{7}+\cos {\displaystyle \frac{3\pi }{7}}}}\\ & =\cos {\displaystyle \frac{\pi }{7}-\cos {\displaystyle \frac{2\pi }{7}+\cos {\displaystyle \frac{3\pi }{7}\times {\displaystyle \frac{\left(2\sin {\displaystyle \frac{2\pi }{7}}\right)}{\left(2\sin {\displaystyle \frac{2\pi }{7}}\right)}}}}}\\ & ={\displaystyle \frac{2\cos {\displaystyle \frac{\pi }{7}\sin {\displaystyle \frac{2\pi }{7}-2\cos {\displaystyle \frac{2\pi }{7}\sin {\displaystyle \frac{2\pi }{7}+2\cos {\displaystyle \frac{3\pi }{7}\sin {\displaystyle \frac{2\pi }{7}}}}}}}}{2\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{3\pi }{7}-\sin (-{\displaystyle \frac{\pi }{7}})-\left(\sin {\displaystyle \frac{4\pi }{7}-\sin {\displaystyle \frac{0\pi }{7}}}\right)+\sin {\displaystyle \frac{5\pi }{7}-\sin {\displaystyle \frac{\pi }{7}}}}}{2\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{3\pi }{7}+\sin {\displaystyle \frac{\pi }{7}-\sin {\displaystyle \frac{4\pi }{7}+\sin {\displaystyle \frac{5\pi }{7}-\sin {\displaystyle \frac{\pi }{7}}}}}}}{2\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{3\pi }{7}-\sin {\displaystyle \frac{4\pi }{7}+\sin {\displaystyle \frac{5\pi }{7}}}}}{2\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{\sin (\pi -{\displaystyle \frac{4\pi }{7}})-\sin {\displaystyle \frac{4\pi }{7}+\sin (\pi -{\displaystyle \frac{2\pi }{7}})}}{2\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{4\pi }{7}-\sin {\displaystyle \frac{4\pi }{7}+\sin {\displaystyle \frac{2\pi }{7}}}}}{2\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{2\pi }{7}}}{2\sin {\displaystyle \frac{2\pi }{7}}}}\\ & ={\displaystyle \frac{1}{2}◼}\end{array}$.

$\begin{array}{l}\\ 32.& \text{Nilai dari}\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{16}}}& & & \text{d}.& {\displaystyle \frac{1}{2}}\\ \\ \text{b}.& {\displaystyle \frac{1}{8}}& \text{c}.& {\displaystyle \frac{1}{4}}& \text{e}.& {\displaystyle 1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Perhat}& \text{ikan bahwa}\\ \sin {\displaystyle {\displaystyle \frac{\pi }{14}}}& =\sin \left({\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14}}\right)=\sin \left({\displaystyle \frac{1}{2}\pi -\frac{6\pi }{14}}\right)\\ & =\cos {\displaystyle \frac{6\pi }{14}}\\ \sin {\displaystyle \frac{3\pi }{14}}& =...=\cos {\displaystyle \frac{4\pi }{14}}\\ \sin {\displaystyle \frac{9\pi }{14}}& =...=\sin {\displaystyle \frac{5\pi }{14}=\cos {\displaystyle \frac{2\pi }{14}}}\end{array}\\ & ...\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & =\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\cos {\displaystyle \frac{2\pi }{14}\times {\displaystyle \frac{2\sin {\displaystyle \frac{2\pi }{14}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\sin {\displaystyle \frac{4\pi }{14}}}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}\\ & \text{silahkan dilanjutkan}\\ & ...\\ & ={\displaystyle \frac{1}{8}}\end{array}$.

$\begin{array}{l}\\ 33.& \text{Nilai dari}\\ & \cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos {\displaystyle \frac{8\pi }{5}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{1}{16}}}& & & \text{d}.& {\displaystyle \frac{1}{16}}\\ \\ \text{b}.& {\displaystyle \frac{1}{8}}& \text{c}.& {\displaystyle 0}& \text{e}.& {\displaystyle \frac{1}{8}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos {\displaystyle \frac{8\pi }{5}}}\\ & =\cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos (\pi +{\displaystyle \frac{3\pi }{5}})}\\ & =\cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}(-\cos {\displaystyle \frac{3\pi }{5}})}\\ & =-\cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos {\displaystyle \frac{3\pi }{5}}}\\ & =-\cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos {\displaystyle \frac{4\pi }{5}}}\\ & =-\cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos {\displaystyle \frac{4\pi }{5}\times {\displaystyle \frac{2\sin {\displaystyle \frac{\pi }{5}}}{2\sin {\displaystyle \frac{\pi }{5}}}}}}\\ & =\frac{-\cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}(\sin \pi -\sin {\displaystyle \frac{3\pi }{5}})}}{2\sin {\displaystyle \frac{\pi }{5}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\sin \frac{3\pi }{5}}}{2\sin {\displaystyle \frac{\pi }{5}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{\pi }{5}\cos {\displaystyle \frac{3\pi }{5}\left(\cos {\displaystyle \frac{2\pi }{5}\sin {\displaystyle \frac{3\pi }{5}}}\right)}}}{2\sin {\displaystyle \frac{\pi }{5}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{\pi }{5}\cos {\displaystyle \frac{3\pi }{5}(\sin \pi -\sin (-{\displaystyle \frac{\pi }{5}}))}}}{4\sin {\displaystyle \frac{\pi }{5}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{\pi }{5}\cos {\displaystyle \frac{3\pi }{5}\sin {\displaystyle \frac{\pi }{5}}}}}{4\sin {\displaystyle \frac{\pi }{5}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{3\pi }{5}\cos {\displaystyle \frac{\pi }{5}\sin {\displaystyle \frac{\pi }{5}}}}}{4\sin {\displaystyle \frac{\pi }{5}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{3\pi }{5}\left(\cos {\displaystyle \frac{\pi }{5}\sin {\displaystyle \frac{\pi }{5}}}\right)}}{4\sin {\displaystyle \frac{\pi }{5}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{3\pi }{5}\left(\sin {\displaystyle \frac{2\pi }{5}-\sin 0}\right)}}{8\sin {\displaystyle \frac{\pi }{5}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{3\pi }{5}\sin {\displaystyle \frac{2\pi }{5}}}}{8\sin {\displaystyle \frac{\pi }{5}}}}\\ & ={\displaystyle \frac{\sin \pi -\sin {\displaystyle \frac{\pi }{5}}}{16\sin {\displaystyle \frac{\pi }{5}}}}\\ & =-{\displaystyle \frac{\sin {\displaystyle \frac{\pi }{5}}}{16\sin {\displaystyle \frac{\pi }{5}}}}\\ & =-{\displaystyle \frac{1}{16}}\end{array}\end{array}$.

$\begin{array}{l}\\ 34.& \text{Nilai dari}\\ & \sin {18}^{\circ }\cos {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{6}}& & & \text{d}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{1}{5}}& \text{c}.& {\displaystyle {\displaystyle \frac{1}{4}}}& \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sin {18}^{\circ }\cos {36}^{\circ }\\ & =\sin {18}^{\circ }\cos {36}^{\circ }\times {\displaystyle \frac{2\cos {18}^{\circ }}{2\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }(\sin {36}^{\circ }+\sin 0^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }\sin {36}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin {72}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin ({90}^{\circ }-{18}^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {18}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Nilai eksak dari}\sin {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}}}& & & \text{d}.& {\displaystyle \frac{\sqrt{5}-1}{4}}\\ \text{b}.& {\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}}& & & \text{e}.& {\displaystyle \frac{\sqrt{5}-1}{2}}\\ \text{c}.& {\displaystyle {\displaystyle \frac{\sqrt{5}+1}{4}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Perhatikan bahwa}△ABC\text{sama kaki}\\ & \text{dengan}AD=DC=CB=1,AC=x\\ & \text{Diketahui pula}CD\text{adalah garis bagi}\\ & \text{serta}ABC\text{sebangun}△BCD\\ & \text{akibatnya}:\\ & \text{perbandingan sisi yang bersesuaian}\\ & \text{akan sama},\text{maka}\\ & {\displaystyle \frac{AB}{BC}={\displaystyle \frac{BC}{AB-AD}}}\\ & \iff {\displaystyle \frac{x}{1}=\frac{1}{x-1}}\\ & \iff x(x-1)=1\\ & \iff x^2-x-1=0\\ & \iff x={\displaystyle \frac{1\pm \sqrt{5}}{2}}\\ & \text{akibatnya}AB=AC={\displaystyle \frac{1+\sqrt{5}}{2}}\\ & \text{Selanjutnya gunakan}\text{aturan sinus}\\ & {\displaystyle \frac{AB}{\sin \mathrm{\angle }C}=\frac{BC}{\sin \mathrm{\angle }A}}\\ & \iff {\displaystyle \frac{AB}{BC}=\frac{\sin \mathrm{\angle }C}{\sin \mathrm{\angle }A}}\\ & \iff {\displaystyle \frac{\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right)}{1}=\frac{\sin {72}^{\circ }}{\sin {36}^{\circ }}}\\ & \iff {\displaystyle \frac{1+\sqrt{5}}{2}={\displaystyle \frac{2\sin {36}^{\circ }\cos {36}^{\circ }}{\sin {36}^{\circ }}}}\\ & \iff {\displaystyle \frac{1+\sqrt{5}}{2}=2\cos {36}^{\circ }}\\ & \iff \cos {36}^{\circ }=\iff {\displaystyle \frac{1+\sqrt{5}}{4}}\\ & \text{Dari fakta di atas kita akan dengan}\\ & \text{mudah menentukan nilai sinusnya}\\ & \text{yaitu dengan menggunakan}\\ & \text{identitas trigonometri berikut}:\\ & {\sin }^2{36}^{\circ }+{\cos }^2{36}^{\circ }=1\\ & \iff {\sin }^2{36}^{\circ }=1-{\cos }^2{36}^{\circ }\\ & \iff \sin {36}^{\circ }=\sqrt{1-{\cos }^2{36}^{\circ }}\\ & \iff =\sqrt{1-{\left({\displaystyle \frac{1+\sqrt{5}}{4}}\right)}^2}\\ & \iff =\sqrt{1-{\displaystyle \frac{6+2\sqrt{5}}{16}}}\\ & \iff =\sqrt{{\displaystyle \frac{10-2\sqrt{5}}{16}}}\\ & \iff ={\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}}\end{array}$