Latihan Soal 5 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 41.& \text{Pada gambar berikut ini, pertidaksamaan}\\ & \text{yang memenuhi adalah}\\ \end{array}$

$.\begin{array}{l}\\ \text{a}.& 2x+y-4\le 0,2x+3y-6\ge 0,x\ge 0,y\ge 0\\ \text{b}.& 2x+y-4\ge 0,2x+3y-6\le 0,x\ge 0,y\ge 0\\ \text{c}.& 2x+y-4\le 0,2x+3y-6\le 0,x\ge 0,y\ge 0\\ \text{d}.& (2x+y-4)(2x+3y-6)\le 0,x\ge 0,y\ge 0\\ \text{e}.& (2x+y-4)(2x+3y-6)\ge 0,x\ge 0,y\ge 0\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{d}}\\ & \text{Misalkan titik potong kedua garis adalah M}\\ & \text{Persamaan garisnya sebelah kiri M}:\\ & (1)4x+2y-8=0\\ & \text{kendala}:2x+y-4\le 0\\ & (2)2x+3y=6\\ & \text{kendala}:2x+3y-6\ge 0\\ & (3)y=0,\text{kendala}:y\ge 0\\ & (4)x=0,\text{kendala}:x\ge 0\\ & \text{Persamaan garisnya sebelah kanan M}:\\ & (5)4x+2y-8=0\\ & \text{kendala}:2x+y-4\ge 0\\ & (6)2x+3y=6\\ & \text{kendala}:2x+3y-6\le 0\\ & (7)y=0,\text{kendala}:y\ge 0\\ & (8)x=0,\text{kendala}:x\ge 0\end{array}$

$\begin{array}{l}\\ 42.& \text{Pada gambar berikut ini, pertidaksamaan}\\ & \text{yang memenuhi adalah}\\ \end{array}$

$.\begin{array}{l}\\ \text{a}.& 2x-y-4\le 0,x-y-3\le 0,x\ge 0,y\ge 0\\ \text{b}.& 2x-y-4\ge 0,x-y-3\ge 0,x\ge 0,y\le 0\\ \text{c}.& 2x-y-4\le 0,x-y-3\ge 0,x\ge 0,y\le 0\\ \text{d}.& (2x-y-4)(x-y-3)\ge 0,x\ge 0,y\le 0\\ \text{e}.& (2x-y-4)(x-y-3)\le 0,x\ge 0,y\le 0\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{d}}\\ & \text{Misalkan titik potong kedua garis adalah M}\\ & \text{Persamaan garisnya di atas M dan model}\\ & \text{matematikanya adalah sebagai berikut}:\\ & (1)-3x+3y=-9\\ & \text{garisnya menjadi}:-3x+3y+9=0\\ & \text{maka}:-x+y+3=0\\ & \text{kendala}:-x+y+3\ge 0,\text{atau}\\ & \text{kendala}:x-y-3\le 0\\ & (2)-4x+2y=-8\\ & \text{garisnya menjadi}:-4x+2y+8=0\\ & \text{maka}:-2x+y+4=0\\ & \text{kendala}:-2x+y+4\ge 0,\text{atau}\\ & \text{kendala}:2x-y-4\le 0\\ & (3)y=0,\text{kendala}:y\ge 0\\ & (4)x=0,\text{kendala}:x\ge 0\end{array}$

$\begin{array}{l}\\ 43.& \text{Seorang penjual hewan kurban memiliki}\\ & \text{15 kandang ternak untuk memelihara sapi}\\ & \text{dan kambing. Setiap kandang hanya berisi}\\ & \text{kambing saja atau sapi saja. Setiap kandang}\\ & \text{dapat menampung sapi sebanyak 20 ekor}\\ & \text{atau kambing sebanyak 38 ekor. Penjual}\\ & \text{hewan kurban tersebut menaksir biaya}\\ & \text{perawatan yang dikeluarkan untuk setiap}\\ & \text{kandang sapi setiap bulannya sebesar}\\ & Rp500.000,00\text{dan kambing}Rp300.000,00.\\ & \text{Sementara itu, jumlah hewan yang}\\ & \text{direncanakan tidak lebih dari 300 ekor}.\\ & \text{Jika banyak kandang yang berisi sapi}\\ & \text{disebut}x\text{dan banyak kandang yang berisi}\\ & \text{kambing disebut}y,\text{sistem pertidaksamaan}\\ & \text{yang harus dipenuhi oleh}x\text{dan}y\text{serta}\\ & \text{fungsi objektif untuk meminimumkan biaya}\\ & \text{perawatan hewan kurban adalah}....\end{array}$

$.\begin{array}{l}\\ \text{a}.& x\ge 0,y\ge 0,20x+38y\le 15,x+y\le 300\\ & \text{minimum}f(x,y)=500.000x+300.000y\\ \text{b}.& x\ge 0,y\ge 0,38x+20y\le 15,x+y\le 300\\ & \text{minimum}f(x,y)=500.000x+300.000y\\ \text{c}.& x\ge 0,y\ge 0,28x+30y\le 300,x+y\le 15\\ & \text{minimum}f(x,y)=500.000x+300.000y\\ \text{d}.& x\ge 0,y\ge 0,38x+20y\le 300,x+y\le 15\\ & \text{minimum}f(x,y)=500.000x+300.000y\\ \text{e}.& x\ge 0,y\ge 0,20x+38y\le 300,x+y\le 15\\ & \text{minimum}f(x,y)=500.000x+300.000y\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{e}}\\ & \text{Misalkan titik potong kedua garis adalah M}\\ & \text{Misalkan sapi}=x,\text{kambing}=y,\text{maka}\\ & (1)\text{Sapi}+\text{Kambing}=15\text{ekor}\\ & \text{persamaan garisnya}:\\ & x+y=15,\text{dan}\\ & \text{kendalanya}:x+y\le 15\\ & (2)\text{Daya tampung kandang}\\ & \text{persamaan garisnya}:\\ & 20x+38y=300\\ & \text{kendala}:20x+38y\le 300\\ & (3)y=0,\text{kendala}:y\ge 0\\ & (4)x=0,\text{kendala}:x\ge 0\\ & (5)\text{Fungsi optimumnya adalah}:\\ & f(x,y)=500.000x+300.000y\end{array}$

$\begin{array}{l}\\ 44.& \text{Suatu perusahaan bangunan merencanakan}\\ & \text{pembangunan paling banyak 150 unit rumah}\\ & \text{untuk disewakan kepada 500 orang. Ada dua}\\ & \text{jenis rumah, yaitu rumah jenis A dengan}\\ & \text{kapasitas 4 orang yang akan disewakan dengan}\\ & \text{harga}Rp2.000.000,00\text{per tahun dan rumah}\\ & \text{jenis B dengan kapasitas 6 orang yang disewakan}\\ & Rp2.500.000,00\text{per tahun. Jika rumah jenis}\\ & \text{A dibuat sebanyak}x\text{unit dan jenis B sebanyak}\\ & y\text{unit},\mathbf{\text{model matematika}}\text{dari masalah tersebut}\\ & \text{adalah}....\end{array}$

$.\begin{array}{l}\\ \text{a}.& x\ge 0,y\ge 0,x+y\le 100,4x+6y\le 500\\ \text{b}.& x\ge 0,y\ge 0,x+y\le 150,4x+6y\le 500\\ \text{c}.& x\ge 0,y\ge 0,x+y\le 200,4x+6y\le 250\\ \text{d}.& x\ge 0,y\ge 0,x+y\le 200,6x+4y\le 250\\ \text{e}.& x\ge 0,y\ge 0,x+y\le 500,6x+4y\le 250\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{b}}\\ & \text{Model matematikanya adalah}:\\ & \text{Misalkan rumah jenis A}=x,\text{jenis B}=y,\text{maka}\\ & (1)\text{Jenis A}+\text{Jenis B}=150\text{unit}\\ & \text{persamaan garisnya}:\\ & x+y=150,\text{dan}\\ & \text{kendalanya}:x+y\le 150\\ & (2)\text{Kapasitas atau daya tampung}\\ & \text{Rumah jenis A muat 4 orang dan jenis B}\\ & \text{6 orang sedangkan targetnya 500 orang, maka}\\ & \text{persamaan garisnya}:\\ & 4x+6y=500\\ & \text{kendala}:4x+6y\le 500\\ & (3)y=0,\text{kendala}:y\ge 0\\ & (4)x=0,\text{kendala}:x\ge 0\\ & (5)\text{Fungsi optimumnya adalah}:\\ & f(x,y)=2.000.000x+2.500.000y\end{array}$

$\begin{array}{l}\\ 45.& \text{Pedagang teh mempunyai lemari yang hanya}\\ & \text{cukup ditempati 40 boks teh. Teh A dibeli}\\ & \text{dengan harga}Rp6.000,00\text{setiap boks dan teh B}\\ & \text{dibeli dengan harga}Rp8.000,00\text{setiap boks}\\ & \text{Jika pedang tersebut mempunyai modal sebesar}\\ & Rp300.000,00\text{untuk membeli}x\text{boks teh A dan}\\ & y\text{boks teh B, maka sistem pertidaksamaan dari}\\ & \text{permasalahan tersebut adalah}....\end{array}$

$.\begin{array}{l}\\ \text{a}.& 3x+4y\ge 150,x+y\ge 40,x\ge 0,y\ge 0\\ \text{b}.& 3x+4y\le 150,x+y\le 40,x\ge 0,y\ge 0\\ \text{c}.& 3x+4y\ge 150,x+y\le 40,x\ge 0,y\ge 0\\ \text{d}.& 6x+8y\le 300,x+y\ge 40,x\ge 0,y\ge 0\\ \text{e}.& 8x+4y\le 300,x+y\le 40,x\ge 0,y\ge 0\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{b}}\\ & \text{Pembahasan diserahkan kepada pembaca}\end{array}$.

$\begin{array}{l}\\ 46.& \text{Pada pertidaksamaan}\\ & 2y\ge x;y\le 2x;2y+x\le 20;x+y\ge 9\\ & \text{Nilai maksimum untuk}3y-x\text{dicapai saat}....\\ \end{array}$

$.\begin{array}{l}\\ \text{a}.& \text{P}\\ \text{b}.& \text{Q}\\ \text{c}.& \text{R}\\ \text{d}.& \text{S}\\ \text{e}.& \text{T}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{c}}\\ & \text{Dengan membuat garis(selidik)}:f(x,y)=3y-x\\ & \text{digeser dari bawah ke atas, maka akan didapatkan}\\ & \text{titik sudut(verteks) yang diinginkan}\\ & \end{array}$

$\begin{array}{l}\\ 47.& \text{Nilai minimum dari}-2x+4y+6\\ & \text{untuk}x\text{dan}y\text{yang memenuhi}\\ & \{\begin{array}{ll}2x+y-20& \le 0\\ 2x-y+10& \ge 0\\ x+y-5& \ge 0\\ x-2y-5& \le 0\\ x\ge 0& \\ y\ge 0& \end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -14\\ \text{b}.& -11\\ \text{c}.& -9\\ \text{d}.& -6\\ \text{e}.& -4\end{array}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{e}}\\ & \text{Diketahui fungsi objektif}:f(x,y)=-2x+4y+6\\ & \text{dan kendala-kendalanya}\\ & \{\begin{array}{ll}2x+y-20& \le 0\\ 2x-y+10& \ge 0\\ x+y-5& \ge 0\\ x-2y-5& \le 0\\ x\ge 0& \\ y\ge 0& \end{array}\\ & \text{maka daerah penyelesaiannya adalah}:\end{array}$

$.\begin{array}{rl}& \text{Dan persamaan garisnya adalah}\\ & \{\begin{array}{ll}{\text{L}}_1\equiv & 2x+y=20\\ {\text{L}}_2\equiv & 2x-y=-10\\ {\text{L}}_3\equiv & x+y=5\\ {\text{L}}_4\equiv & x-2y=5\end{array}\\ & \text{Perpotongan untuk garis}{\text{L}}_1\mathrm{\&}{\text{L}}_2\\ & \text{akan didapatkan titik C}\left({\displaystyle \frac{5}{2},15}\right)\\ & \text{Perpotongan untuk garis}{\text{L}}_1\mathrm{\&}{\text{L}}_4\\ & \text{akan didapatkan titik B}\left({\displaystyle 9,2}\right)\end{array}$

$.\begin{array}{|l|}\hline \begin{array}{rl}\text{untuk}& \text{mendapatkan titik potong}\\ \text{garis}:& {\text{L}}_1\mathrm{\&}{\text{L}}_2\text{adalah sebagai berikut}:\\ & \{\begin{array}{ll}2x+y& =20\\ 2x-y& =-10\end{array}\\ & -------{.}^{-}\\ & 2y=30\\ & y=15\Rightarrow x={\displaystyle \frac{5}{2}}\\ \text{Sehing}& \text{ga didapatkan titik}\left({\displaystyle \frac{5}{2},15}\right)\end{array}\\ \begin{array}{rl}\text{untuk}& \text{mendapatkan titik potong}\\ \text{garis}:& {\text{L}}_1\mathrm{\&}{\text{L}}_4\text{adalah sebagai berikut}:\\ & \{\begin{array}{ll}2x+y& =20\\ x-2y& =5\end{array}\\ & \{\begin{array}{ll}4x+2y& =40\\ x-2y& =5\end{array}\\ & -------{.}^{+}\\ & 5x=45\\ & x=9\Rightarrow y=2\\ \text{Sehing}& \text{ga didapatkan titik}(9,2)\end{array}\\ \hline\end{array}$

$.\text{Selanjutnya}$

$.\begin{array}{|ccc|}\hline \text{Verteks}& \text{Nilai}:& \text{Keterangan}\\ & -2x+4y+6& \\ \text{A}(5,0)& -2(5)+4.0+6=-4& \text{Minimum}\\ \text{B}(9,2)& -2(9)+4.2+6=-4& \text{Minimum}\\ \text{C}\left({\displaystyle \frac{5}{2},15}\right)& -2\left({\displaystyle \frac{5}{2}}\right)+4.15+6=61& \text{Maksimum}\\ \text{D}(0,10)& -2.0+4.10+6=46& \\ \text{E}(0,5)& -2.0+4.5+6=26& \\ \hline\end{array}$

$\begin{array}{l}\\ 48.& \text{Nilai minimum}f(x,y)=3+4x-5y\\ & \text{untuk}x\text{dan}y\text{yang memenuhi}\\ & \{\begin{array}{ll}-x+y& \le 1\\ x+2y& \ge 5\\ 2x+y& \le 10\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -19\\ \text{b}.& -6\\ \text{c}.& -5\\ \text{d}.& -3\\ \text{e}.& 23\end{array}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{c}}\\ & \text{Diketahui fungsi objektif}:f(x,y)=3+4x-5y\\ & \text{dan kendala-kendalanya}\\ & \{\begin{array}{ll}-x+y& \le 1\\ x+2y& \ge 5\\ 2x+y& \le 10\end{array}\\ & \text{maka daerah penyelesaiannya adalah}:\end{array}$

$.\begin{array}{|ccc|}\hline \text{Verteks}& \text{Nilai}:& \text{Keterangan}\\ & 3+4x-5y& \\ \text{A}(1,2)& 3+4.1-5.2=-3& \\ \text{B}(3,4)& 3+4.3-5.4=-5& \text{Minimum}\\ \text{C}(5,0)& 3+4.5-5.0=23& \text{Maksimum}\\ \hline\end{array}$

$\begin{array}{l}\\ 49.& \text{Fungsi}f(x,y)=10x+15y\\ & \text{untuk}x\text{dan}y\text{yang memenuhi}\\ & \{\begin{array}{ll}x& \ge 0\\ y& \ge 0\\ x& \le 800\\ x& +y\le 1000\end{array}\\ & \text{mempunyai nilai maksimum}....\\ & \begin{array}{l}\\ \text{a}.& 9.000\\ \text{b}.& 11.000\\ \text{c}.& 13.000\\ \text{d}.& 15.000\\ \text{e}.& 16.000\end{array}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{c}}\\ & \text{Diketahui fungsi objektif}:f(x,y)=10x+15y\\ & \text{dan kendala-kendalanya}\\ & \{\begin{array}{ll}x& \ge 0\\ y& \ge 0\\ x& \le 800\\ x& +y\le 1000\end{array}\\ & \text{maka daerah penyelesaiannya adalah}:\end{array}$

$.\begin{array}{|ccc|}\hline \text{Verteks}& \text{Nilai}:& \text{Keterangan}\\ & f(x,y)=10x+15y& \\ \text{A}(800,0)& 800.10+0=8000& \text{Minimum}\\ \text{B}(800,200)& 800.10+15.200=11.000& \\ \text{C}(400,600)& 10.400+15.600=13.000& \text{Maksimum}\\ \text{D}(0,600)& 3+0+15.600=9.000& \\ \hline\end{array}$

$\begin{array}{l}\\ 50.& \text{Nilai maksimum fungsi sasaran}\\ & f(x,y)=4x+5y\\ & \text{untuk}x\text{dan}y\text{yang memenuhi}\\ & \{\begin{array}{ll}x& \ge 0\\ y& \ge 0\\ (& 2x+y-4)(2x+3y-6)\le 0\end{array}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 11\\ \text{b}.& 12\\ \text{c}.& 16\\ \text{d}.& 20\\ \text{e}.& 24\end{array}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{d}}\\ & \text{Diketahui fungsi objektif}:f(x,y)=4x+5y\\ & \text{dan kendala-kendalanya}\\ & \{\begin{array}{ll}x& \ge 0\\ y& \ge 0\\ (& 2x+y-4)(2x+3y-6)\le 0\end{array}\\ & \text{maka daerah penyelesaiannya adalah}:\end{array}$

$.\begin{array}{rl}\text{untuk}& \text{mendapatkan titik potongnya}\\ & \{\begin{array}{ll}2x+y& =4\\ 2x+3y& =6\end{array}\\ & -------{.}^{-}\\ & -2y=-2\\ & y=1\Rightarrow x={\displaystyle \frac{3}{2}}\\ \text{sehing}& \text{ga akan didapatkan}\\ \text{titik p}& \text{otongnya adalah}:\left({\displaystyle \frac{3}{2},1}\right)\\ \text{Selanj}& \text{utnya, kita dapat menentukan}\\ \text{nilai}& \text{maksimunya dengan bantuan tabel berikut}\end{array}$

$.\begin{array}{|ccc|}\hline \text{Verteks}& \text{Nilai}:& \text{Keterangan}\\ & f(x,y)=4x+5y& \\ (2,0)& 4.2+0=8& \text{Minimum}\\ (3,0)& 4.3+0=12& \\ \left({\displaystyle \frac{3}{2},1}\right)& 4.\left({\displaystyle \frac{3}{2}}\right)+5.1=11& \\ (0,2)& 0+52=10& \\ (0,4)& 0+5.4=20& \text{Maksimum}\\ \hline\end{array}$

Latihan Soal 4 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 31.& \text{Daerah yang diarsir berikut adalah himpunan}\\ & \text{penyelesaian pertidaksamaan dari}....\end{array}$

$.\begin{array}{l}\\ & \begin{array}{l}\\ \text{a}.& y\ge 0,x-2y\ge -2,3x+4y\le 12\\ \text{b}.& y\ge 0,x-2y\ge -2,3x+4y\ge 12\\ \text{c}.& y\ge 0,-2x+y\ge -2,4x+3y\le 12\\ \text{d}.& x\ge 0,-2x+y\le -2,4x+3y\ge 12\\ \text{e}.& x\ge 0,x-2y\le -2,3x+4y\le 12\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \text{Cukup jelas. Anda bisa mengecek dengan titik uji}\end{array}$

$\begin{array}{l}\\ 32.& \text{Daerah yang diarsir berikut adalah himpunan}\\ & \text{penyelesaian pertidaksamaan dari}....\end{array}$

$.\begin{array}{l}\\ & \begin{array}{l}\\ \text{a}.& x\ge 0,4x+y\ge 4,x+y\le 2\\ \text{b}.& x\ge 0,4x+y\le 4,x+y\ge 2\\ \text{c}.& x\ge 0,4x+y>4,x+y<2\\ \text{d}.& x\ge 0,x+4y>4,x+y<2\\ \text{e}.& x\ge 0,x+4y\le 4,x+y\ge 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Cukup jelas}\end{array}$

$\begin{array}{l}\\ 33.& \text{Pada gambar berikut ini, yang merupakan}\\ & \text{himpunan penyelesaian sistem pertidaksamaan}\\ \\ & \{\begin{array}{ll}2x+y& \le 24\\ x+2y& \ge 12\\ x-y& \ge -2\end{array}\\ \\ & \text{adalah}....\\ & \end{array}$

$.\begin{array}{l}\\ & \begin{array}{l}\\ \text{a}.& \text{I}\\ \text{b}.& \text{II}\\ \text{c}.& \text{III}\\ \text{d}.& \text{IV}\\ \text{e}.& \text{V}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \text{Pembahasan diserahkan kepada pembaca}\end{array}$

$\begin{array}{l}\\ 34.& \text{Pada gambar berikut ini, yang merupakan}\\ & \text{himpunan penyelesaian sistem pertidaksamaan}\\ \\ & \{\begin{array}{ll}x+2y& \ge 6\\ 4x+5y& \le 20\\ 2x+y& \ge 6\end{array}\\ \\ & \text{adalah}....\end{array}$

$.\begin{array}{l}\\ & \begin{array}{l}\\ \text{a}.& \text{I}\\ \text{b}.& \text{II}\\ \text{c}.& \text{III}\\ \text{d}.& \text{IV}\\ \text{e}.& \text{V}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Pembahasan diserahkan kepada pembaca}\end{array}$

$\begin{array}{l}\\ 35.& \text{Pada gambar berikut ini, yang merupakan}\\ & \text{himpunan penyelesaian sistem pertidaksamaan}\\ \\ & \{\begin{array}{ll}2x+y& \ge 4\\ x+y& \ge 3\\ x-4y& \ge 4\end{array}\\ \\ & \text{adalah}....\end{array}$

$.\begin{array}{l}\\ & \begin{array}{l}\\ \text{a}.& \text{I}\\ \text{b}.& \text{II}\\ \text{c}.& \text{III}\\ \text{d}.& \text{IV}\\ \text{e}.& \text{V}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \text{Pembahasan diserahkan kepada pembaca}\end{array}$.

$\begin{array}{l}\\ 36.& \text{Pada gambar berikut ini, yang merupakan}\\ & \text{himpunan penyelesaian sistem pertidaksamaan}\\ \\ & \{\begin{array}{ll}-x+2y& \le 2\\ 4x+3y& \le 12\\ x\ge 0& \\ y\ge 0& \end{array}\\ \\ & \text{adalah}....\end{array}$

$.\begin{array}{l}\\ & \begin{array}{l}\\ \text{a}.& \text{I}\\ \text{b}.& \text{II}\\ \text{c}.& \text{III}\\ \text{d}.& \text{I}\text{dan}\text{IV}\\ \text{e}.& \text{II}\text{dan}\text{II}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \text{Pembahasan diserahkan kepada pembaca}\end{array}$

$\begin{array}{l}\\ 37.& \text{Pada gambar berikut ini, yang merupakan}\\ & \text{himpunan penyelesaian sistem pertidaksamaan}\\ \\ & \{\begin{array}{ll}5x+y& \ge 10\\ 2x+y& \le 8\\ y\ge 2& \end{array}\\ \\ & \text{adalah}....\end{array}$

$.\begin{array}{l}\\ & \begin{array}{l}\\ \text{a}.& \text{I}\\ \text{b}.& \text{II}\\ \text{c}.& \text{III}\\ \text{d}.& \text{IV}\\ \text{e}.& \text{V}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \text{Pembahasan diserahkan kepada pembaca}\end{array}$

$\begin{array}{l}\\ 38.& \text{Pada gambar berikut ini, yang merupakan}\\ & \text{himpunan penyelesaian sistem pertidaksamaan}\\ \\ & \{\begin{array}{ll}x+y& \ge 4\\ x+2y& \le 6\\ y\ge 1& \end{array}\\ \\ & \text{adalah}....\end{array}$

$.\begin{array}{l}\\ & \begin{array}{l}\\ \text{a}.& \text{I}\\ \text{b}.& \text{II}\\ \text{c}.& \text{III}\\ \text{d}.& \text{IV}\\ \text{e}.& \text{V}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \text{Pembahasan juga diserahkan kepada pembaca}\end{array}$

$\begin{array}{l}\\ 39.& (\text{SPMB 2003})\\ & \text{Daerah yang diarsir pada ilustrasi gambar berikut}\\ & \text{adalah himpunan semua}(x,y)\text{yang memenuhi}\end{array}$

$.\begin{array}{l}\\ \text{a}.& 2x+y\le 30,3x+4y\le 60,x\ge 0,y\ge 0\\ \text{b}.& 2x+y\ge 30,3x+4y\ge 60,x\ge 0,y\ge 0\\ \text{c}.& x+2y\ge 30,3x+4y\ge 60,x\ge 0,y\ge 0\\ \text{d}.& x+2y\le 30,4x+3y\le 60,x\ge 0,y\ge 0\\ \text{e}.& 2x+y\ge 30,4x+3y\le 60,x\ge 0,y\ge 0\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{a}}\\ & \text{Persamaan garisnya adalah}:\\ & (1)15x+20y=300\\ & \text{kendala}:3x+4y\le 60\\ & (2)30x+15y=450\\ & \text{kendala}:2x+y\le 30\\ & (3)y=0,\text{kendala}:y\ge 0\\ & (4)x=0,\text{kendala}:x\ge 0\end{array}$

$\begin{array}{l}\\ 40.& (\text{SPMB 2004})\\ & \text{Daerah yang diarsir pada ilustrasi gambar berikut}\\ & \text{adalah himpunan penyelesaian yang dipenuhi oleh}\end{array}$

$.\begin{array}{l}\\ \text{a}.& 6x+5y-30\le 0,x+6y-6\le 0,x-y\le 0\\ \text{b}.& 6x+5y-30\ge 0,x+6y-6\le 0,x-y\le 0\\ \text{c}.& 6+5y-30\le 0,x+6y-6\ge 0,x-y\ge 0\\ \text{d}.& 6x+5y-30\le 0,x+6y-6\ge 0,x-y\le 0\\ \text{e}.& 6x+5y-30\ge 0,x+6y-6\ge 0,x-y\ge 0\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\mathbf{\text{c}}\\ & \text{Perhatikan bahwa kendala-kendalanya}:\\ & \begin{array}{|ccc|}\hline \begin{array}{rl}6x+5y& =6\times 5\\ 6x+5y& =30\end{array}& \begin{array}{rl}x+6y& =1\times 6\\ x+6y& =6\end{array}& \begin{array}{rl}x& =y\end{array}\\ \mathbf{\text{Sebelah kiri}}& \mathbf{\text{Sebelah kanan}}& \\ \begin{array}{rl}6x+5y& \le 30\\ 6x+5y-30& \le 0\end{array}& \begin{array}{rl}x+6y& \ge 6\\ x+6y-6& \ge 0\end{array}& \begin{array}{rl}x& \ge y\\ x-y& \ge 0\end{array}\\ \hline\end{array}\end{array}$

Latihan Soal 3 Persiapan PAS Gasal Matematika Wajib Kelas XI

$\begin{array}{l}\\ 21.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 3+6+12+24+...+\left({3.2}^{n-1}\right)=3.(2^n-1)\\ & \text{Jika}S(n)\text{benar, untuk}n=k+1,\text{maka}\\ & \text{ruas kiri persamaan tersebut dapat dituliskan}\\ & \text{dengan}....\\ & \begin{array}{l}\\ \text{a}.& 3+6+12+24+...+{3.2}^{k+1}\\ \text{b}.& 3+6+12+24+...+{3.2}^{k-1}\\ \text{c}.& 3+6+12+24+...+{3.2}^{k-1}+{3.2}^k\\ \text{d}.& 3+6+12+24+...+{3.2}^{k-1}+{3.2}^{k+1}\\ \text{e}.& 3+6+12+24+...+{3.2}^k+{3.2}^{k+1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& 3+6+12+24+...+{3.2}^{n-1}=3.(2^n-1)\\ & 3+6+12+24+...+{3.2}^{k-1}+{3.2}^k=3.(2^{k+1}-1)\end{array}\end{array}$.

$\begin{array}{l}\\ 22.& \mathbf{\text{(EBTANAS 1999)}}\\ & \text{Nilai dari}{\displaystyle \sum _{k=1}^{100}5k-\sum _{k=1}^{100}(2k-1)\text{adalah}....}\\ & \text{a}.30.900\\ & \text{b}.30.500\\ & \text{c}.16.250\\ & \text{d}.15.450\\ & \text{e}.15.250\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \text{Diketahi}\\ & \begin{array}{rl}{\displaystyle \sum _{k=1}^{100}5k-\sum _{k=1}^{100}(2k-1)}& =\sum _{k=1}^{100}(5k-2k+1)\\ & ={\displaystyle \sum _{k=1}^{100}(3k+1)}\\ & =3{\displaystyle \sum _{k=1}^{100}k+1.100}\\ & =3\left({\displaystyle \frac{100}{2}(1+100)}\right)+100\\ & =3.(5.050)+100\\ & =15.150+100\\ & =15.250\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \mathbf{\text{(EBTANAS 2000)}}\\ & \text{Diketahui}{\displaystyle \sum _{k=5}^{25}(2-pk)=0,\text{maka nilai}}\\ & {\displaystyle \sum _{k=5}^{25}pk=....}\\ & \text{a}.20\\ & \text{b}.28\\ & \text{c}.30\\ & \text{d}.42\\ & \text{e}.112\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \sum _{k=5}^{25}(2-pk)={\displaystyle \sum _{k=5}^{25}2-\sum _{k=5}^{25}pk}}& =0\\ {\displaystyle \sum _{k=5-4}^{25-4}2-\sum _{k=5}^{25}pk}& =0\\ {\displaystyle \sum _{k=5}^{25}pk}& ={\displaystyle \sum _{k=1}^{21}2}\\ & =21.2\\ & =42\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \mathbf{\text{(EBTANAS 2000)}}\\ & \text{Nilai dari}{\displaystyle \sum _{k=1}^7{\left({\displaystyle \frac{1}{2}}\right)}^{k+1}=....}\\ & \text{a}.{\displaystyle \frac{127}{1024}}\\ \\ & \text{b}.{\displaystyle \frac{127}{256}}\\ \\ & \text{c}.{\displaystyle \frac{255}{512}}\\ \\ & \text{d}.{\displaystyle \frac{127}{128}}\\ \\ & \text{e}.{\displaystyle \frac{255}{256}}\\ \\ & \text{Jawab}:\text{b}\\ & \begin{array}{rl}& {\displaystyle \sum _{k=1}^7{\left({\displaystyle \frac{1}{2}}\right)}^{k+1}}\\ & ={\left({\displaystyle \frac{1}{2}}\right)}^2+{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^5+{\left(\frac{1}{2}\right)}^6+{\left(\frac{1}{2}\right)}^7+{\left({\displaystyle \frac{1}{2}}\right)}^8\\ & ={\displaystyle \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}}\\ & ={\displaystyle \frac{64+32+16+8+4+2+1}{256}}\\ & ={\displaystyle \frac{127}{256}}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \mathbf{\text{EBTANAS 1999}}\\ & \text{Diketahui jumlah n suku pertama }\\ & \text{deret aritmetika dinyatakan sebagai}\\ & S_n=n^2+2n.\text{Beda dari deret tersebut }\\ & \text{adalah}....\\ & \text{a}.3\\ & \text{b}.2\\ & \text{c}.1\\ & \text{d}.-2\\ & \text{e}.-3\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Diketahui bahwa}{S}_n=n^2+2n,\\ & \text{dengan}\{\begin{array}{ll}{S}_1& =U_1=a\\ S_2& =U_1+U_2\\ S_3& =U_1+U_2+U_3\\ & ⋮\\ S_n& =U_1+U_2+U_3+\cdots +U_n\end{array}\\ & \begin{array}{rl}\text{Beda}=b& =U_2-U_1\\ & =(S_2-S_1)-S_1\\ & =S_2-2S_1\\ & =(2^2+2.(2))-2(1^2+2.(1))\\ & =(4+4)-2(1+2)=8-6\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 26.& \mathbf{\text{(UMPTN 1994)}}\\ & \text{Diketahui jumlah n suku pertama suatu }\\ & \text{deret dinyatakan sebagai}{S}_n=12n-n^2.\\ & \text{Suku kelima dari deret tersebut adalah}....\\ & \text{a}.-1\\ & \text{b}.1\\ & \text{c}.-3\\ & \text{d}.3\\ & \text{e}.0\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \text{Diketahui bahwa}\\ & S_n=12n-n^2\\ & \begin{array}{rl}{U}_5& =S_5-S_4\\ & =(12.(5)-(5{)}^2)-(12.(4)-(4{)}^2)\\ & =(60-25)-(48-16)\\ & =3\end{array}\end{array}$.

$\begin{array}{l}\\ 27.& \mathbf{\text{(UMPTN 1997)}}\\ & \text{Diketahui}{U}_n\text{adalah suku ke - n }\\ & \text{deret aritmetika dengan}\\ & U_1+U_2+U_3=-9\text{dan}\\ & U_3+U_4+U_5=15.\\ & \text{Maka jumlah lima suku pertama}\\ & \text{deret aritmetika tersebut adalah}....\\ & \text{a}.4\\ & \text{b}.5\\ & \text{c}.6\\ & \text{d}.15\\ & \text{e}.24\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Diketahui bahwa}\\ & \begin{array}{rl}& U_1+U_2+U_3=-9,\\ & \iff a+(a+b)+(a+2b)=-9\\ & \iff 3a+3b=-9\\ & U_3+U_4+U_5=15,\\ & \iff (a+2b)+(a+3b)+(a+4b)=15\\ & \iff 3a+9b=15{}_{-}\\ & -----------------\\ & -6b=-24\Rightarrow b=4\\ & \text{sehingga akan diperoleh}a=-7\\ & \text{Selanjutnya}\\ & S_5={\displaystyle \frac{5}{2}(U_1+U_5)}\\ & ={\displaystyle \frac{5}{2}(a+a+(5-1)b)}\\ & ={\displaystyle \frac{5}{2}(-7-7+4.4)}\\ & ={\displaystyle \frac{5}{2}(2)}\\ & =5\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& \mathbf{\text{(UN 2013)}}\\ & \text{Diketahui suatu barisan aritmetika dengan }\\ & \text{suku ketiga adalah 4 dan suku ketujuhnya }\\ & \text{adalah 16. Jumlah 10 suku pertama dari }\\ & \text{deret tersebut adalah}....\\ & \text{a}.115\\ & \text{b}.125\\ & \text{c}.130\\ & \text{d}.135\\ & \text{e}.{\displaystyle 140}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{U}_3=a+2b& =4\\ U_7=a+6b& =16{}_{-}\\ ------& --\\ -4b& =-12\\ b& =3\\ \text{Sehingga}& \text{didapatkan}\\ \text{nilai}a=& 4-2b\\ =& 4-2.3\\ =& -2\end{array}\\ & \begin{array}{rl}& \text{Maka}\text{jumlah 10 suku pertama }\\ & \text{deret tersebut adalah}\\ & S_n={\displaystyle \frac{n}{2}(2a+(n-1)b)}\\ & S_{10}={\displaystyle \frac{10}{2}(2.(-2)+(10-1).3)}\\ & =5(-4+27)\\ & =5(23)\\ & =115\end{array}\end{array}$.

$\begin{array}{l}\\ 29.& \mathbf{\text{(UN 2014)}}\\ & \text{Diketahui tempat duduk gedung pertunjukan }\\ & \text{film diatur mulai dari baris depan ke belakang }\\ & \text{dengan banyak banyak baris dibelakang lebih }\\ & \text{4 kursi dari baris di depannya.}\\ & \text{Jika dalam gedung pertunjukan terdapat 15}\\ & \text{baris kursi dan baris terdepan ada 20 kursi, }\\ & \text{maka kapasitas gedung pertunjukan tersebut }\\ & \text{adalah}....\text{kursi}\\ & \text{a}.1200\\ & \text{b}.800\\ & \text{c}.720\\ & \text{d}.600\\ & \text{e}.300\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Diketahui}& \{\begin{array}{ll}a& =U_1=20\\ b& =4\\ n& =15\\ S_n& ={\displaystyle \frac{n}{2}(2a+(n-1)b)}\end{array}\\ & \\ & \end{array}\\ & \begin{array}{rl}{S}_n& ={\displaystyle \frac{n}{2}(2a+(n-1)b)}\\ & ={\displaystyle \frac{15}{2}(2(20)+(15-1).4)}\\ & =15(20+28)\\ & =15(48)\\ & =750\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \mathbf{\text{(UN 2015)}}\\ & \text{Diketahui suatu barisan aritmetika dengan }\\ & \text{suku ke-3 adalah 2 dan suku ke-8 adalah -13}.\\ & \text{Jumlah 20 suku pertama dari deret tersebut }\\ & \text{adalah}....\\ & \text{a}.-580\\ & \text{b}.-490\\ & \text{c}.-440\\ & \text{d}.-410\\ & \text{e}.-380\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{U}_3=a+2b& =2\\ U_8=a+7b& =-13{}_{-}\\ ------& ---\\ -5b& =15\\ b& =-3\\ \text{Sehingga}& \text{didapatkan}\\ \text{nilai}a=& 2-2b\\ =& 2-2.(-3)\\ =& 8\end{array}\\ & \begin{array}{rl}& \text{Maka}\text{jumlah 20 suku pertama }\\ & \text{deret tersebut adalah}\\ & S_n={\displaystyle \frac{n}{2}(2a+(n-1)b)}\\ & S_{20}={\displaystyle \frac{20}{2}(2.(8)+(20-1).(-3))}\\ & =10(16-57)\\ & =10(-41)\\ & =-410\\ & \\ & \end{array}\end{array}$

DAFTAR PUSTAKA

1. Budhi, W.S. 2018. Bupena Matematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: ERLANGGA.

DAFTAR PUSTAKA WEB

1. Thohir, A. https://ahmadthohir1089.wordpress.com/2016/01/11/insyaallah-44/

Latihan Soal 2 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 11.& \text{Perhatikanlah pernyataan-pernyataan berikut}\\ & (1){\displaystyle \sum _{i=1}^5(5i+2)=4{\displaystyle \sum _{i=1}^{5}i+10}}\\ & (2){\displaystyle \sum _{i=1}^5(5i^2-i)=5{\displaystyle \sum _{i=1}^5{i}^2-\sum _{i=1}^{5}i}}\\ & (3){\displaystyle \sum _{i=1}^5(3i-4)=3{\displaystyle \sum _{i=1}^5{i}^2-4}}\\ & (4){\displaystyle \sum _{i=1}^5(i+7i^2)={\displaystyle \sum _{i=1}^{5}i-7\sum _{i=1}^{5}i}}\\ & \text{Pernyataan yang tepat ditunjukkan oleh}....\\ & \begin{array}{l}\\ \text{a}.& (1)\text{dan}(2)\\ \text{b}.& (1)\text{dan}(3)\\ \text{c}.& (1)\text{dan}(4)\\ \text{d}.& (2)\text{dan}(3)\\ \text{e}.& (2)\text{dan}(4)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}(1)& {\displaystyle \sum _{i=1}^5(5i+2)=4{\displaystyle \sum _{i=1}^{5}i+\sum _{i=1}^{5}2}}\\ & =4{\displaystyle \sum _{i=1}^{5}i+5\times 2}\\ & =4{\displaystyle \sum _{i=1}^{5}i+10}\\ (2)& {\displaystyle \sum _{i=1}^5(5i^2-i)=5{\displaystyle \sum _{i=1}^5{i}^2-\sum _{i=1}^{5}i}}\\ (3)& {\displaystyle \sum _{i=1}^5(3i-4)=3{\displaystyle \sum _{i=1}^5{i}^2-\sum _{i=1}^{5}4}}\\ & =3{\displaystyle \sum _{i=1}^5{i}^2-5\times 4}\\ & =3{\displaystyle \sum _{i=1}^5{i}^2-20}\\ (4)& {\displaystyle \sum _{i=1}^5(i+7i^2)={\displaystyle \sum _{i=1}^{5}i+7\sum _{i=1}^{5}i}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Hasil dari}{\displaystyle \sum _{i=1}^4{i}^2+\sum _{i=5}^6{i}^2}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 86}\\ \text{b}.& {\displaystyle 91}\\ \text{c}.& {\displaystyle 95}\\ \text{d}.& {\displaystyle 101}\\ \text{e}.& {\displaystyle 105}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \sum _{i=1}^4{i}^2+\sum _{i=5}^6{i}^2}& ={\displaystyle \sum _{i=1}^6{i}^2}\\ & =1^2+2^2+3^2+4^2+5^2+6^2\\ & =1+4+9+16+25+36\\ & =91\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Hasil dari}{\displaystyle \sum _{i=2}^5(4i^2-2i)\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 144}\\ \text{b}.& {\displaystyle 148}\\ \text{c}.& {\displaystyle 154}\\ \text{d}.& {\displaystyle 164}\\ \text{e}.& {\displaystyle 188}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \sum _{i=2}^5(4i^2-2i)}\\ & =({4.2}^2-2.2)+({4.3}^2-2.3)+({4.4}^2-2.4)+({4.5}^2-2.5)\\ & =12+30+56+90\\ & =188\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Bentuk}{11}^n-1\text{dengan}n\text{bilangan asli}\\ & \text{akan habis dibagi oleh}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 7}\\ \text{b}.& {\displaystyle 9}\\ \text{c}.& {\displaystyle 10}\\ \text{d}.& {\displaystyle 11}\\ \text{e}.& {\displaystyle 13}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Bentuk}& {11}^n-1\\ \text{untuk}& n=1\\ & ={11}^1-1\\ & =10\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Rumus yang tepat untuk pola}12,13,14,15,...\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle U_n=n+9}\\ \text{b}.& {\displaystyle U_n=n+10}\\ \text{c}.& {\displaystyle U_n=n+11}\\ \text{d}.& {\displaystyle U_n=2n+10}\\ \text{e}.& {\displaystyle U_n=2n+11}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Bentuk}& 12,13,14,15,...\\ \text{untuk}& U_n=pn+q\\ 12& =p+q\\ 13& =2p+q\\ \text{akan}& \text{didapatkan}\\ & \{\begin{array}{ll}p& =1\\ q& =11\end{array}\\ \text{Sehing}& \text{ga}\\ U_n& =n+11\end{array}\end{array}$.

$\begin{array}{l}\\ 16.& \text{Diketahui}1+5+9+...+(4n-1)=2n^2-n\\ & \text{dengan}n\text{bilangan asli}.\text{Jika}m<k\text{dengan}\\ & m,k\text{bilangan asli juga},\text{maka}\\ & (4m-3)+(4m+1)+...+(4k-3)=....\\ & \begin{array}{l}\\ \text{a}.& (k-m)(2k+2m-2)\\ \text{b}.& (k-m+1)(2k+2m-3)\\ \text{c}.& (k-m+1)(2k-2m+1)\\ \text{d}.& (k-m+1)(2k^2+2m^2-3)\\ \text{e}.& (k-m{)}^2(2k-2m+4)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& 1+5+9+...+(4m-3)+(4m+1)+...+(4k-3)\\ & =\underset{2k^2-k}{\underbrace{1+5+...+(4k-3)}}-\underset{2(m-1{)}^2-(m-1)}{\underbrace{1+5+...+(4(m-1)-3)}}\\ & =2k^2-k-(2(m-1{)}^2-(m-1))\\ & =2k^2-k-2(m-1{)}^2+(m-1)\\ & =2k^2-k-2(m^2-2m+1)+m-1\\ & =2k^2-k-2m^2+4m-2+m-1\\ & =2k^2-k-2m^2+5m-3\\ & =(k-m+1)(2k+2m-3)\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Diketahui}{2}^1+2^2+2^3+...+2^n=2^{n+1}-2\\ & \text{dengan}n\text{bilangan asli}.\text{Jika}k\text{bilangan asli},\\ & \text{maka}{2}^2+2^3+2^4+...+2^k+2^{k+1}=....\\ & \begin{array}{l}\\ \text{a}.& (k-m)(2k+2m-2)\\ \text{b}.& (k-m+1)(2k+2m-3)\\ \text{c}.& (k-m+1)(2k-2m+1)\\ \text{d}.& (k-m+1)(2k^2+2m^2-3)\\ \text{e}.& (k-m{)}^2(2k-2m+4)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 2^2+2^3+2^4+...+2^k+2^{k+1}\\ & =2^1+2^2+2^3+2^4+...+2^k+2^{k+1}-2^1\\ & =\underset{2^{k+1+1}-2}{\underbrace{2^1+2^2+2^3+2^4+...+2^k+2^{k+1}}}-2^1\\ & =2^{k+2}-2-2\\ & =2^{k+2}-4\\ & =2^k{.2}^2-4\\ & =2^k\times 4-4\\ & =4(2^k-1)\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 2+5+10+17+...+(n^2+1)={\displaystyle \frac{1}{6}(n+1)(2n^2+n+6)}\\ & \text{Jika}S(n)\text{benar, untuk}n=k,\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 2+5+10+17+...+(k^2+1)\\ & ={\displaystyle \frac{1}{6}(k+1)(2k^2+k+6)}\\ \text{b}.& 2+5+10+17+...+(n^2+1)\\ & ={\displaystyle \frac{1}{6}(k+1)(2k^2+k+6)}\\ \text{c}.& 2+5+10+17+...+(k^2+2)\\ & ={\displaystyle \frac{1}{6}(k+2)(2k^2+5k+9)}\\ \text{d}.& (k^2+1)={\displaystyle \frac{1}{6}(k+1)(2k^2+k+6)}\\ \text{e}.& (n^2+2)={\displaystyle \frac{1}{6}(n+1)(2n^2+5n+9)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Cukup jelas}\\ & \text{Tinggal mensubstitusikan dari}\\ & \text{tiap}n\text{diganti}k\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 12+17+22+...+(5n+7)={\displaystyle \frac{1}{2}(n+1)(5n+14)}\\ & \text{Jika}S(n)\text{benar, untuk}n=k,\text{maka benar}\\ & \text{untuk}n=k+1.\text{Pernyataan ini dapat}\\ & \text{dinyatakan dengan}....\\ & \begin{array}{l}\\ \text{a}.& 12+17+22+...+(5k+7)={\displaystyle \frac{1}{2}(k+1)(5k+14)}\\ \text{b}.& 12+17+22+...+(5k+7)={\displaystyle \frac{1}{2}(k+1)(5k+19)}\\ \text{c}.& 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+1)(5k+19)}\\ \text{d}.& 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+2)(5k+14)}\\ \text{e}.& 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+2)(5k+19)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 12+17+22+...+(5k+7)={\displaystyle \frac{1}{2}(k+1)(5k+14)}\\ & 12+17+22+...+(5(k+1)+7)\\ & ={\displaystyle \frac{1}{2}((k+1)+1)(5(k+1)+14)}\\ & 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+2)(5k+19)}\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 4+5+6+7+...+(n+3)<5n^2\\ & \text{Jika}S(n)\text{benar, untuk}n=k+1,\text{maka}\\ & \text{pernyataan ini dapat ditulis dengan}....\\ & \begin{array}{l}\\ \text{a}.& 4+5+6+...+(k+4)<5k^2\\ \text{b}.& 4+5+6+...+(k+3)<5k^2\\ \text{c}.& 4+5+6+...+(k+3)<5(k+1{)}^2\\ \text{d}.& 4+5+6+...+(k+4)<5(k^2+2k+1)\\ \text{e}.& 4+5+6+...+(k+4)<5(k+1)(k-1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 4+5+6+...+(n+3)<5n^2\\ & \text{Saat}n=k+1,\text{maka}\\ & 4+5+6+...+((k+1)+3)<5(k+1{)}^2\\ & =4+5+6+...+(k+4)<5(k^2+2k+1)\end{array}\end{array}$.

Latihan Soal 1 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{l}\\ 1.& \text{Hasil dari}{\displaystyle \sum _{i=1}^{6}16i\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 306\\ \text{b}.& 314\\ \text{c}.& 326\\ \text{d}.& 336\\ \text{e}.& 402\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \sum _{i=1}^{6}16i}& =16.1+16.2+16.3+16.4+16.5+16.6\\ & =16+32+48+64+80+96\\ & =336\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Hasil dari}{\displaystyle \sum _{i=2}^9{i}^2\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 274\\ \text{b}.& 278\\ \text{c}.& 280\\ \text{d}.& 284\\ \text{e}.& 286\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \sum _{i=2}^9{i}^2}& =2^2+3^2+4^2+5^2+..+9^2\\ & =4+9+16+25+...+81\\ & =284\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Poa bilangan}12,14,16,18,20,...,(2n+10).\\ & \text{Nilai suku ke-100 adalah}....\\ & \begin{array}{l}\\ \text{a}.& 180\\ \text{b}.& 194\\ \text{c}.& 198\\ \text{d}.& 208\\ \text{e}.& 210\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{U}_n& =2n+10\\ U_{100}& =2\times 100+10\\ & =210\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Diketahui bahwa jika}\\ & 31+39+47+\cdots +8n+23=4n^2+27n\\ & \text{dengan}k,n\in \mathbb{N}\text{maka}\\ & 31+39+47+\cdots +8n+23+8k+31=....\\ & \begin{array}{l}\\ \text{a}.& 4k^2+27k\\ \text{b}.& 4k^2+35k\\ \text{c}.& 4k^2+35k+31\\ \text{d}.& 4k^2+35k+1\\ \text{e}.& 4k^2+35k+54\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \underset{4k^2+27k}{\underbrace{31+39+47+\cdots +8k+23}}+8k+31\\ & =4k^2+27k+8k+31\\ & =4k^2+35k+31\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Dengan Induksi Matematika untuk}n\in \mathbb{N}\\ & \text{dapat dibuktikan bahwa}n(n+1)(n+2)\\ & \text{akan habis dibagi oleh}\\ & \begin{array}{l}\\ \text{a}.& 4\\ \text{b}.& 5\\ \text{c}.& 6\\ \text{d}.& 7\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}P(n)& =n(n+1)(n+2)\\ P(1)& =1(1+1)(1+2)=1.2.3\\ & \text{adalah bilangan yang habis dibagi 6}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Dengan Induksi Matematika untuk}n\in \mathbb{N}\\ & \text{dapat dibuktikan juga bahwa}{7}^n-2^n\\ & \text{akan habis dibagi oleh}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}P(n)& =7^n-2^n\\ P(1)& =7^1-2^1\\ & =7-2=5\\ & \text{adalah bilangan yang habis dibagi 5}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Diketahui bahwa}P(n)\text{rumus dari}\\ & 3+6+9+\cdots +3n={\displaystyle \frac{3}{2}n(n+1)}\\ & \text{maka langkah pertama dengan induksi matematika}\\ & \text{dalam pembuktian rumus tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& P(n)\text{benar untuk}n=-1\\ \text{b}.& P(n)\text{benar untuk}n=1\\ \text{c}.& P(n)\text{benar untuk}n\text{bilangan bulat}\\ \text{d}.& P(n)\text{benar untuk}n\text{bilangan rasional}\\ \text{d}.& P(n)\text{benar untuk}n\text{bilangan real}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Langkah}& \text{awal yang harus ditunjukkan adalah}\\ n=1& \text{atau}P(1)\text{harus benar, yaitu}:\\ P(1)& =3.1(\mathit{\text{ruas kiri}})={\displaystyle \frac{3}{2}.1.(1+1)(\mathit{\text{ruas kanan}})=3}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Bila kita hendak membuktikan}{\displaystyle \sum _{i=1}^n={\displaystyle \frac{1}{2}n(n+1)}}\\ & \text{dengan induksi matematika}\\ & \text{maka untuk langkah}n=k+1\\ & \text{bentuk yang harus ditunjukkan adalah}...\\ & \begin{array}{l}\\ \text{a}.& 1+2+3+\cdots +n={\displaystyle \frac{1}{2}n(n+1)}\\ \text{b}.& 1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+1)}\\ \text{c}.& 1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+2)}\\ \text{d}.& 1+2+3+\cdots +k+(k+1)={\displaystyle \frac{1}{2}(k+1)(k+2)}\\ \text{e}.& 1+2+3+\cdots +k+(k+1)={\displaystyle \frac{1}{2}(k+2)(k+3)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}P(n)& =1+2+3+\cdots +n={\displaystyle \frac{1}{2}n(n+1)}\\ P(k)& =1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+1)}\\ P(k+1)& =1+2+3+\cdots +k+(k+1)\\ & =\underset{{\displaystyle \frac{1}{2}k(k+1)}}{\underbrace{1+2+3+\cdots +k}}+(k+1)\\ & ={\displaystyle \frac{1}{2}k(k+1)+(k+1)=(k+1)\left({\displaystyle \frac{1}{2}k+1}\right)}\\ & =(k+1){\displaystyle \frac{1}{2}(k+2)}\\ & ={\displaystyle \frac{1}{2}(k+1)(k+2)}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}P(n)={\displaystyle \frac{n-1}{n+3},\text{maka}P(k+1)}\\ & \text{dinyatakan dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{k-1}{k+3}}\\ \text{b}.& {\displaystyle \frac{k-1}{k+4}}\\ \text{c}.& {\displaystyle \frac{k}{k+4}}\\ \text{d}.& {\displaystyle \frac{k+1}{k+4}}\\ \text{e}.& {\displaystyle \frac{k+1}{k+5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}P(n)=& {\displaystyle \frac{n-1}{n+3}}\\ P(k+1)& ={\displaystyle \frac{k+1-1}{k+1+3}}\\ & ={\displaystyle \frac{k}{k+4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jika}P(n)={\displaystyle \frac{n^2+1}{4},\text{maka}}\\ & \text{pernyataan untuk}P(k+1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{k^2+2k+1}{4}}\\ \text{b}.& {\displaystyle \frac{k^2+2k+2}{4}}\\ \text{c}.& {\displaystyle \frac{k^2+2k+2}{5}}\\ \text{d}.& {\displaystyle \frac{k^2+2k+3}{5}}\\ \text{e}.& {\displaystyle \frac{k^2+2k+3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}P(n)& ={\displaystyle \frac{n^2+1}{4}}\\ P(k+1)& ={\displaystyle \frac{(k+1{)}^2+1}{4}}\\ & ={\displaystyle \frac{k^2+2k+2}{4}}\end{array}\end{array}$

Latihan Soal 11 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{l}\\ 101.& \text{Nilai dari}\sin {1020}^{\circ }=....\\ & \text{a}.-1\\ & \text{b}.{\displaystyle -\frac{1}{2}\sqrt{3}}\\ & \text{c}.{\displaystyle -\frac{1}{2}}\\ & \text{d}.{\displaystyle \frac{1}{2}}\\ & \text{e}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\sin {1020}^{\circ }& =\sin (3\times {360}^{\circ }-{60}^{\circ })\\ & =\sin (0^{\circ }-{60}^{\circ })\\ & =\sin (-{60}^{\circ })\\ & =-\sin {60}^{\circ }\\ & =-{\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 102.& \text{Nilai dari}\cot (-1290{)}^{\circ }=....\\ & \text{a}.-\sqrt{3}\\ & \text{b}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ & \text{c}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ & \text{d}.{\displaystyle -\frac{1}{2}}\\ & \text{e}.{\displaystyle \frac{1}{2}}\\ & \text{Jawab}:\text{a}\\ & \begin{array}{rl}\cot (-1290{)}^{\circ }& =-\cot (3\times {360}^{\circ }+{210}^{\circ })\\ & =-\cot (0^{\circ }+{210}^{\circ })\\ & =-\cot \left({210}^{\circ }\right)\\ & =-\frac{1}{\tan {210}^{\circ }}\\ & =-\frac{1}{\tan ({180}^{\circ }+{30}^{\circ })}\\ & =-\frac{1}{\tan {30}^{\circ }}\\ & =-{\displaystyle \sqrt{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 103.& \text{Nilai dari}\\ & \sin {240}^{\circ }+\sin {225}^{\circ }+\cos {315}^{\circ }=....\\ & \begin{array}{l}\\ \text{a}.-\sqrt{3}& & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{b}.{\displaystyle -\frac{1}{2}\sqrt{3}}& & \text{e}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \text{c}.{\displaystyle -\frac{1}{2}}\end{array}\\ & \text{Jawab}:\text{b}\\ & \begin{array}{rl}& \sin {240}^{\circ }+\sin {225}^{\circ }+\cos {315}^{\circ }\\ & =\sin ({180}^{\circ }+{60}^{\circ })+\sin ({180}^{\circ }+{45}^{\circ })+\cos ({360}^{\circ }-{45}^{\circ })\\ & =-\sin {60}^{\circ }+[-\sin {45}^{\circ }]+\cos {45}^{\circ }\\ & =(-\frac{1}{2}\sqrt{3})+(-\frac{1}{2}\sqrt{2})+\frac{1}{2}\sqrt{2}\\ & =-\frac{1}{2}\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 104.& \text{Nilai dari}\\ & {\displaystyle \frac{\sin {30}^{\circ }+\sin {150}^{\circ }+\cos {330}^{\circ }}{\tan {45}^{\circ }+\cos {210}^{\circ }}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}}\\ \text{b}.{\displaystyle \frac{1-\sqrt{3}}{1+\sqrt{3}}}\\ \text{c}.{\displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}}\\ \text{d}.{\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}}\\ \text{e}.{\displaystyle \frac{1+2\sqrt{3}}{1-2\sqrt{3}}}\end{array}\\ & \text{Jawab}:\text{d}\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {30}^{\circ }+\sin {150}^{\circ }+\cos {330}^{\circ }}{\tan {45}^{\circ }+\cos {210}^{\circ }}}\\ & ={\displaystyle \frac{\sin {30}^{\circ }+\sin ({180}^{\circ }-{30}^{\circ })+\cos ({360}^{\circ }-{30}^{\circ })}{\tan {45}^{\circ }+\cos ({180}^{\circ }+{30}^{\circ })}}\\ & ={\displaystyle \frac{\sin {30}^{\circ }+\sin {30}^{\circ }+\cos {30}^{\circ }}{\tan {45}^{\circ }-\cos {30}^{\circ }}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}\sqrt{3}}}{1-{\displaystyle \frac{1}{2}\sqrt{3}}}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{1}{2}\sqrt{3}}}{1-{\displaystyle \frac{1}{2}\sqrt{3}}}}\\ & ={\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 105.& \text{Nilai dari}\\ & {\displaystyle \frac{\sin {270}^{\circ }\times \cos {135}^{\circ }\times \tan {135}^{\circ }}{\sin {150}^{\circ }\times \cos {225}^{\circ }}=....}\\ & \begin{array}{l}\\ \text{a}.-2& & \text{d}.1\\ \text{b}.{\displaystyle -\frac{1}{2}}& \text{c}.{\displaystyle \frac{1}{2}}& \text{e}.2\end{array}\\ & \text{Jawab}:\text{e}\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {270}^{\circ }\times \cos {135}^{\circ }\times \tan {135}^{\circ }}{\sin {150}^{\circ }\times \cos {225}^{\circ }}}\\ & ={\displaystyle \frac{\sin {270}^{\circ }\times \cos ({180}^{\circ }-{45}^{\circ })\times \tan ({180}^{\circ }-{45}^{\circ })}{\sin ({180}^{\circ }-{30}^{\circ })\times \cos ({180}^{\circ }+{45}^{\circ })}}\\ & ={\displaystyle \frac{-1\times (-\cos {45}^{\circ })\times (-\tan {45}^{\circ })}{\sin {30}^{\circ }\times (-\cos {45}^{\circ })}}\\ & ={\displaystyle \frac{-1\times (-{\displaystyle \frac{1}{2}\sqrt{2}})\times -1}{{\displaystyle \frac{1}{2}\times (-\frac{1}{2}\sqrt{2})}}}\\ & ={\displaystyle \frac{-\frac{1}{2}\sqrt{2}}{-\frac{1}{4}\sqrt{2}}}\\ & ={\displaystyle 2}\end{array}\end{array}$

$\begin{array}{l}\\ 106.& \text{Perhatikanlah gambar kurva berikut ini}\end{array}$.

$\begin{array}{l}\\ .& \begin{array}{l}\\ \text{a}.& y=-2\cos 2x\\ \text{b}.& y=2\cos {\displaystyle \frac{3}{2}x}\\ \text{c}.& y=-2\cos {\displaystyle \frac{3}{2}x}\\ \text{d}.& y=2\sin {\displaystyle \frac{3}{2}x}\\ \text{e}.& y=-2\sin {\displaystyle \frac{3}{2}x}\\ \\ & (\mathbf{\text{SIMAK UI 2009 Mat Das}})\end{array}\\ & \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \text{Dari gambar tampak jelas bahwa}\\ & \text{grafik di atas atas adalah grafik}\mathbf{\text{fungsi cosinus}}\\ & \text{dengan amplitudo 2 dan periodenya}:{\displaystyle \frac{3}{2}\pi }\\ & \text{Maka persamaan grafiknya adalah}:\\ & y=2\cos {\displaystyle \frac{3}{2}\pi }\\ & \text{Karena posisinya terbalik, maka}\\ & y=-2\cos {\displaystyle \frac{3}{2}\pi }\end{array}\end{array}$.

$\begin{array}{l}\\ 107.& \text{Nilai minimum jika}\\ & f(x)={(2004\cos 2005x-2006)}^2+2007\\ & \text{adalah}=....\\ & \begin{array}{l}\\ \text{a}.& 2005\\ \text{b}.& 2007\\ \text{c}.& 2011\\ \text{d}.& 2013\\ \text{e}.& \text{tidak ada satupun jawaban dari a sampai d}\\ \\ & (\mathbf{\text{NUS Mathematics A Level}})\end{array}\\ & \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& f(x)={(2004\cos 2005x-2006)}^2+2007\\ & \text{Supaya bernilai minimum, }\\ & \text{maka nilai}\cos 500x=1,\\ & \text{ingat nilai}-1\le \cos n\pi \le 1\\ & \text{maka},\\ & f_{min}={(2004.1-2006)}^2+2007\\ & =(-2{)}^2+2007\\ & =4+2007\\ & =2011\end{array}\end{array}$.

$\begin{array}{l}\\ 108.& \text{Penyelesaian persamaan}\\ & {\cos }^{2}x-2\cos x=4\sin x-2\sin x\cos x\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \pi -{\cot }^{-1}\left({\displaystyle \frac{1}{2}}\right)\\ \text{b}.& \pi +{\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)\\ \text{c}.& \pi -{\cot }^{-1}(-1)\\ \text{d}.& \pi +{\tan }^{-1}(-{\displaystyle \frac{1}{2}})\\ \text{e}.& \pi -{\tan }^{-1}\left({\displaystyle \frac{1}{4}}\right)\end{array}\\ & \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa},\\ & {\cos }^{2}x-2\cos x=4\sin x-2\sin x\cos x\\ & \cos x(\cos x-2)=2\sin x(2-\cos x)\\ & \cos x(\cos x-2)=-2\sin x(\cos x-2)\\ & (\cos x+2\sin x)(\cos x-2)=0\\ & (\cos x+2\sin x)=0\text{atau}(\cos x-2)=0\\ & 2\sin x=-\cos x\mathbf{\text{(mm)}}\text{atau}\cos x=2\mathbf{\text{(tm)}}\\ & \text{maka}\\ & {\displaystyle \frac{\sin x}{\cos x}=-{\displaystyle \frac{1}{2}\iff \tan x=-{\displaystyle \frac{1}{2}}}}\\ & x={\tan }^{-1}(-{\displaystyle \frac{1}{2}})+k.\pi ,k\in \mathbb{Z}\end{array}\end{array}$.

$\begin{array}{l}\\ 109.& \text{Jika diketahui bahwa}\\ & \sin \beta -\tan \beta -2\cos \beta +2=0\\ & \text{dengan}0<\beta <{\displaystyle \frac{\pi }{2},}\\ & \text{maka himpunan harga}\sin \beta =....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{2}{5}\sqrt{5}}\right\}\\ \text{b}.& \left\{0\right\}\\ \text{c}.& \left\{{\displaystyle \frac{2}{5}\sqrt{5},0}\right\}\\ \text{d}.& \left\{{\displaystyle \frac{1}{5}\sqrt{5}}\right\}\\ \text{e}.& \left\{{\displaystyle \frac{1}{5}\sqrt{5},0}\right\}\\ \\ & \mathbf{\text{(SIMAK UI 2009)}}\end{array}\\ & \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \sin \beta -\tan \beta -2\cos \beta +2=0\\ & \sin \beta -{\displaystyle \frac{\sin \beta }{\cos \beta }-2\cos \beta +2=0}\\ & \sin \beta \cos \beta -\sin \beta -2{\cos }^2\beta +2\cos \beta =0\\ & \sin \beta (\cos \beta -1)-2\cos \beta (\cos \beta -1)=0\\ & (\sin \beta -2\cos \beta )(\cos \beta -1)=0\\ & (\sin \beta -2\cos \beta )=0\mathbf{\text{(mm)}}\\ & \text{atau}(\cos \beta -1)=0\mathbf{\text{(tmm)}}\\ & \text{maka},\\ & (\sin \beta -2\cos \beta )=0\\ & \sin \beta =2\cos \beta \\ & {\displaystyle \frac{\sin \beta }{\cos \beta }=2}\\ & \tan \beta =2={\displaystyle \frac{2}{1},\text{buatlah ilustrasi}}\\ & \text{dengan membuat segitiga siku-siku}.\\ & \text{Sehingga akan didapatkan nilai}\\ & \sin \beta ={\displaystyle \frac{2}{5}\sqrt{5}}\end{array}\end{array}$.

$\begin{array}{l}\\ 110.& \text{Diketahui bahwa}\\ & \sin \theta -\cos \theta ={\displaystyle \frac{\sqrt{5}-\sqrt{3}}{2}\text{dan}}\\ & {\cos }^3\theta -{\sin }^3\theta ={\displaystyle \frac{1}{a}(b\sqrt{5}-c\sqrt{3}),}\\ & \text{dengan}a,b,c\text{adalah bilangan asli, maka}\\ & (1)b-c>0\\ & (2)a-b=7\\ & (3)a-3b+c=0\\ & (4)a+b+c=12\\ & \begin{array}{l}\\ \text{a}.& (1),(2).\text{dan}(3)\text{benar}\\ \text{b}.& (1),\text{dan}(3)\text{benar}\\ \text{c}.& (2),\text{dan}(4)\text{benar}\\ \text{d}.& \text{hanya}(4)\text{yang benar}\\ \text{e}.& \text{semuanya benar}\\ \\ & (\mathbf{\text{SIMAK UI 2015 Mat IPA}})\end{array}\\ & \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \sin \theta -\cos \theta ={\displaystyle \frac{\sqrt{5}-\sqrt{3}}{2}}\\ & {\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta ={\displaystyle \frac{8-2\sqrt{5}}{4}}\\ & 1-2\sin \theta \cos \theta ={\displaystyle \frac{8-2\sqrt{5}}{4}}\\ & \sin \theta \cos \theta ={\displaystyle \frac{\sqrt{5}-2}{4}}\\ & \text{maka},\\ & {\cos }^3\theta -{\sin }^3\theta \\ & =(\cos \theta -\sin \theta )({\cos }^2\theta +\sin \theta \cos \theta +{\sin }^2\theta )\\ & =\left({\displaystyle \frac{\sqrt{3}-\sqrt{5}}{2}}\right)(1+{\displaystyle \frac{\sqrt{5}-2}{4}})\\ & ={\displaystyle \frac{1}{8}(\sqrt{3}-\sqrt{5})(2+\sqrt{5})}\\ & ={\displaystyle \frac{1}{8}(2\sqrt{3}+3\sqrt{5}-2\sqrt{5}-5\sqrt{3})}\\ & ={\displaystyle \frac{1}{8}(\sqrt{5}-3\sqrt{3})={\displaystyle \frac{1}{a}(b\sqrt{5}-c\sqrt{3})\{\begin{array}{c}a=8\\ b=1\\ c=3\end{array}}}\\ & \text{sehingga}\\ & a-b=8-1=7\\ & a+b+c=8+1+3=12\end{array}\end{array}$

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

$\begin{array}{l}\\ 91.& \text{Bentuk sederhana dari}\\ & 4\sin \left({\displaystyle \frac{1}{4}\pi +x}\right)\cos \left({\displaystyle \frac{1}{4}\pi -x}\right)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 2+2\sin 2x}& & & \text{d}.& 2+2\sin x\\ \text{b}.& {\displaystyle 2+\sin 2x}& & & \text{e}.& 2+\sin x\\ \text{c}.& {\displaystyle 2\sin 2x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& 4\sin \left({\displaystyle \frac{1}{4}\pi +x}\right)\cos \left({\displaystyle \frac{1}{4}\pi -x}\right)\\ & =2(\sin \left({\displaystyle \frac{1}{2}\pi }\right)+\sin (2x))\\ & =2(1+\sin 2x)\\ & =2+2\sin 2x\end{array}\end{array}$.

$\begin{array}{l}\\ 92.& \text{Bentuk sederhana dari}\\ & 2\sin \left({\displaystyle \frac{3}{4}\pi +x}\right).\cos \left({\displaystyle \frac{\pi }{4}+x}\right)=....\\ & \begin{array}{l}\\ \text{a}.1-\sin 2x& & \text{d}.\cos 2x\\ \\ \text{b}.1+\sin 2x& \text{c}.1-\cos 2x& \text{e}.-\cos 2x\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}& 2\sin \left({\displaystyle \frac{3}{4}\pi +x}\right).\cos \left({\displaystyle \frac{\pi }{4}+x}\right)\\ & =2\sin ({135}^{\circ }+x).\cos ({45}^{\circ }+x)\\ & =2\sin ({90}^{\circ }+{45}^{\circ }+x).\cos ({45}^{\circ }+x)\\ & =2\cos ({45}^{\circ }+x).\cos ({45}^{\circ }+x)\\ & =2{\cos }^2({45}^{\circ }+x)\\ & =2{(\cos {45}^{\circ }\cos x-\sin {45}^{\circ }\sin x)}^2\\ & =2{(\frac{1}{2}\sqrt{2}.\cos x-{\displaystyle \frac{1}{2}\sqrt{2}.\sin x})}^2\\ & ={\displaystyle {(\cos x-\sin x)}^2}\\ & ={\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x\\ & =1-\sin 2x\end{array}\end{array}$.

$\begin{array}{l}\\ 93.& \text{Bentuk sederhana dari}\\ & 2\cos (x+{\displaystyle \frac{\pi }{4}}).\cos \left({\displaystyle \frac{3}{4}\pi -x}\right)=....\\ & \begin{array}{l}\\ \text{a}.\cos (2x+\pi )& & \text{d}.-\cos (2x-{\displaystyle \frac{\pi }{2}})\\ \text{b}.\cos (2x+{\displaystyle \frac{\pi }{2}})& & \text{e}.\sin 2x-1\\ \text{c}.\cos 2x\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 2\cos (x+{\displaystyle \frac{\pi }{4}}).\cos \left({\displaystyle \frac{3}{4}\pi -x}\right)\\ & =\cos (x+{\displaystyle \frac{\pi }{4}+{\displaystyle \frac{3}{4}\pi -x}})+\cos (x+{\displaystyle \frac{\pi }{4}-\left({\displaystyle \frac{3}{4}\pi -x}\right)})\\ & =\cos \pi +\cos (2x-\frac{2}{4}\pi )\\ & =-1+\cos \left({\displaystyle \frac{1}{2}\pi -2x}\right),\\ & \text{ingat bahwa}\cos (-\alpha )=\cos \alpha \\ & =-1+\sin 2x\\ & =\sin 2x-1\end{array}\end{array}$.

$\begin{array}{l}\\ 94.& \text{Nilai dari}\\ & \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{3}{8}}}& & & \text{d}.& {\displaystyle \frac{3}{8}}\\ \\ \text{b}.& {\displaystyle -\frac{1}{8}}& & & \text{e}.& {\displaystyle \frac{5}{8}}\\ \text{c}.& {\displaystyle \frac{1}{8}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & =\sqrt{3}\sin {80}^{\circ }\sin {20}^{\circ }(-\sin {40}^{\circ })\\ & =-\sqrt{3}\sin {80}^{\circ }\sin {40}^{\circ }\sin {20}^{\circ }\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{2}(\cos {60}^{\circ }-\cos {20}^{\circ })})\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{4}+{\displaystyle \frac{\cos {20}^{\circ }}{2}}})\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{2}\sqrt{3}\sin {80}^{\circ }\cos {20}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {100}^{\circ }+\sin {60}^{\circ })}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {80}^{\circ }+{\displaystyle \frac{1}{2}\sqrt{3}})}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }+{\displaystyle \frac{1}{8}\sqrt{9}}}}\\ & ={\displaystyle \frac{3}{8}}\end{array}\end{array}$.

$\begin{array}{l}\\ 95.& \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  $\cos {\displaystyle \frac{\pi }{7}-\cos {\displaystyle \frac{2\pi }{7}+\cos {\displaystyle \frac{3\pi }{7}={\displaystyle \frac{1}{2}}}}}$.

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

$\begin{array}{l}\\ 96.& \text{Nilai dari}\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{16}}}& & & \text{d}.& {\displaystyle \frac{1}{2}}\\ \\ \text{b}.& {\displaystyle \frac{1}{8}}& \text{c}.& {\displaystyle \frac{1}{4}}& \text{e}.& {\displaystyle 1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Perhat}& \text{ikan bahwa}\\ \sin {\displaystyle {\displaystyle \frac{\pi }{14}}}& =\sin \left({\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14}}\right)=\sin \left({\displaystyle \frac{1}{2}\pi -\frac{6\pi }{14}}\right)\\ & =\cos {\displaystyle \frac{6\pi }{14}}\\ \sin {\displaystyle \frac{3\pi }{14}}& =...=\cos {\displaystyle \frac{4\pi }{14}}\\ \sin {\displaystyle \frac{9\pi }{14}}& =...=\sin {\displaystyle \frac{5\pi }{14}=\cos {\displaystyle \frac{2\pi }{14}}}\end{array}\\ & ...\\ & \sin {\displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}}\\ & =\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\cos {\displaystyle \frac{2\pi }{14}\times {\displaystyle \frac{2\sin {\displaystyle \frac{2\pi }{14}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}}}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{6\pi }{14}\cos {\displaystyle \frac{4\pi }{14}\sin {\displaystyle \frac{4\pi }{14}}}}}{2\sin {\displaystyle \frac{2\pi }{14}}}}\\ & \text{silahkan dilanjutkan}\\ & ...\\ & ={\displaystyle \frac{1}{8}}\end{array}$.

$\begin{array}{l}\\ 97.& \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}$.

$\begin{array}{l}\\ 98.& \text{Nilai dari}\sin {\displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{5}{16}}& & \text{d}.{\displaystyle \frac{2}{16}}\\ \\ \text{b}.{\displaystyle \frac{4}{16}}& \text{c}.{\displaystyle \frac{3}{16}}& \text{e}.{\displaystyle \frac{1}{16}}\end{array}\\ \\ & \mathbf{\text{(Olimpiade Sains PORSEMA NU 2012)}}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \sin {\displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}}\\ & ={\displaystyle \frac{1}{4}\left(2\sin {\displaystyle \frac{11\pi }{24}.\sin \frac{\pi }{24}.2\sin \frac{7\pi }{24}.\sin \frac{5\pi }{24}}\right)}\\ & ={\displaystyle \frac{1}{4}[(\cos \left(\frac{10\pi }{24}\right)-\cos \left(\frac{12\pi }{24}\right))\times (\cos \left(\frac{2\pi }{24}\right)-\cos \left(\frac{12\pi }{24}\right))]}\\ & ={\displaystyle \frac{1}{4}[(\cos {75}^{\circ }-\cos {90}^{\circ })\times (\cos {15}^{\circ }-\cos {90}^{\circ })]}\\ & ={\displaystyle \frac{1}{4}[\cos {75}^{\circ }.\cos {15}^{\circ }]}\\ & ={\displaystyle \frac{1}{8}[\cos {90}^{\circ }+\cos {60}^{\circ }]}\\ & ={\displaystyle \frac{1}{8}(0+\frac{1}{2})}\\ & ={\displaystyle \frac{1}{16}}\end{array}\end{array}$.

$\begin{array}{l}\\ 99.& \text{Nilai dari}\\ & \sin {18}^{\circ }\cos {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{6}}& & & \text{d}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{1}{5}}& \text{c}.& {\displaystyle {\displaystyle \frac{1}{4}}}& \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sin {18}^{\circ }\cos {36}^{\circ }\\ & =\sin {18}^{\circ }\cos {36}^{\circ }\times {\displaystyle \frac{2\cos {18}^{\circ }}{2\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }(\sin {36}^{\circ }+\sin 0^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {36}^{\circ }\sin {36}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin {72}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\sin ({90}^{\circ }-{18}^{\circ })}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{\cos {18}^{\circ }}{4\cos {18}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 100.& \text{Nilai eksak dari}\sin {36}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}}}& & & \text{d}.& {\displaystyle \frac{\sqrt{5}-1}{4}}\\ \text{b}.& {\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}}& & & \text{e}.& {\displaystyle \frac{\sqrt{5}-1}{2}}\\ \text{c}.& {\displaystyle {\displaystyle \frac{\sqrt{5}+1}{4}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah ilustrasi gambar berikut}\end{array}$.

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