Contoh Soal Induksi Matematika

\begin{array}{ll}\\ \fbox{1}.&\textrm{Diketahui bahwa jika}\: \: 31+39+47+\cdots +8n+23=4n^{2}+27n\: \: \textrm{dengan}\: \: k,n\in \mathbb{N}\\ &\textrm{maka}\: \: 31+39+47+\cdots +8n+23+8k+31=....\\ &\begin{array}{lllllll}\\ \textrm{a}.&4k^{2}+27k&&&\textrm{d}.&4k^{2}+35k+1\\ \textrm{b}.&4k^{2}+35k&\textrm{c}.&4k^{2}+35k+31&\textrm{e}.&4k^{2}+35k+54\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\begin{aligned}\underset{4k^{2}+27k}{\underbrace{31+39+47+\cdots +8k+23}}+8k+31&=4k^{2}+27k+8k+31\\ &=4k^{2}+35k+31 \end{aligned} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Dengan Induksi Matematika untuk}\: \: n\in \mathbb{N}\: \: \textrm{dapat dibuktikan bahwa}\: \: n(n+1)(n+2)\\ &\textrm{akan habis dibagi oleh}\\ &\begin{array}{lllllll}\\ \textrm{a}.&4&&&\textrm{d}.&7\\ \textrm{b}.&5&\textrm{c}.&6&\textrm{e}.&8\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}P(n)&=n(n+1)(n+2)\\ P(1)&=1(1+1)(1+2)=1.2.3\\ &\textrm{adalah bilangan yang habis dibagi 6} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Dengan Induksi Matematika untuk}\: \: n\in \mathbb{N}\: \: \textrm{dapat dibuktikan juga bahwa}\: \: 7^{n}-2^{n}\\ &\textrm{akan habis dibagi oleh}\\ &\begin{array}{lllllll}\\ \textrm{a}.&2&&&\textrm{d}.&5\\ \textrm{b}.&3&\textrm{c}.&4&\textrm{e}.&6\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}P(n)&=7^{n}-2^{n}\\ P(1)&=7^{1}-2^{1}\\ &=7-2=5\\ &\textrm{adalah bilangan yang habis dibagi 5} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Diketahui bahwa}\: \: P(n)\: \: \textrm{rumus dari}\: \: 3+6+9+\cdots +3n=\displaystyle \frac{3}{2}n(n+1)\\ &\textrm{maka langkah pertama dengan induksi matematika}\\ &\textrm{dalam pembuktian rumus tersebut adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.&P(n)\: \: \textrm{benar untuk}\: \: n=-1\\ \textrm{b}.&P(n)\: \: \textrm{benar untuk}\: \: n=1\\ \textrm{c}.&P(n)\: \: \textrm{benar untuk}\: \: n\: \: \textrm{bilangan bulat}\\ \textrm{d}.&P(n)\: \: \textrm{benar untuk}\: \: n\: \: \textrm{bilangan rasional}\\ \textrm{d}.&P(n)\: \: \textrm{benar untuk}\: \: n\: \: \textrm{bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Langkah}&\: \textrm{awal yang harus ditunjukkan adalah}\\ n=1&\: \: \textrm{atau}\: \: P(1)\: \: \textrm{harus benar, yaitu}:\\ P(1)&=3.1(\textit{ruas kiri})=\displaystyle \frac{3}{2}.1.(1+1)(\textit{ruas kanan})=3 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Bila kita hendak membuktikan}\: \: \sum_{i=1}^{n}=\displaystyle \frac{1}{2}n(n+1)\: \: \textrm{dengan induksi matematika}\\ &\textrm{maka untuk langkah}\: \: n=k+1\: , \: \textrm{bentuk yang harus ditunjukkan adalah}\\ &\begin{array}{lll}\\ \textrm{a}.&1+2+3+\cdots +n=\displaystyle \frac{1}{2}n(n+1)\\ \textrm{b}.&1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+1)\\ \textrm{c}.&1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+2)\\ \textrm{d}.&1+2+3+\cdots +k+(k+1)=\displaystyle \frac{1}{2}(k+1)(k+2)\\ \textrm{e}.&1+2+3+\cdots +k+(k+1)=\displaystyle \frac{1}{2}(k+2)(k+3)\\ \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}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{aligned} \end{array}.

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insyaAllah

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Program Linear Kelas XI ( K13 Revisi )

Sebelumnya kita buka arsip lama

Program Linear adalah suatu metode atau progaram untuk menentukan nilai optimal (maksimum atau minimum) dari beberapa pertidaksamaan linear yang diketahui.

Dalam program linear terdapat dua bagian yaitu fungsi kendala (batasan-batasan berupa pertidaksamaan) dan fungsi Objektif (sasaran / tujuan).

Sebagai tambahan sebagai kemudahan dalam mengingat kaitannya dalam materi ini khususnya dalam menentukan himpunan penyelesaian pertidaksamaan 2 variabel(peubah)

A. Menentukan Persamaan Garis Sebelum Menjadi Pertidaksamaan

\begin{array}{|c|c|}\hline \textrm{Gambar (I)}&\textrm{Gambar (II)}\\\hline &\\ &\displaystyle \frac{y-b}{n-b}=\displaystyle \frac{x-a}{m-a}\\ &\\ \textbf{ax+by}=\textbf{ab}&\textrm{atau}\\ &\\ &y=\displaystyle \frac{n-b}{m-a}\left ( x-a \right )+b\\ &\\\hline \end{array}.

B. Menentukan Daerah Pertidaksamaan

Adalah  koefisien x bertanda positif, maka

\begin{cases} \textrm{Daerah yang diarsir adalah \textbf{sebelah kiri} garis}\: \: \left (< \textrm{atau} \leq \right ) \\\\ \textrm{Daerah yang diarsir adalah \textbf{sebelah kanan} garis}\: \: \left (> \textrm{atau} \geq \right ) \end{cases}.

Coba perhatikanlah ilustrasi berikut!

C. Langkah-Langkah Penyelesaian Program Linear

Hal penting dalam menyelesaikan program linear adalah

Model Matematika dan Penentuan Nilai Optimum Fungsi Objektif

adalapun langkah-langkahnya sebagai berikut:

  • Membuat model matematika(menerjemahkan persoalan ke dalam bahasa matematika)
  • Menyelesaikan sistem pertidaksamaan linear (dua variabel) dengan mengarsir daerah yang memenuhi pertidaksamaan (tulisan yang bergaris miring terserah selera pembaca)
  • menentukan titik-titik sudut (Verteks / titik ekstrem )
  • Menentukan penyelesaian Optimasi dari fungsi objektif (kadang ditulis sebagai fungsi sasaran / tujuan) f(x,y)=ax+by baik dengan  metode uji titik sudut (Verteks / titik ekstrem) atau garis selidik.

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

\begin{array}{ll}\\ 1.&(\textrm{SPMB 2003})\\ &\textrm{Daerah yang diarsir pada ilustrasi gambar berikut adalah himpunan semua (x,y)}\\ &\textrm{yang memenuhi} \end{array}.

.\, \quad\begin{array}{ll}\\ \textrm{A}.&2x+y\leq 30,\: 3x+4y\leq 60,\: x\geq 0,\: y\geq 0\\ \textrm{B}.&2x+y\geq 30,\: 3x+4y\geq 60,\: x\geq 0,\: y\geq 0\\ \textrm{C}.&x+2y\geq 30,\: 3x+4y\geq 60,\: x\geq 0,\: y\geq 0\\ \textrm{D}.&x+2y\leq 30,\: 4x+3y\leq 60,\: x\geq 0,\: y\geq 0\\ \textrm{E}.&2x+y\geq 30,\: 4x+3y\leq 60,\: x\geq 0,\: y\geq 0 \end{array}.

.\, \quad\begin{aligned}&\textrm{Jawab}:\quad \textbf{A}\\ &\textrm{Persamaan garisnya adalah}:\\ &\begin{array}{|c|c|}\hline \begin{array}{ll|llllllllll}\\ &Y&&&&\\ &&&&&&\\ &15&&&&\\ &&&&&\\ &&&&&&&&&X\\\hline &0&&&&&&20&&\\ \multicolumn{4}{l}{.}&&& \end{array}&\begin{array}{ll|llllllll}\\ &Y&&&&\\ &&&&&&\\ &30&&&&\\ &&&&&\\ &&&&&&&X\\\hline &0&&&15&&&\\ \multicolumn{4}{l}{.}&&& \end{array}\\\hline \begin{aligned}15x+20y&=15\times 20\\ 3x+4y&=60 \end{aligned}&\begin{aligned}30x+15y&=30\times 15\\ 2x+y&=30 \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{aligned}&\textrm{arsiran sebelah kiri}\\ &\textrm{garis masing-masing yang tersebut}\\ &\textrm{sehingga pertidaksamaan akan berupa} \end{aligned}}\\\hline 3x+4y\leq 60&2x+y\leq 30\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 2.&(\textrm{SPMB 2004})\\ &\textrm{Daerah yang diarsir pada ilustrasi gambar berikut adalah himpunan penyelesaian}\\ &\textrm{yang dimenuhi oleh} \end{array}.

.\, \quad\begin{array}{ll}\\ \textrm{A}.&6x+5y-30\leq 0,\: x+6y-6\leq 0,\: x-y\leq 0\\ \textrm{B}.&6x+5y-30\geq 0,\: x+6y-6\leq 0,\: x-y\leq 0\\ \textrm{C}.&6+5y-30\leq 0,\: x+6y-6\geq 0,\: x-y\geq 0\\ \textrm{D}.&6x+5y-30\leq 0,\: x+6y-6\geq 0,\: x-y\leq 0\\ \textrm{E}.&6x+5y-30\geq 0,\: x+6y-6\geq 0,x-y\geq 0 \end{array}.

.\, \quad\begin{aligned}&\textrm{Jawab}:\quad \textbf{C}\\ &\textrm{Perhatikan bahwa}:\\ &\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\textrm{Masing-masing garisnya adalah}:}\\\hline \begin{aligned}6x+5y&=6\times 5\\ 6x+5y&=30 \end{aligned}&\begin{aligned}x+6y&=1\times 6\\ x+6y&=6 \end{aligned}&\begin{aligned}x&=y\\ \end{aligned}\\\hline \multicolumn{3}{|c|}{\textrm{Daerah yang diarsir adalah}}\\\hline \textbf{Sebelah kiri}&\multicolumn{2}{|c|}{\textbf{Sebelah kanan}}\\\hline \begin{aligned}6x+5y&\leq 30\\ 6x+5y-30&\leq 0 \end{aligned}&\begin{aligned}x+6y&\geq 6\\ x+6y-6&\geq 0 \end{aligned}&\begin{aligned}x&\geq y\\ x-y&\geq 0 \end{aligned}\\\hline \end{array} \end{aligned}.

\begin{tabular}{lp{20.0cm}}\\ 3.&Seorang penjahit memiliki persediaan 16 m kain sutera, 15 m katun, dan 11 m rayon yang hendak dibuat dua buah model dengan rincian sebagai berikut.\\ &Model A memerlukan 2 m sutera, 1 m katon, dan 1 m rayon per unit.\\ &Model B memerlukan 1 m sutera, 3 m katun, dan 2 m rayon perunit.\\ &Jika keuntungan pakaian model A adalah Rp3.000,00 per unitnya dan model B akan memberikan keuntungan per unitnya Rp5.000,00 , maka tentukanlah berapa banyak masing-masing pakaian harus dibuat didapatkan keuntungan yang sebesar-besarnya! \end{tabular}.

.\, \quad\begin{aligned}&\textrm{Jawab}:\\ &\textrm{Persoalan tersebut di atas bila dituliskan ke dalam tabel adalah sebagai berikut}:\\ &\begin{array}{|l|c|c|c|}\hline \textrm{Bahan Kain}&\textrm{Model A} &\textrm{Model B}&\textrm{Tersedia}\\\hline \textrm{Sutera}&2&1&16\\\hline \textrm{Katun}&1&3&15\\\hline \textrm{Rayon}&1&2&11\\\hline \textrm{Keuntungan}&3.000&5.000&\\\hline \end{array}\\ &\textrm{Dan untuk \textbf{model matematika}nya adalah}:\\ &\begin{array}{|c|c|l|}\hline \multicolumn{3}{|c|}{\textbf{Proses Penyelesaian persoalan Program Linear}}\\\hline \textbf{Permasalahan}&\textbf{Fungsi Tujuan}&\textbf{Kendala-Kendala}\\\hline \begin{aligned}&\textrm{Menentukan keuntungan}\\ &\textrm{sebesar-besarnya} \end{aligned}&f(x,y)=3000x+5000y&\textcircled{1}\quad 2x+y\leq 16\\ &&\textcircled{2}\quad x+3y\leq 15\\ &&\textcircled{3}\quad x+2y\leq 11\\ &&\textcircled{4}\quad x\geq 0\\ &&\textcircled{5}\quad y\geq 0\\\hline \end{array}\\ & \end{aligned}.

.\, \quad\begin{aligned}&\textrm{Langkah berikutnya adalah menentukan titik \textbf{verteks/ektrim}}:\\ &\textrm{Perhatikan ilustrasi gambar berikut}\\ &\textrm{daerah yang tidak berarsir adalah \textbf{daerah/wilayah penyelesaian}} \end{aligned}.

.\, \quad\begin{aligned}&\textrm{Dengan bantuan ilustrasi gambar kita mendapatkan koordinat \textbf{titik-titik pojok (verteks/ektrem)}}:\\ &\textrm{yaitu A(8,0), B(7,2), C(3,4), dan D(0,5) khusus titik (0,0) tidak diperlukan }\\ &\textrm{karena yang diinginkan adalah nilai maksimal (keuntungan terbesar)}\\ &\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\textbf{f(x,y) = 3000x+5000y}}\\\hline \textrm{Titik Vertek}&\textrm{Nilai Optimum}&\textrm{Keterangan}\\\hline A(8,0)&f(A)=f(8,0)=3000(8)+5000(0)=24000&\textbf{Minimum}\\\hline B(7,2)&f(B)=f(7,2)=3000(7)+5000(2)=31000&\textbf{Maksimum}\\\hline C(3,4)&f(C)=f(3,4)=3000(3)+5000(4)=29000&\\\hline D(0,5)&f(D)=f(0,5)=3000(0)+5000(5)=25000&\\\hline \end{array}\\ &\textrm{Jadi, supaya penjahit medapat keuntungan \textbf{sebesar-besarnya} (Rp31.000,00), maka ia}\\ &\textrm{harus \textbf{membuat 7 unit} model A dan \textbf{2 unit} model B} \end{aligned}.

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

\begin{array}{ll}\\ 1.&\textrm{Tunjukkan pada \textbf{bidang Cartesius} wilayah himpunan penyelesaian dari}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x< -5&\textrm{e}.&x> 2&\textrm{i}.&\left | y \right |> 1\\ \textrm{b}.&x\leq -5&\textrm{f}.&y\leq -3&\textrm{j}.&\left | y \right |\geq 4\\ \textrm{c}.&y> 7&\textrm{g}.&\left | x \right |< 2&\textrm{k}.&\left | x \right |+\left | y \right |< 4\\ \textrm{d}.&y\geq 7&\textrm{h}.&\left | x \right |\leq 2&\textrm{l}.&\left | x \right |+\left | y \right |\geq 4 \end{array} \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Tunjukkan pada \textbf{bidang Cartesius} wilayah himpunan penyelesaian dari}\\ &\begin{array}{llllllll}\\ \textrm{a}.&0\leq x< 3&\textrm{e}.&0\leq y\leq 3\\ \textrm{b}.&-5<x\leq -3&\textrm{f}.&-3\leq y\leq 0\\ \textrm{c}.&0\leq x\leq 3&\textrm{g}.&-3\leq 2x< 0\\ \textrm{d}.&-3\leq x\leq 0&\textrm{h}.&0\leq 2y< 3 \end{array} \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Tunjukkan pada \textbf{bidang Cartesius} wilayah penyelesaian dari sistem pertidaksamaan}\\ &\begin{array}{llllllllll}\\ \textrm{a}.&\begin{cases} 2x+5y\geq 10 \\ x\geq 2 \\ y\leq 1 \end{cases}&\textrm{b}.&\begin{cases} x+y\leq 6 \\ 2x+3y\leq 12\\ x\geq 0 \\ y\geq 0 \end{cases}&\textrm{c}.&\begin{cases} x+y\leq 6 \\ 5x+9y\leq 45\\ 2x+y\leq 10\\ x\geq 0 \\ y\geq 0 \end{cases} \end{array} \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Tentukann nilai optimum(minimum dan maksimum) dari fungsi objektif}\: \: f(x,y)=7x+2y\\ &\textrm{dengan wilayah kendala}\\ &\begin{cases} 2y-x\geq 0 \\ x+y\leq 8\\ 7x+2y\leq 14\\ x\geq 0,\: y\geq 0\\ x,y\in \mathbb{R} \end{cases} \end{array}.

\begin{array}{ll}\\ 5.&\textrm{Wilayah yang diarsir berikut di bawah adalah wilayah himpunan penyelesaaian, tentukanlah}\\ &\textrm{a}.\quad \textrm{sistem pertidaksamaan tersebut}\\ &\textrm{b}.\quad\textrm{nilai maksimum dan minimum bila fungsi objektifnya}\: \: f(x,y)=x+4y \end{array}.

Sumber Referensi.

  1. Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika 3A SMA Kelas XII Semester Pertama. Jakarta: Yudistira.
  2. Muis, A. 2009. Perang Siasat Matematika Dasar (Cetakan Kesembilanbelas). Bantul: KREASI WACANA.
  3. Sharma, dkk. 2017. Jelajah Matematika 2 SMA Kelas XI Program Wajib. Jakarta: YUDISTIRA
  4. Siswanto, Umi Supraptinah. 2009. Matematika Inovatif 3 : Konsep dan Aplikasinya untuk Kelas XII SMA dan MA Program Ilmu Pengetahuan Sosial. Jakarta: Pusat Perbukuan-Departemen Pendidikan NAsional.
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Identitas Trigonometri (Kelas XI Peminatan Mat & IA K13 Revisi)

Identitas Trigonometri adalah persamaan-persamaan yang berlaku untuk semua nilai pengganti variabelnya yang mengandung perbandingan trigonometri.

Sebelumnya silahkan lihat kembali

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

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

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

Bukti:

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

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

Bukti:

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

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

Bukti:

\begin{aligned}3&.\textrm{j}.\quad \cot ^{2}\alpha \left ( \displaystyle \frac{\sec \alpha -1}{1+\sin \alpha } \right )+\sec ^{2}\alpha \left ( \displaystyle \frac{\sin \alpha -1}{1+\sec \alpha } \right )\\ &=\left ( \csc ^{2}\alpha -1 \right ) \left ( \displaystyle \frac{\sec \alpha -1}{1+\sin \alpha } \right )+\sec ^{2}\alpha \left ( \displaystyle \frac{\sin \alpha -1}{1+\sec \alpha } \right )\\ &=\left ( \displaystyle \frac{1}{\sin ^{2}\alpha } -1 \right ) \left ( \displaystyle \frac{\displaystyle \frac{1}{\cos \alpha } -1}{1+\sin \alpha } \right )+\displaystyle \frac{1}{\cos ^{2}\alpha } \left ( \displaystyle \frac{\sin \alpha -1}{1+\displaystyle \frac{1}{\cos \alpha } } \right )\\ &=\left ( \displaystyle \frac{1-\sin ^{2}\alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{\displaystyle \frac{1-\cos \alpha }{\cos \alpha }}{1+\sin \alpha } \right )+\displaystyle \frac{1}{\cos ^{2}\alpha } \left ( \displaystyle \frac{\sin \alpha -1}{\displaystyle \frac{\cos \alpha +1}{\cos \alpha } } \right )\\ \end{aligned}.

\begin{aligned} .\: &=\left ( \displaystyle \frac{\cos ^{2}\alpha }{\sin ^{2}\alpha .\cos \alpha } \right ) \left ( \displaystyle \frac{1-\cos \alpha }{1+\sin \alpha } \right )+\displaystyle \frac{\cos \alpha }{\cos ^{2}\alpha } \left ( \displaystyle \frac{\sin \alpha -1}{\cos \alpha +1} \right )\\ .\: &=\left ( \displaystyle \frac{\cos \alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{1-\cos \alpha }{1+\sin \alpha } \right )\times \left ( \displaystyle \frac{1-\sin \alpha }{1-\sin \alpha } \right )+\displaystyle \frac{1 }{\cos \alpha } \left ( \displaystyle \frac{\sin \alpha -1}{\cos \alpha +1} \right )\times \left ( \displaystyle \frac{\cos \alpha -1}{\cos \alpha -1} \right )\\ .\: &=\left ( \displaystyle \frac{\cos \alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{1-\sin \alpha -\cos \alpha +\cos \alpha \sin \alpha }{1-\sin^{2} \alpha } \right )+\displaystyle \frac{1 }{\cos \alpha } \left ( \displaystyle \frac{\sin \alpha \cos \alpha -\sin \alpha -\cos \alpha +1}{\cos ^{2}\alpha -1} \right )\\ \end{aligned}.

\begin{aligned} .\: &=\left ( \displaystyle \frac{\cos \alpha }{\sin ^{2}\alpha } \right ) \left ( \displaystyle \frac{1-\sin \alpha -\cos \alpha +\cos \alpha \sin \alpha }{\cos^{2} \alpha } \right )+\displaystyle \frac{1 }{\cos \alpha } \left ( \displaystyle \frac{\sin \alpha \cos \alpha -\sin \alpha -\cos \alpha +1}{-\sin ^{2}\alpha } \right )\\ &=\left ( \displaystyle \frac{1}{\sin ^{2}\alpha \cos \alpha }-\displaystyle \frac{1}{\sin ^{2}\alpha \cos \alpha } \right ) \left ( 1-\sin \alpha -\cos \alpha +\cos \alpha \sin \alpha \right )\\ .\: &=0\qquad \blacksquare \end{aligned}.

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

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

Sumber Referensi

  1. Kurnia, N.,  dkk. 2017. Jelajah Matematika 2 SMA Kelas XI Peminatan MIPA(Edisi Revisi 2016). Jakarta: Yudistira.
  2. Tampomas, H. 1999. Seribu Pena Matematika SMU Jilid 1 Kelas 1. Jakarta: ERLANGGA.
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Lanjutan: Rumus Jumlah dan Selisih (Kelas XI Peminatan Mat & IA K13 Revisi)

B. Rumus Trigonometri Jumlah dan Selisih Dua Sudut

Arsip lama

  • di sini 1  (dari contoh pengembangan soal materi KTSP kelas x smt 1)
  • di sini 2 (dari materi KTSP kelas XI smt 1)

Sebagai pengingat saja

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

C. Rumus Jumlah dan Selisih untuk Sinus dan Cosinus

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

Sedikit gambaran untuk rumus no. 3, yaitu :

\begin{array}{|cll|c|}\hline &\begin{aligned}\cos \left ( \alpha +\gamma \right )&=\cos \alpha \cos \beta -\sin \alpha \sin \gamma \\ \cos \left ( \alpha -\gamma \right )&=\cos \alpha \cos \beta +\sin \alpha \sin \gamma \\ \end{aligned}&&\\\cline{2-2} &&+&\textrm{Misalkan}\\ &\begin{aligned}\cos \left ( \alpha +\gamma \right )+\cos \left ( \alpha -\gamma \right )&=2\cos \alpha \cos \gamma \end{aligned}&&X=\alpha +\gamma \\ &&&Y=\alpha -\gamma\\ &\textbf{Proses perubahannya adalah sebagai berikut}&&\\\cline{1-3} &\underset{\textrm{perhatikan}}{\underbrace{\begin{matrix} \begin{array}{ccc}\\ &X=\alpha +\gamma &\\ &Y=\alpha -\gamma &+\\\cline{2-2}\\ &X+Y=2\alpha &\\ &2\alpha =X+Y&\\ &\alpha =\displaystyle \frac{X+Y}{2}& \end{array} & \textrm{dan} & \begin{array}{ccc}\\ &X=\alpha +\gamma &\\ &Y=\alpha -\gamma &-\\\cline{2-2}\\ &X-Y=2\gamma &\\ &2\gamma =X-Y&\\ &\gamma =\displaystyle \frac{X-Y}{2}& \end{array} \end{matrix}}} &&\\\cline{1-3} &\textbf{Sehingga rumus akan menjadi}&&\\ &\cos X+\cos Y =2\cos \displaystyle \frac{\left ( X+Y \right )}{2}\cos \frac{\left ( X- Y\right )}{2}&&\\\hline \end{array}.

Untuk rumus yang lain diperoleh dengan cara yang semisal ilustrasi di atas tentunya dengan penyesuaian.

D. Rumus Perkalian Trigonometri

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

E. Persamaan Trigonometri Bentuk  a\sin x+b\cos x=k\cos \left ( x-\alpha \right ).

Perhatikanlah bentuk berikut

\begin{aligned}a\sin x+b\cos x&=k\cos \left ( x-\alpha \right )\\ a\sin x+b\cos x&=k\left ( \cos x\cos \alpha +\sin x \sin \alpha \right )\\ a\sin x+b\cos x&=k\cos x\cos \alpha +k\sin x \sin \alpha\\ \textrm{sehingga didap}&\textrm{at kesamaan yang menghasilkan}\\ \bullet \quad a&=k\cos \alpha \\ \bullet \quad b&=k\sin \alpha \\ \textrm{Selanjutnya jik}&\textrm{a dikuadratkan masing-masing lalu dijumlahkan}\\ a^{2}+b^{2}&=k^{2}\cos ^{2}\alpha +k^{2}\sin ^{2}\alpha \\ a^{2}+b^{2}&=k^{2}\left ( \sin ^{2}\alpha +\cos ^{2}\alpha \right )\\ a^{2}+b^{2}&=k^{2} \\ k^{2}&=a^{2}+b^{2}\\ k&=\sqrt{a^{2}+b^{2}} , \textrm{ambil}\: \: k\: \: \textrm{positif saja}\\ \textrm{Selanjutnya jik}&\textrm{a}\\ \displaystyle \frac{k\sin \alpha }{k\cos \alpha }&=\displaystyle \frac{b}{a}\\ \tan \alpha &=\displaystyle \frac{b}{a}\\ &\begin{array}{|c|c|c|c|}\hline \textrm{Nilai}\: \: a\: \: \textrm{dan}\: \: b&\displaystyle \frac{b}{a}&\tan \alpha &\textrm{Kuadran sudut}\: \: \alpha \\\hline a> 0\: \: \textrm{dan}\: \: b> 0&> 0&> 0&\textrm{I}\\ a< 0\: \: \textrm{dan}\: \: b> 0& < 0&< 0&\textrm{II}\\ a< 0\: \: \textrm{dan}\: \: b< 0&> 0&> 0&\textrm{III}\\ a> 0\: \: \textrm{dan}\: \: b< 0&< 0&< 0&\textrm{IV}\\\hline \end{array} \end{aligned}.

\begin{array}{|c|}\hline \begin{aligned}&\\ \textbf{CATATAN}:&\\ \textrm{Bentuk}\: &\\ &a\cos x^{\circ}+b\sin x^{\circ}\\ \textrm{dapat d}&\textrm{inyatakan dalam tipe}\\ &\begin{cases} k \cos \left ( x+\alpha \right )^{\circ} \\ k \cos \left ( x-\alpha \right )^{\circ} \quad\Rightarrow \qquad \textrm{yang sedang kita bahas di atas}\\ k \sin \left ( x+\alpha \right )^{\circ} \\ k \sin \left ( x-\alpha \right )^{\circ} \end{cases}\\ & \end{aligned}\\\hline \end{array}.

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

\begin{array}{ll}\\ 1.&\textrm{Tunjukkanlah}\\ &\begin{array}{ll}\\ \textrm{a}.&\sin 2\alpha =2\sin \alpha \cos \alpha \\ \textrm{b}.&\cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha =2\cos ^{2}\alpha -1=1-2\sin ^{2}\alpha \\ \textrm{c}.&\tan 2\alpha =\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\\ \textrm{d}.&\sin 3\alpha=3\sin \alpha -4\sin ^{3}\alpha \\ \textrm{e}.&\cos 3\alpha =4\cos ^{3}\alpha -3\cos \alpha \\ \textrm{f}.&\tan 3\alpha =\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \end{array} \end{array}.

Bukti:

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \sin \left ( \alpha +\gamma \right )&=\sin \alpha \cos \gamma +\cos \alpha \sin \gamma \\ \textrm{dengan}\: &\textrm{mengubah}\: \: \gamma =\alpha\\ \sin \left ( \alpha +\alpha \right )&=\sin \alpha \cos \alpha +\cos \alpha \sin \alpha \\ \sin 2\alpha &=2\sin \alpha \cos \alpha \qquad \blacksquare \end{aligned}&\begin{aligned}\textrm{b}.\quad \cos \left ( \alpha +\gamma \right )&=\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \\ \textrm{dengan}\: &\textrm{mengubah}\: \: \gamma =\alpha\\ \cos \left ( \alpha +\alpha \right )&=\cos \alpha \cos \alpha -\sin \alpha \sin \alpha \\ \cos 2\alpha &=\cos ^{2}\alpha -\sin ^{2}\alpha \qquad \blacksquare \end{aligned}\\\cline{1-1} \begin{aligned}\textrm{c}.\quad \tan \left ( \alpha +\gamma \right )&=\displaystyle \frac{\tan \alpha +\tan \gamma }{1-\tan \alpha \tan \gamma }\\ \textrm{dengan}\: &\textrm{mengganti}\: \: \alpha =\gamma \\ \tan \left ( \alpha +\alpha \right )&=\displaystyle \frac{\tan \alpha +\tan \alpha }{1-\tan \alpha \tan \alpha }\\ \tan 2\alpha &=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\qquad \blacksquare \\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{ingat}&\: \textrm{bahwa}\\ &\begin{cases} 1 &=\sin ^{2}\alpha +\cos ^{2}\alpha \\ \sin ^{2}\alpha & =1-\cos ^{2}\alpha \\ \cos ^{2}\alpha &= 1-\sin ^{2}\alpha \end{cases}\\ \textrm{sehin}&\textrm{gga persamaan}\\ \cos 2\alpha &=\cos ^{2}\alpha -\sin ^{2}\alpha\\ &=\cos ^{2}\alpha -\left ( 1-\cos ^{2}\alpha \right )\\ &=2\cos ^{2}\alpha -1\qquad \blacksquare \\ &=2\left ( 1-\sin ^{2}\alpha \right )-1\\ &=2-2\sin ^{2}\alpha -1\\ &=1-2\sin ^{2}\alpha \qquad \blacksquare \end{aligned}\\\hline \end{array}.

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{e}.\quad \cos 3\alpha &=\cos \left ( 2\alpha +\alpha \right )\\ &=\cos 2\alpha \cos \alpha -\sin 2\alpha \sin \alpha \\ &=\left ( 2\cos ^{2}\alpha -1 \right )\cos \alpha -\left ( 2\sin \alpha \cos \alpha \right )\sin \alpha \\ &=2\cos ^{3}\alpha -\cos \alpha -2\sin ^{2}\alpha \cos \alpha \\ &=2\cos ^{3}\alpha -\cos \alpha -2\left ( 1-\cos ^{2}\alpha \right ) \cos \alpha\\ &=2\cos ^{3}\alpha -\cos \alpha -2\cos \alpha +2\cos ^{2}\alpha\\ &=4\cos ^{3}\alpha -3\cos \alpha \qquad \blacksquare\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{f}.\quad \tan 3\alpha &=\tan \left ( 2\alpha +\alpha \right )\\ &=\displaystyle \frac{\tan 2\alpha +\tan \alpha }{1-\tan 2\alpha \tan \alpha }\\ &=\displaystyle \frac{\left ( \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \right )+\tan \alpha }{1-\left ( \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \right ).\tan \alpha }\\ &=\displaystyle \frac{\displaystyle \frac{2\tan \alpha +\tan \alpha -\tan ^{3}\alpha }{1-\tan ^{2}\alpha }}{\displaystyle \frac{\left ( 1-\tan ^{2}\alpha \right )-2\tan ^{2}\alpha }{1-\tan ^{2}\alpha }}\\ &=\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha }\qquad \blacksquare \end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Buktikanlah bahwa}\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sin \left ( 90^{\circ}-\beta \right )=\cos \beta &\textrm{i}.&\sin \left ( 270^{\circ}-\beta \right )=-\cos \beta&\textrm{q}.&\sin 15^{\circ}=\displaystyle \frac{\sqrt{3}-1}{2\sqrt{2}}\\ \textrm{b}.&\cos \left ( 90^{\circ}-\beta \right )=\sin \beta &\textrm{j}.&\cot \left ( 270^{\circ}-\beta \right )=\tan \beta&\textrm{r}.&\cos 15^{\circ}=\displaystyle \frac{1}{4}\sqrt{2}\left ( \sqrt{3}+1 \right )\\ \textrm{c}.&\tan \left ( 90^{\circ}-\beta \right )=\cot \beta &\textrm{k}.&\sin \left ( 360^{\circ}-\beta \right )=-\sin \beta&\textrm{s}.&\tan 15^{\circ}=2-\sqrt{3} \\ \textrm{d}.&\sin \left ( 180^{\circ}-\beta \right )=\sin \beta &\textrm{l}.&\cos \left ( 360^{\circ}-\beta \right )=\cos \beta&\textrm{t}.&\displaystyle \frac{\sin \left ( \alpha +\beta \right )}{\cos \alpha \cos \beta }=\tan \alpha +\tan \beta \\ \textrm{e}.&\cos \left ( 180^{\circ}-\beta \right )=-\cos \beta &\textrm{m}.&\cot \left ( 360^{\circ}-\beta \right )=-\cot \beta&\textrm{u}.&\displaystyle \frac{\sin \left ( \alpha -\beta \right )}{\sin \left ( \alpha +\beta \right ) }=\frac{\tan \alpha -\tan \beta }{\tan \alpha +\tan \beta }\\ \textrm{f}.&\cot \left ( 180^{\circ}-\beta \right )=-\cot \beta &\textrm{n}.&\sin \left ( -\beta \right )=-\sin \beta &\textrm{v}.&\displaystyle \frac{\cos \left ( \alpha +\beta \right )}{\cos\alpha \cos \beta }=1-\tan \alpha \tan \beta \\ \textrm{g}.&\sin \left ( 180^{\circ}+\beta \right )=-\sin \beta &\textrm{o}.&\cos \left ( -\beta \right )=\cos \beta &\textrm{w}.&\displaystyle \frac{\cos \left ( \alpha +\beta \right )}{\cos \left ( \alpha -\beta \right ) }=\frac{1-\tan \alpha \tan \beta }{1+\tan \alpha \tan \beta } \\ \textrm{h}.&\csc \left ( 180^{\circ}+\beta \right )=-\csc \beta &\textrm{p}.&\tan \left ( -\beta \right )=-\tan \beta &\textrm{x}.&\displaystyle \frac{\sin 3\gamma +\sin \gamma }{\cos 3\gamma +\cos \gamma }=\tan 2\gamma\\ \end{array} \end{array}.

Bukti:

Sebagai pengingat

\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \cdots \alpha &0^{0}&30^{0}&45^{0}&60^{0}&90^{0}&180^{0}&270^{0}&360^{0}\\\hline \sin \alpha &0&\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0&-1&0\\\hline \cos \alpha &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1&0&1\\\hline \tan \alpha &0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&TD&0&TD&0\\\hline \end{array}.

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \sin \left ( 90^{\circ}-\beta \right )&=\sin 90^{\circ}\cos \beta -\cos 90^{\circ}\sin \beta \\ &=1.\cos \beta -0.\sin \beta \\ &=\cos \beta\qquad \blacksquare \\ & \end{aligned}&\begin{aligned}\textrm{n}.\quad \sin \left ( -\beta \right )&=\sin \left ( 0^{\circ}-\beta \right )\\ &=\sin 0^{\circ}.\cos \beta -\cos 0^{\circ}\sin \beta \\ &=0-1.\sin \beta \\ &=-\sin \beta\qquad \blacksquare \end{aligned}\\\hline \begin{aligned}\textrm{h}.\quad \csc \left ( 180^{\circ}+\beta \right )&=\displaystyle \frac{1}{\sin \left ( 180^{\circ}+\beta \right )}\\ &=\displaystyle \frac{1}{\sin 180^{\circ}\cos \beta +\cos 180^{\circ}\sin \beta }\\ &=\displaystyle \frac{1}{0.\cos \beta +(-1).\sin \beta }\\ &=-\displaystyle \frac{1}{\sin \beta }\\ &=-\csc \beta\qquad \blacksquare \\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{s}.\quad\tan 15^{\circ}&=\tan \left ( 45^{\circ}-30^{\circ} \right )\\ &=\displaystyle \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}}\\ &=\displaystyle \frac{1-\displaystyle \frac{1}{\sqrt{3}}}{1+\displaystyle \frac{1}{\sqrt{3}}}\\ &=\left ( \displaystyle \frac{1-\displaystyle \frac{1}{\sqrt{3}}}{1+\displaystyle \frac{1}{\sqrt{3}}} \right )\times \left ( \displaystyle \frac{\sqrt{3}}{\sqrt{3}} \right )\\ &=\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}\times \left (\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}-1} \right )\\ &=\displaystyle \frac{3-2\sqrt{3}+1}{3-1}\\ &=\displaystyle \frac{4-2\sqrt{3}}{2}\\ &=2-\sqrt{3}\qquad \blacksquare \end{aligned}\\\hline \end{array}.

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{t}.\quad \displaystyle \frac{\sin \left ( \alpha +\beta \right )}{\cos \alpha \cos \beta }&=\displaystyle \frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta }\\ &=\displaystyle \frac{\sin \alpha \cos \beta }{\cos \alpha \cos \beta }+\displaystyle \frac{\cos \alpha \sin \beta }{\cos \alpha \cos \beta }\\ &=\displaystyle \frac{\sin \alpha }{\cos \alpha}+\displaystyle \frac{\sin \beta }{\cos \beta }\\ &=\tan \alpha +\tan \beta\qquad \blacksquare \end{aligned}&\begin{aligned}\textrm{v}.\quad \displaystyle \frac{\cos \left ( \alpha +\beta \right )}{\cos \alpha \cos \beta }&=\displaystyle \frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta }\\ &=\displaystyle \frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }-\displaystyle \frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }\\ &=1-\displaystyle \frac{\sin \alpha }{\cos \alpha}\times \displaystyle \frac{\sin \beta }{\cos \beta }\\ &=1-\tan \alpha \tan \beta\qquad \blacksquare \end{aligned}\\\hline \end{array}.

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{x}.\quad \displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }&=\displaystyle \frac{2\sin \displaystyle \frac{\left (3\alpha +\alpha \right )}{2}\cos \displaystyle \frac{\left ( 3\alpha -\alpha \right )}{2}}{2\cos \displaystyle \frac{\left ( 3\alpha +\alpha \right )}{2}\cos \displaystyle \frac{\left ( 3\alpha -\alpha \right )}{2}}\\ &=\displaystyle \frac{\sin 2\alpha }{\cos 2\alpha }\\ &=\tan 2\alpha \qquad \blacksquare\\ &\\ &\\ &\quad\qquad\qquad\textbf{atau}\Rightarrow\qquad \\ &\textrm{mungkin lumayan rumit}\\ &\\ & \end{aligned}&\begin{aligned}\textrm{x}.\quad \displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }&=\displaystyle \frac{3\sin \alpha -4\sin ^{3}\alpha +\sin \alpha }{4\cos ^{3}\alpha -3\cos \alpha +\cos \alpha }\\ &=\displaystyle \frac{4\sin \alpha -4\sin ^{3}\alpha }{4\cos ^{3}\alpha -2\cos \alpha }\\ &=\displaystyle \frac{4\sin \alpha \left ( 1-\sin ^{2}\alpha \right )}{2\cos \alpha \left ( 2\cos ^{2}\alpha -1 \right )}=\displaystyle \frac{4\sin \alpha }{2\cos \alpha }\times \displaystyle \frac{\cos ^{2}\alpha }{\cos 2\alpha }\\ &=\displaystyle \frac{2\tan \alpha \cos ^{2}\alpha }{\cos 2\alpha }\\ &=\displaystyle \frac{2\tan \alpha \cos ^{2}\alpha }{\cos^{2}\alpha -\sin ^{2}\alpha }\\ &=\displaystyle \frac{2\tan \alpha }{\displaystyle \frac{\cos ^{2}\alpha }{\cos ^{2}\alpha }-\displaystyle \frac{\sin ^{2}\alpha }{\cos ^{2}\alpha }}=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\\ &=\tan 2\alpha \qquad \blacksquare \end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Sederhanakanlah bentuk berikut ini}\\ &\begin{array}{ll}\\ \textrm{a}.&\sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )\\ \textrm{b}.&\sin \left ( x+\displaystyle \frac{1}{4}m \right )+\sin \left ( x-\displaystyle \frac{1}{4}m \right )\\ \textrm{c}.&\cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )\\ \textrm{d}.&\cos \left ( \displaystyle \frac{5}{4}m+3x \right )+\cos \left ( \displaystyle \frac{5}{4}m-3x \right ) \end{array} \end{array}.

Jawab:

\begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad \sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )&=2\cos \displaystyle \frac{\left (2x+60^{\circ} \right )+\left ( 2x-60^{\circ} \right )}{2}\sin \displaystyle \frac{\left (2x+60^{\circ} \right )-\left ( 2x-60^{\circ} \right )}{2}\\ &=2\cos \displaystyle \frac{4x}{2}\sin \displaystyle \frac{120^{\circ}}{2}\\ &=2\cos 2x\sin 60^{\circ}\\ &=2\cos 2x.\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\sqrt{3}\cos 2x \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad \cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )&=-2\sin \displaystyle \frac{\left (2x+4y \right )+\left ( 2x-4y \right )}{2}\sin \displaystyle \frac{\left (2x+4y \right )-\left ( 2x-4y \right )}{2}\\ &=-2\sin \displaystyle \frac{4x}{2}\sin \displaystyle \frac{8y}{2}\\ &=-2\sin 2x\sin 4y\\ \end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Carilah nilai}\: \: k\: \: \textrm{dan}\: \: \: \alpha \: \: \left ( k>0,\: \: 0^{\circ}\leq \alpha \leq 360^{\circ} \right )\: \: \textrm{untuk tiap pasangan persaamaan berikut ini}!\\ &\begin{array}{llll}\\ \textrm{a}.&k\cos \alpha =2\: ;\: k\sin \alpha =2&\textrm{f}.&k\cos \alpha =-\sqrt{3}\: ;\: k\sin \alpha =1\\ \textrm{b}.&k\cos \alpha =2\: ;\: k\sin \alpha =-2&\textrm{g}.&k\cos \alpha =-\sqrt{3}\: ;\: k\sin \alpha =-1\\ \textrm{c}.&k\cos \alpha =-2\: ;\: k\sin \alpha =2&\textrm{h}.&k\cos \alpha =1\: ;\: k\sin \alpha =\sqrt{3}\\ \textrm{d}.&k\cos \alpha =-2\: ;\: k\sin \alpha =-2&\textrm{h}.&k\cos \alpha =1\: ;\: k\sin \alpha =-\sqrt{3}\\ \textrm{e}.&k\cos \alpha =3\: ;\: k\sin \alpha =\sqrt{3}&\textrm{h}.&k\cos \alpha =3\: ;\: k\sin \alpha =-\sqrt{3} \end{array} \end{array}.

Jawab:

\begin{array}{|c|c|c|c|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\begin{cases} b & =y=k\sin \alpha \\ a & =x=k\cos \alpha \end{cases}\\ & \end{aligned}}\\\hline \textrm{No}&\multicolumn{2}{|c|}{\tan \alpha =\displaystyle \frac{b}{a}}&k=\sqrt{b^{2}+a^{2}}\\\hline 4.\textrm{a}&\begin{aligned}\tan \alpha &=\displaystyle \frac{2}{2}=1\\ &(\textrm{kuadran 1}) \end{aligned}&\begin{aligned}\tan \alpha &=1=\tan 45^{\circ}\\ \alpha &=45^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{2^{2}+2^{2}}=\sqrt{2.2^{2}}=2\sqrt{2}\end{aligned}\\\hline 4.\textrm{b}&\begin{aligned}\tan \alpha &=\displaystyle \frac{-2}{2}=-1\\ &(\textrm{kuadran 4}) \end{aligned}&\begin{aligned}\tan \alpha &=-1=\tan \left ( -45^{\circ} \right )\\ &=\tan \left ( 360^{\circ}-45^{\circ} \right )=\tan 315^{\circ}\\ \alpha &=315^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(-2)^{2}+2^{2}}=\sqrt{2.2^{2}}=2\sqrt{2}\end{aligned}\\\hline 4.\textrm{c}&\begin{aligned}\tan \alpha &=\displaystyle \frac{2}{-2}=-1\\ &(\textrm{kuadran 2}) \end{aligned}&\begin{aligned}\tan \alpha &=-1=\tan \left ( -45^{\circ} \right )\\ &=\tan \left ( 180^{\circ}-45^{\circ} \right )=\tan 135^{\circ}\\ \alpha &=135^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{2^{2}+(-2)^{2}}=\sqrt{2.2^{2}}=2\sqrt{2}\end{aligned}\\\hline 4.\textrm{d}&\begin{aligned}\tan \alpha &=\displaystyle \frac{-2}{-2}=1\\ &(\textrm{kuadran 3}) \end{aligned}&\begin{aligned}\tan \alpha &=1=\tan \left ( 45^{\circ} \right )\\ &=\tan \left ( 180^{\circ}+45^{\circ} \right )=\tan 225^{\circ}\\ \alpha &=225^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{2.2^{2}}=2\sqrt{2}\end{aligned}\\\hline  \end{array}.

\begin{array}{|c|c|c|c|}\hline \multicolumn{4}{|c|}{\begin{aligned}&\\ &\begin{cases} b & =y=k\sin \alpha \\ a & =x=k\cos \alpha \end{cases}\\ & \end{aligned}}\\\hline \textrm{No}&\multicolumn{2}{|c|}{\tan \alpha =\displaystyle \frac{b}{a}}&k=\sqrt{b^{2}+a^{2}}\\\hline 4.\textrm{e}&\begin{aligned}\tan \alpha &=\displaystyle \frac{\sqrt{3}}{3}\\ &(\textrm{kuadran 1}) \end{aligned}&\begin{aligned}\tan \alpha &=\displaystyle \frac{\sqrt{3}}{3}=\tan \left ( 30^{\circ} \right )\\ \alpha &=30^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(\sqrt{3})^{2}+(3)^{2}}=\sqrt{3+9}\\ &=\sqrt{12}=\sqrt{4.3}=\sqrt{2^{2}.3}=2\sqrt{3}\end{aligned}\\\hline 4.\textrm{f}&\begin{aligned}\tan \alpha &=\displaystyle \frac{1}{-\sqrt{3}}\\ &(\textrm{kuadran 2}) \end{aligned}&\begin{aligned}\tan \alpha &=-\displaystyle \frac{\sqrt{3}}{3}=\tan \left ( -30^{\circ} \right )\\ \tan \alpha &=\tan \left ( 180^{\circ}-30^{\circ} \right )\\ &=\tan 150^{\circ}\\ \alpha &=150^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(1)^{2}+\left ( -\sqrt{3} \right )^{2}}=\sqrt{1+3}\\ &=\sqrt{4}=\sqrt{2^{2}}=2\end{aligned}\\\hline 4.\textrm{g}&\begin{aligned}\tan \alpha &=\displaystyle \frac{-1}{-\sqrt{3}}\\ &(\textrm{kuadran 3}) \end{aligned}&\begin{aligned}\tan \alpha &=\displaystyle \frac{\sqrt{3}}{3}=\tan \left ( 30^{\circ} \right )\\ \tan \alpha &=\tan \left ( 180^{\circ}+30^{\circ} \right )\\ &=\tan 210^{\circ}\\ \alpha &=210^{\circ} \end{aligned}&\begin{aligned}k&=\sqrt{(-1)^{2}+\left ( -\sqrt{3} \right )^{2}}=\sqrt{1+3}\\ &=\sqrt{4}=\sqrt{2^{2}}=2\end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ 5.&\textrm{Tunjukkan bahwa}:\\ &\begin{array}{llll}\\ \textrm{a}.&3\cos x^{\circ} -3\sin x^{\circ} =3\sqrt{2}\cos \left ( x^{\circ}-\displaystyle \frac{7}{4}\pi \right )\\ \textrm{b}.&3\cos x^{\circ} -3\sin x^{\circ} =3\sqrt{2}\cos \left ( x^{\circ}+\displaystyle \frac{1}{4}\pi \right )\\ \textrm{c}.&3\cos x^{\circ} -3\sin ^{\circ} =3\sqrt{2}\sin \left ( x^{\circ}-\displaystyle \frac{5}{4}\pi \right )\\ \textrm{d}.&3\cos x^{\circ} -3\sin x^{\circ} =3\sqrt{2}\sin \left ( x^{\circ}+\displaystyle \frac{3}{4}\pi \right )\\ \end{array} \end{array}.

Bukti:

\begin{aligned}\textrm{a}.\quad\textrm{Diketahui bahwa}:&\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos \left ( x^{\circ}-\alpha \right )^{\circ}\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos x^{\circ}\cos \alpha +k\sin x^{\circ}\sin \alpha \\ \textrm{diperoleh persam}&\textrm{aan}\begin{cases} b=y=k\sin \alpha & =-3\\ a=x=k\cos \alpha & =3 \end{cases}\\ k&=\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+(-3)^{2}}=\sqrt{2.3^{2}}=3\sqrt{2}\\ \tan \alpha &=\displaystyle \frac{y}{x}=\displaystyle \frac{-3}{3}=-1\quad (\textbf{kuadran 4})\\ &=\tan \left ( -45^{\circ} \right )=\tan \left ( -\displaystyle \frac{1}{4}\pi \right )=\tan \left (2 \pi -\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{7}{4}\pi \\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos \left ( x-\alpha \right )^{\circ}=3\sqrt{2}\cos \left ( x^{\circ}-\displaystyle \frac{7}{4}\pi \right )\qquad \blacksquare \end{aligned}.

\begin{aligned}\textrm{b}.\quad\textrm{Diketahui bahwa}:&\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos \left ( x^{\circ}+\alpha \right )^{\circ}\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos x^{\circ}\cos \alpha -k\sin x^{\circ}\sin \alpha \\ \textrm{diperoleh persam}&\textrm{aan}\begin{cases} b=y=k\sin \alpha & =3\\ a=x=k\cos \alpha & =3 \end{cases}\\ k&=\sqrt{a^{2}+b^{2}}=\sqrt{3^{2}+3^{2}}=\sqrt{2.3^{2}}=3\sqrt{2}\\ \tan \alpha &=\displaystyle \frac{y}{x}=\displaystyle \frac{3}{3}=1\quad (\textbf{kuadran 1})\\ &=\tan \left ( 45^{\circ} \right )=\tan \left ( \displaystyle \frac{1}{4}\pi \right )\\ 3\cos x^{\circ} -3\sin x^{\circ} &=k\cos \left ( x+\alpha \right )^{\circ}=3\sqrt{2}\cos \left ( x^{\circ}+\displaystyle \frac{1}{4}\pi \right )\qquad \blacksquare \end{aligned}.

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

Yang belum dibahas dan atau belum ditunjukkan buktinya, silahkan gunakan sebagai latihan mandiri

Sumber Referensi

  1. Kurnia, N.,  dkk. 2017. Jelajah Matematika 2 SMA Kelas XI Peminatan MIPA(Edisi Revisi 2016). Jakarta: Yudistira.
  2. Sembiring, S., Zulkifli, M., Marsito, dan Rusdi, I. 2017. Matematika untuk siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam(Edisi Revisi 2016). Bandung: SEWU.
  3. Wirodikromo, S. 2003. Matematika 2000 untuk SMU Jilid 5 Kelas 3. Jakarta: Erlangga.

 

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Trigonometri (Kelas XI Peminatan Mat & IA K13 Revisi)

Arsip lama

A. Persamaan Trigonometri

Untuk persamaan trigonometriny adalah sebagai berikut:

\begin{array}{|c|l|l|}\hline \multicolumn{3}{|c|}{\textbf{Persamaan Trigonometri}}\\\hline \textrm{No}&\qquad\qquad \textrm{Persamaannya}&\quad\qquad\qquad \textrm{atau}\\\hline 1&\begin{aligned}\sin x&=\sin \alpha \\ x&=\begin{cases} &=\alpha +k.360^{\circ} \\ &=\left ( 180^{\circ}-\alpha \right )+k.360^{\circ} \end{cases}\\ \textrm{deng}&\textrm{an}\\ k&\in \mathbb{Z} \end{aligned}&\begin{aligned}\sin x&=\sin \alpha \\ x&=\begin{cases} &=\alpha +k.2\pi \\ &=\left ( \pi -\alpha \right )+k.2\pi \end{cases}\\ \textrm{deng}&\textrm{an}\\ k&\in \mathbb{Z} \end{aligned}\\\hline 2&\begin{aligned}\cos x&=\cos \alpha \\ x&=\begin{cases} &=\alpha +k.360^{\circ} \\ &=-\alpha +k.360^{\circ} \end{cases}\\ \textrm{deng}&\textrm{an}\\ k&\in \mathbb{Z} \end{aligned}&\begin{aligned}\cos x&=\cos \alpha \\ x&=\begin{cases} &=\alpha +k.2\pi \\ &=-\alpha +k.2\pi \end{cases}\\ \textrm{deng}&\textrm{an}\\ k&\in \mathbb{Z} \end{aligned}\\\hline 3&\begin{aligned}\tan x&=\tan \alpha \\ x&=\alpha +k.180^{\circ}\\ \textrm{deng}&\textrm{an}\\ k&\in \mathbb{Z} \end{aligned}&\begin{aligned}\tan x&=\tan \alpha \\ x&=\alpha +k.\pi \\ \textrm{deng}&\textrm{an}\\ k&\in \mathbb{Z} \end{aligned}\\\hline \end{array}.

Berikut beberapa grafik fungsi trigonometri

A. 1. Grafik Fungsi Sinus

A. 2. Grafik Fungsi Cosinus

A. 3. Grafik Fungsi Tangen

Sebagai pengingat untuk nilai-nilai sudut istimewa dari grafik di atas adalah sebagai berikut:

\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \cdots \alpha &0^{0}&30^{0}&45^{0}&60^{0}&90^{0}&180^{0}&270^{0}&360^{0}\\\hline \sin \alpha &0&\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0&-1&0\\\hline \cos \alpha &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1&0&1\\\hline \tan \alpha &0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&TD&0&TD&0\\\hline \end{array}.

Hal-hal yang perlu diperhatikan di antaranya:

\begin{array}{|c|c|c|c|c|}\hline \textrm{No}&\textrm{Fungsi Trigonometri}&\textrm{Periode}&\textrm{Nilai}&\textrm{Amplitudo}\\\hline 1&y=A\sin nx&\displaystyle \frac{360^{\circ}}{n}\: \: \textrm{atau}\: \: \frac{2\pi }{n}&-A\leq A\sin nx\leq A&\\\cline{1-4} 2&y=A\cos nx&\displaystyle \frac{360^{\circ}}{n}\: \: \textrm{atau}\: \: \frac{2\pi }{n}&-A\leq A\cos nx\leq A&\frac{1}{2}\left ( Max-Min \right )\\\cline{1-4} 3&y=A\tan nx&\displaystyle \frac{180^{\circ}}{n}\: \: \textrm{atau}\: \: \frac{\pi }{n}&-\infty \leq A\tan nx\leq \infty &\\\hline 4&\textbf{sebagai ilustrasi}&\multicolumn{3}{|c|}{\begin{aligned}y&=A\sin (nx-\alpha )\\ &\textrm{dapat diperoleh dari fungsi}\\ &y=A\sin nx\: \: \textrm{dengan menggesernya ke \textbf{kanan}}\\ &\textrm{sejauh}\: \: \displaystyle \frac{\alpha }{n}.\\ &\textrm{Dan jika sebaliknya}\: \: \left ( y=A\sin (nx+\alpha ) \right )\\ &\textrm{maka menggesernya ke sebelah \textbf{kiri}}\\ &\textrm{demikian juga untuk fungsi yang lainnya} \end{aligned}}\\\hline \end{array}.

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

\begin{array}{ll}\\ 1.&\textrm{Tentukanlah himpunan penyelesaian dari persamaan-persamaan trigonometri berikut}\\ &\textrm{ini untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\begin{array}{lllllll}\\ \textrm{a}.& \sin x=\displaystyle \frac{1}{2}&\textrm{f}.& \tan x=-\displaystyle \frac{1}{3}\sqrt{3}&\textrm{k}.& \sin 2x=\displaystyle \frac{1}{2}\\ \textrm{b}.& \cos x=\displaystyle \frac{1}{2}\sqrt{3}&\textrm{g}& 2\cos x=-\sqrt{3}&\textrm{l}.& \cos 2x=-\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{c}.& \tan x=\sqrt{3}&\textrm{h}& 3\tan x=\sqrt{3}&\textrm{m}.& \tan 2x=\sqrt{3}\\ \textrm{d}.& \sin x=-1&\textrm{i}.& \sin x=\sin 46^{\circ}&\textrm{n}.& \sin \left ( 2x-30^{\circ} \right )=\sin 45^{\circ}\\ \textrm{e}.& \cos x=-\displaystyle \frac{1}{2}\sqrt{2}&\textrm{j}.& \cos x=\cos 93^{\circ}&\textrm{o}.& \sin \left ( 2x+60^{\circ} \right )=\sin 90^{\circ}\\ \end{array}\\ \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad\sin x&=\displaystyle \frac{1}{2}\\ \sin x&=\sin 30^{\circ}\\ x&=\begin{cases} 30^{\circ} & +k.360^{\circ}\\ \left (180^{\circ}-30^{\circ} \right ) & +k.360^{\circ} \end{cases}\\ k=0&\: \: \textrm{diperoleh}:\\ x&=\begin{cases} 30^{\circ} & \\ 150^{\circ} & \end{cases}\\ k=1&\: \: \textrm{tidak ada yang memenuhi}\\ \textrm{HP}&=\left \{ 30^{\circ},150^{\circ} \right \} \end{aligned}&\begin{aligned}\textrm{d}.\quad\sin x&=-1\\ \sin x&= \sin 270^{\circ}\\ x&=\begin{cases} 270^{\circ} & +k.360^{\circ} \\ \left ( 180^{\circ}-270^{\circ} \right ) & +k.360^{\circ} \end{cases}\\ k=0&\: \: \textrm{diperoleh}\\ x&=\begin{cases} 270^{\circ} & \\ -90^{\circ} & \textrm{tidak memenuhi} \end{cases}\\ k=1&\: \: \textrm{tidak memenuhi semuanya}\\ \textrm{HP}&=\left \{ 270^{\circ} \right \} \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{n}.\quad\sin \left ( 2x-30^{\circ} \right )&=\sin 45^{\circ}\\ \left ( 2x-30^{\circ} \right )&=\begin{cases} 45^{\circ} & +k.360^{\circ} \\ \left ( 180^{\circ}-45^{\circ} \right ) & +k.360^{\circ} \end{cases}\\ 2x&=\begin{cases} 45^{\circ}+30^{\circ} &+k.360^{\circ} \\ 135^{\circ}+30^{\circ} &+ k.360^{\circ} \end{cases}\\ x&=\begin{cases} 37,5^{\circ} & +k.180^{\circ} \\ 82,5^{\circ} & +k.180^{\circ} \end{cases}\\ k=0&\: \: \textrm{diperoleh}\\ x&=\begin{cases} 37,5^{\circ} & \\ 82,5^{\circ} & \end{cases}\\ k=1&\: \: \textrm{diperoleh}\\ x&=\begin{cases} 37,5^{\circ}+180^{\circ} &=217,5^{\circ} \\ 82,5^{\circ}+180^{\circ} &=262,5^{\circ} \end{cases}\\ k=2&\: \: \textrm{tidak ada yang memenuhi}\\ \textrm{HP}&=\left \{ 37,5^{\circ},82,5^{\circ},217,5^{\circ},262,5^{\circ} \right \} \end{aligned} }\\\hline \end{array}.

Yang belum dibahas, silahkan gunakan sebagai latihan mandiri

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah himpunan penyelesaian dari persamaan-persamaan trigonometri berikut}\\ &\textrm{ini untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\begin{array}{llll}\\ \textrm{a}.& \sin \left ( x+30^{\circ} \right )-\sin 2x=0&\textrm{d}.& \sqrt{1-\sin ^{2}x}=\cos 90^{\circ}\\ \textrm{b}.& \cot x-\sin x\cos x=0&\textrm{e}.&\sqrt{4-4\cos ^{2}x}=\displaystyle \frac{1}{\sqrt{2}}\\ \textrm{c}.& 1-\cos 2x=0&\textrm{f}.&\sqrt{\sin x\sqrt{\sin x\sqrt{\sin x\sqrt{\cdots }}}}=\cos x\\ \end{array}\\ \end{array}..

Jawab:

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \sin \left ( x+30^{\circ} \right )&-\sin 2x=0\\ \sin 2x&=\sin \left ( x+30^{\circ} \right )\\ 2x_{1,2}&=\begin{cases} \left ( x_{1}+30^{\circ} \right ) &+k.360^{\circ} \\ 180^{\circ}-\left ( x_{2}+30^{\circ} \right ) & +k.360^{\circ} \end{cases}\\ 2x_{1}-x_{1}&=30^{\circ}+k360^{\circ}\Leftrightarrow x_{1}=30^{\circ}+k.360^{\circ}\\ 2x_{2}+x_{2}&=150^{\circ}+k.360^{\circ}\Leftrightarrow 3x_{2}=150^{\circ}+k.360^{\circ}\\ \textrm{Sehingga},&\\ \Leftrightarrow x_{1}&=30^{\circ}+k.360^{\circ}\\ \Leftrightarrow x_{2}&=50^{\circ}+k.120^{\circ}\\ &\\ & \end{aligned}&\begin{aligned}k=0&\\ x_{1}&=30^{\circ}\\ x_{2}&=50^{\circ}\\ k=1&\\ x_{1}&=390^{\circ}\quad \textrm{tidak memenuhi=tm}\\ x_{2}&=170^{\circ}\\ k=2&\\ x_{1}&=30^{\circ}+2.360^{\circ}=750^{\circ}\: \: (\textrm{tm})\\ x_{2}&=290^{\circ}\\ \textrm{yang lain}&\textrm{ jelas tidak ada yang memenuhi}\\ \textbf{HP}&=\left \{ 30^{\circ},50^{\circ},170^{\circ},290^{\circ} \right \} \end{aligned}\\\hline \end{array}.

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{e}.\quad \sqrt{4-4\cos ^{2}x}&=\displaystyle \frac{1}{\sqrt{2}}\\ 4-4\cos ^{2}x&=\displaystyle \frac{1}{2}\\ 4\left ( 1-\cos ^{2}x \right )&=\displaystyle \frac{1}{2}\\ 4\sin ^{2}x&=\displaystyle \frac{1}{2}\\ \sin ^{2}x&=\displaystyle \frac{1}{8}\\ \sin x&=\pm \sqrt{\displaystyle \frac{1}{8}} \end{aligned}&\begin{aligned}x&=\begin{cases} \sin ^{-1}\left ( \sqrt{\displaystyle \frac{1}{8}} \right ) & +k.360^{\circ}\Leftrightarrow x_{1,2}=\begin{cases} \sin ^{-1}\left ( \sqrt{\displaystyle \frac{1}{8}} \right ) & +k.360^{\circ} \\ 180^{\circ}-\left ( \sin ^{-1}\left ( \sqrt{\displaystyle \frac{1}{8}} \right ) \right ) & +k.360^{\circ} \end{cases} \\ \sin ^{-1}\left ( \sqrt{\displaystyle -\frac{1}{8}} \right ) & +k.360^{\circ}\Leftrightarrow x_{3,4}=\begin{cases} \sin^{-1} \left ( -\sqrt{\displaystyle \frac{1}{8}} \right ) & +k.360^{\circ} \\ 180^{\circ}-\sin^{-1} \left ( - \sqrt{\displaystyle \frac{1}{8}}\right ) & +k.360^{\circ} \end{cases} \end{cases} \end{aligned}\\\hline \end{array}.

HP silahkan tentukan sendiri dapat dilakukan dengan bantuan tabel nilai trigonometri.

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{f}.\quad \sqrt{\sin x\sqrt{\sin x\sqrt{\sin x\sqrt{...}}}}&=\cos x\\ \sin x\sqrt{\sin x\sqrt{\sin x\sqrt{...}}}&=\cos^{2} x\\ \sin x\underset{\cos x}{\underbrace{\sqrt{\sin x\sqrt{\sin x\sqrt{...}}}}}&=\cos^{2} x\\ \sin x.\cos x&=\cos ^{2}x\\ \sin x&=\cos x\\ & \end{aligned}&\begin{aligned}\displaystyle \frac{\sin x}{\cos x}&=1\\ \tan x&=\tan 45^{\circ}\\ x&=45^{\circ}+k.180^{\circ}\\ k=0&\\ x&=45^{\circ}\\ k=1&\\ x&=45^{\circ}+180^{\circ}=225^{\circ}\\ k=2&\quad (\textrm{tidak ad yang memenuhi})\\ \textbf{HP}&=\left \{45^{\circ},225^{\circ} \right \} \end{aligned}\\\hline \end{array}.

\LARGE{\fbox{\LARGE{\fbox{LATIHAN SOAL}}}};

Yang belum dibahas, silahkan gunakan sebagai latihan mandiri

Sumber Referensi

  1. Kurnia, N.,  dkk. 2017. Jelajah Matematika 2 SMA Kelas XI Peminatan MIPA(Edisi Revisi 2016). Jakarta: Yudistira.
  2. Sembiring, S., Zulkifli, M., Marsito, dan Rusdi, I. 2017. Matematika untuk siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam(Edisi Revisi 2016). Bandung: SEWU.
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Induksi Matematika Kelas XI (K13 Revisi)

Sebelumnya silahkan Anda buka arsip lama

\begin{tabular}{|llp{11,0cm}|}\hline \multicolumn{3}{|c|}{\textbf{Prinsip Induksi Matematika}}\\\hline \multicolumn{3}{|l|}{\textrm{Misalkan}\: \: \textbf{P(n)}\: \: adalah suatu pernyataan untuk bilangan asli,}\\ \multicolumn{3}{|l|}{Maka pernyataan\: \: \textbf{P(n)}\: \: dianggap benar jika memenuhi kriteria berikut:}\\ Langkah&\textbf{basis}&:\: Rumus/Pernyataan dibuktikan benar untuk n = 1,\quad \textbf{P(1)}\\ Langkah&\textbf{induksi}&:\: Rumus/Pernyataan diasumsikan berlaku untuk n = k,\quad \textbf{P(k)}. selanjutnya rumus/pernyataan dibuktikan benar untuk n = k+1,\quad \textbf{P(k+1)}\\ \multicolumn{2}{|l}{Kesimpulan(\textbf{konklusi})}&:\: Rumus/Pernyataan (\textbf{P(n)})\: berlaku untuk setiap n bilangan asli (sesuaikan dengan keadaan).\\\hline \end{tabular}.

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

  1. Buktikan dengan induksi matematika untuk  n  bilangan asli   1+2+3+...+n=\displaystyle \frac{n(n+1)}{2}.
  2. Buktikan dengan induksi matematika untuk  n  bilangan asli  1+3+5+7+...+(2n-1)=n^{2}.
  3. Buktikan dengan induksi matematika untuk n  bilangan  asli berlaku 2n-3<2^{n-2}.
  4. Buktikanlah dengan induksi matematika untuk  n  bilangan asli  berlaku   n<2^{n}.
  5. Dengan induksi matematika untuk  n  bilangan asli buktikan bahwa  3^{3n+3}-26n-27  habis dibagi oleh 13.
  6. Buktikanlah bahwa  \displaystyle \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...+\frac{1}{n^{2}}\leq 2-\displaystyle \frac{1}{n},\quad n\in \mathbb{N}.

Bukti:

Untuk  Soal yang no. 1

\begin{array}{lll}\\ \textrm{Langkah basis}&:&\textrm{Rumus benar untuk \textit{n} = 1 , karena}\: 1=\displaystyle \frac{1(1+1)}{2}\\ \textrm{Langkah induksi}&:&\textrm{Misalkan rumus berlaku untuk \textit{n = k} , maka}\\ &&1+2+3+\cdots +(k)=\displaystyle \frac{k(k+1)}{2}.\\ &&\textrm{Akan ditunjukkan rumus berlaku untuk \textit{n = k}+1, yaitu}\\ &&1+2+3+\cdots +(k)+(k+1)=\displaystyle \frac{(k+1)(k+2)}{2}\\ &&\underset{\displaystyle \frac{k(k+1)}{2}}{\underbrace{1+2+3+\cdots +(k)}}+(k+1)=\displaystyle \frac{(k+1)(k+2)}{2}\\ && \displaystyle \frac{k(k+1)}{2}+(k+1)=\displaystyle \frac{(k+1)(k+2)}{2}\\ && \displaystyle (k+1)\left ( \frac{k}{2}+1 \right )=\displaystyle \frac{(k+1)(k+2)}{2}\\ && \displaystyle (k+1)\frac{k+2}{2} =\displaystyle \frac{(k+1)(k+2)}{2}\\ && \displaystyle \frac{(k+1)(k+2)}{2}=\displaystyle \frac{(k+1)(k+2)}{2},\qquad \left (\textrm{ruas kiri = ruas kanan} \right )\\ &&\textrm{Karena ruas kiri = ruas kanan, maka rumus berlaku untuk \textit{n = k}}+1\\ \textrm{Kesimpulan}&:&\textrm{Jadi},\: \: 1+2+3+\cdots +n=\displaystyle \frac{n(n+1)}{2},\: \: \textrm{berlaku untuk}\: \: \forall \: n\in \mathbb{N}. \end{array}.

Untuk soal no. 2

\begin{array}{lll}\\ \textrm{Langkah basis}&:&\textrm{Rumus benar untuk \textit{n} = 1 , karena}\: (2.1-1)=1^{2}\\ \textrm{Langkah induksi}&:&\textrm{Misalkan rumus berlaku untuk \textit{n = k} , maka}\\ &&1+3+5+\cdots +(2k-1)=k^{2}.\\ &&\textrm{Akan ditunjukkan rumus berlaku untuk \textit{n = k}+1, yaitu}\\ &&1+3+5+\cdots +(2k-1)+(2(k+1)-1)=(k+1)^{2}\\ &&\underset{\displaystyle k^{2}}{\underbrace{1+3+5+\cdots +(2k-1)}}+(2(k+1)-1)=(k+1)^{2}\\ &&\qquad\qquad\quad k^{2}+\: \: \quad\quad\quad\quad\quad (2(k+1)-1)=(k+1)^{2}\\ &&\qquad\qquad\quad k^{2}+\: \: \quad\quad\quad\quad\quad (2k+2-1)\: \: \: \: =(k+1)^{2}\\ &&\qquad\qquad\quad k^{2}+2k+1 \quad\quad\quad\qquad\qquad\qquad=(k+1)^{2}\\ &&\qquad\qquad\quad\quad\quad\quad\quad (k+1)^{2}=(k+1)^{2},\qquad \left (\textrm{ruas kiri = ruas kanan} \right )\\ &&\textrm{Karena ruas kiri = ruas kanan, maka rumus berlaku untuk \textit{n = k}}+1\\ \textrm{Kesimpulan}&:&\textrm{Jadi},\: \: 1+3+5+\cdots +(2n-1)=n^{2},\: \: \textrm{berlaku untuk}\: \: \forall \: n\in \mathbb{N}. \end{array}.

Untuk soal no. 3

\begin{array}{lll}\\ \textrm{Langkah basis}&:&\textrm{Rumus benar untuk \textit{n} = 1 , karena}\: 2.1-3<2^{1-2}\\ &&\textrm{demikian juga untuk}\: \: n=2.\\ \textrm{Langkah induksi}&:&\textrm{Misalkan rumus berlaku untuk \textit{n = k}}\geq 3 , \textrm{maka}\\ &&(2k-3)<2^{k-2}.\\ &&\textrm{Akan ditunjukkan rumus berlaku untuk \textit{n = k}+1, yaitu}\\ &&2(k+1)-3=2k+2-3=(2k-3)+2<2^{k-2}+2\\ &&\textrm{sehingga}\\ &&(2k-3)+2<2^{k-2}+2<2^{k-2}+2^{k-2},\quad \textrm{untuk}\quad k\geq 3\\ &&(2k-3)+2<2.2^{k-2}\\ &&(2k-3)+2<2^{(k+1)-2}\\ &&\textrm{maka rumus berlaku untuk \textit{n = k}}+1\\ \textrm{Kesimpulan}&:&\textrm{Jadi},\: \: 2n-3<2^{n-2},\: \: \textrm{berlaku untuk}\: \: \forall \: n\in \mathbb{N}. \end{array}.

Untuk soal  no. 4

\begin{array}{lll}\\ \textrm{Langkah basis}&:&\textrm{Rumus benar untuk \textit{n} = 1 , karena}\: 1<2^{1}\\ \textrm{Langkah induksi}&:&\textrm{Misalkan rumus berlaku untuk \textit{n = k}} , \textrm{maka}\\ &&k<2^{k}.\\ &&\textrm{Akan ditunjukkan rumus berlaku untuk \textit{n = k}+1, yaitu}\\ &&(k+1)<2^{(k+1)}\\ &&\textrm{selanjutnya}\\ &&(k+1)<2^{k}+1 ,\quad \textrm{masing-masing ruas dikali dengan 2}\\ &&2(k+1)<2.2^{k}+2\\ &&2k+2<2^{k+1}+2\\ &&2k<2^{k+1}\\ &&k+k<2^{k+1},\quad \textrm{karena}\: \: k>1,\: \: \textrm{maka}\\ &&k+1<2^{k+1}\\ &&\textrm{sehingga rumus berlaku untuk \textit{n = k}}+1\\ \textrm{Kesimpulan}&:&\textrm{Jadi},\: \: n<2^{n},\: \: \textrm{berlaku untuk}\: \: \forall \: n\in \mathbb{N}. \end{array}.

Untuk soal no. 5

\begin{array}{lll}\\ \textrm{Langkah basis}&:&\textrm{Rumus benar untuk \textit{n} = 1 , karena}\: 3^{3.1+3}-26.1-27=\left ( 3^{3} \right )^{2}-26-27\\ &&=27.27-27-26=27(27-1)-26=26.27-26=26(27-1)=13.2.26\\ &&\textrm{adalah bilangan yang habis dibagi oleh 13}\\ \textrm{Langkah induksi}&:&\textrm{Misalkan rumus berlaku untuk \textit{n = k}} , \textrm{maka}\\ &&3^{3k+3}-26k-27, \: \: \textrm{misal kita sederhanakan bentuk ini dengan}\: \: 13\textbf{m}\\ &&\textrm{Akan ditunjukkan rumus berlaku untuk \textit{n = k}+1, yaitu}\\ &&3^{3(k+1)+3}-26(k+1)-27=3^{3k+3+3}-26k-26-27\\ &&=3^{3k+3}.3^{3}-26k-53=27.3^{3k+3}-26k-53=(26+1).3^{3k+3}-26k-53\\ &&=26.3^{3k+3}+3^{3k+3}-26k-27-26=26\left ( 3^{3k+3}-1 \right )+\underset{13\textbf{m}}{\underbrace{3^{3k+3}-26k-27}}\\ &&=13\left (2\left ( 3^{3k+3}-1 \right ) +\textbf{m} \right ),\: \: \textrm{juga kelipatan}\: \: 13\\ &&\textrm{sehingga rumus berlaku untuk \textit{n = k}}+1\\ \textrm{Kesimpulan}&:&\textrm{Jadi},\: \: 3^{3n+3}-26n-27\: \: \textrm{habis dibagi 13, berlaku untuk}\: \: \forall \: n\in \mathbb{N}. \end{array}.

Untuk soal no. 6

\begin{array}{lll}\\ \textrm{Langkah basis}&:&\textrm{Untuk \textit{n} = 1 diperoleh}\: \displaystyle \frac{1}{1^{2}}\leq 2-\frac{1}{1}\\ &&\textrm{Sehingga}\: \: P(1)\: \: \textrm{benar}\\ \textrm{Langkah induksi}&:&\textrm{Misalkan rumus berlaku untuk \textit{n = k}},\: \textrm{maka}\\ &&\displaystyle \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...+\frac{1}{k^{2}}\leq 2-\displaystyle \frac{1}{k}.\\ &&\textrm{Akan ditunjukkan rumus berlaku untuk \textit{n = k}+1, yaitu}\\ &&\displaystyle \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}\leq 2-\displaystyle \frac{1}{k}+\frac{1}{(k+1)^{2}}=2-\left ( \displaystyle \frac{1}{k}-\frac{1}{(k+1)^{2}} \right )\\ &&=2-\left (\displaystyle \frac{(k+1)^{2}-k}{k(k+1)^{2}} \right )=2-\left ( \displaystyle \frac{k^{2}+k+1}{k(k+1)^{2}} \right )\\ &&=2-\left ( \displaystyle \frac{k(k+1)+1}{k(k+1)^{2}} \right )\leq 2-\left ( \displaystyle \frac{k(k+1)}{k(k+1)^{2}} \right )=2-\displaystyle \frac{1}{(k+1)}\\ &&\textrm{maka rumus berlaku untuk \textit{n = k}}+1\\ \textrm{Kesimpulan}&:&\textrm{Jadi},\: \: \displaystyle \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...+\frac{1}{n^{2}}\leq 2-\displaystyle \frac{1}{n},\: \: \textrm{berlaku untuk}\: \: \forall \: n\in \mathbb{N}. \end{array}.

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

Buktikanlah dengan induksi matematika pernyataan-pernyataan berikut untuk  n  bilangan asli!

\begin{array}{ll}\\ 1.&2+4+6+8+...+2n=n^{2}+n\\ 2.&1^{2}+2^{2}+3^{2}+...+n^{2}=\displaystyle \frac{1}{6}n(n+1)(2n+1)\\ 3.&1^{3}+2^{3}+3^{3}+...+n^{3}=\displaystyle \frac{1}{4}n^{2}(n+1)^{2}\\ 4.&1.2+2.3+3.4+...+n(n+1)=\displaystyle \frac{1}{3}n(n+1)(n+2)\\ 5.&\displaystyle \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}=\displaystyle \frac{n}{n+1}\\ 6.&\displaystyle \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{(2n-1)(2n+1)}=\displaystyle \frac{n}{2n+1}\\ 7.&\displaystyle \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}\\ 8.&1+\displaystyle \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n+1}-2\\ 9.&n^{3}-n\quad \textrm{habis dibagi oleh}\: \: 3\\ 10.&n^{5}-n\quad \textrm{habis dibagi oleh}\: \: 5\\ 11.&5^{n}+6.7^{n}+1\quad \textrm{habis dibagi oleh}\: \: 4\\ 12.&5^{2n}-1\quad \textrm{habis dibagi oleh}\: \: 3\\ 13.&3^{n}-1\geq 2^{n}\\ 14.&2n+7< (n+3)^{2}\\ 15.&2+4+6+8+...+2n\leq 2^{n}\\ 16.&\left ( 3+\sqrt{5} \right )^{n}+\left ( 3-\sqrt{5} \right )^{n}\quad \textrm{habis dibagi oleh}\: \: 2^{n} \end{array}.

Sumber Referensi

  1. Budhi, Wono Setya. 2018. Bupena Mathematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: Erlangga.
  2. Kemendikbud. 2017. Matematika untuk SMA/MA/SMK Kelas XI Edisi Revisi. Jakarta: Kementerian Pendidikan dan Kebudayaan.
  3. Tim Matematika ITB. 2007. Program Pembinaan Kompetensi Siswa Bidang Matematika Tahap 1. Bandung: LPPM ITB
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