Contoh Soal 2 Induksi Matematika (Matematika Wajib Kelas XI)

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

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

$\begin{array}{ll}\\ 8.&\textrm{Bila kita hendak membuktikan}\: \: \displaystyle \sum_{i=1}^{n}=\displaystyle \frac{1}{2}n(n+1)\\ &\textrm{dengan induksi matematika}\\ &\textrm{maka untuk langkah}\: \: n=k+1\\ &\textrm{bentuk yang harus ditunjukkan adalah}...\\ &\begin{array}{lll}\\ \textrm{a}.&1+2+3+\cdots +n=\displaystyle \frac{1}{2}n(n+1)\\ \textrm{b}.&1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+1)\\ \textrm{c}.&1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+2)\\ \color{red}\textrm{d}.&1+2+3+\cdots +k+(k+1)=\displaystyle \frac{1}{2}(k+1)(k+2)\\ \textrm{e}.&1+2+3+\cdots +k+(k+1)=\displaystyle \frac{1}{2}(k+2)(k+3)\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}P(n)&=1+2+3+\cdots +n=\displaystyle \frac{1}{2}n(n+1)\\ P(k)&=1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+1)\\ P(k+1)&=1+2+3+\cdots +k+(k+1)\\ &=\underset{\displaystyle \frac{1}{2}k(k+1)}{\underbrace{1+2+3+\cdots +k}}+(k+1)\\ &=\displaystyle \frac{1}{2}k(k+1)+(k+1)=(k+1)\left ( \displaystyle \frac{1}{2}k+1 \right )\\ &=(k+1)\displaystyle \frac{1}{2}(k+2)\\ &=\displaystyle \frac{1}{2}(k+1)(k+2) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: P(n)=\displaystyle \frac{n-1}{n+3},\: \textrm{maka}\: \: P(k+1)\\ & \textrm{dinyatakan dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{k-1}{k+3}\\ \textrm{b}.&\displaystyle \frac{k-1}{k+4}\\ \color{red}\textrm{c}.&\displaystyle \frac{k}{k+4}\\ \textrm{d}.&\displaystyle \frac{k+1}{k+4}\\ \textrm{e}.&\displaystyle \frac{k+1}{k+5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}P(n)=&\displaystyle \frac{n-1}{n+3}\\ P(k+1)&=\displaystyle \frac{k+1-1}{k+1+3}\\ &=\displaystyle \frac{k}{k+4} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: P(n)=\displaystyle \frac{n^{2}+1}{4},\: \: \textrm{maka}\\ &\textrm{pernyataan untuk}\: \: P(k+1)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{k^{2}+2k+1}{4}\\ \color{red}\textrm{b}.&\displaystyle \frac{k^{2}+2k+2}{4}\\ \textrm{c}.&\displaystyle \frac{k^{2}+2k+2}{5}\\ \textrm{d}.&\displaystyle \frac{k^{2}+2k+3}{5}\\ \textrm{e}.&\displaystyle \frac{k^{2}+2k+3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}P(n)&=\displaystyle \frac{n^{2}+1}{4}\\ P(k+1)&=\displaystyle \frac{(k+1)^{2}+1}{4}\\ &=\displaystyle \frac{k^{2}+2k+2}{4} \end{aligned} \end{array}$

Contoh Soal 1 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{ll}\\ 1.&\textrm{Hasil dari}\: \: \displaystyle \sum_{i=1}^{6}16i\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&306\\ \textrm{b}.&314\\ \textrm{c}.&326\\ \color{red}\textrm{d}.&336\\ \textrm{e}.&402 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\displaystyle \sum_{i=1}^{6}16i&=16.1+16.2+16.3+16.4+16.5+16.6\\ &=16+32+48+64+80+96\\ &=336 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Hasil dari}\: \: \displaystyle \sum_{i=2}^{9}i^{2}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&274\\ \textrm{b}.&278\\ \textrm{c}.&280\\ \color{red}\textrm{d}.&284\\ \textrm{e}.&286 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\displaystyle \sum_{i=2}^{9}i^{2}&=2^{2}+3^{2}+4^{2}+5^{2}+..+9^{2}\\ &=4+9+16+25+...+81\\ &=284 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Poa bilangan}\: \: 12,14,16,18,20,...,(2n+10).\\ &\textrm{Nilai suku ke-100 adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.&180\\ \textrm{b}.&194\\ \textrm{c}.&198\\ \textrm{d}.&208\\ \color{red}\textrm{e}.&210\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}U_{n}&=2n+10\\ U_{100}&=2\times 100+10\\ &=210 \end{aligned} \end{array}$

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

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

Contoh Soal 6 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{ll}\\ 26.&\textrm{Jika nilai}\: \: \cot A+\cos A=x\\ &\textrm{dan}\: \: \cot A-\cos A=y,\\ &\textrm{maka nilai}\: \: \left ( x^{2}-y^{2} \right )=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&4\sqrt{xy}\\ \textrm{b}.&2\sqrt{xy}\\ \textrm{c}.&xy\\ \textrm{d}.&2\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\color{magenta}xy&=\left (\cot A+\cos A \right )\left ( \cot A-\cos A \right )\\ &=\cot ^{2}A-\cos ^{2}A\\ &=\displaystyle \frac{\cos^{2}A }{\sin ^{2}A}-\cos ^{2}A\\ &=\displaystyle \frac{\cos^{2}A }{\sin ^{2}A}-\cos ^{2}A\times \frac{\sin ^{2}A}{\sin ^{2}A}\\ &=\displaystyle \frac{\cos ^{2}A}{\sin ^{2}A}\left ( 1-\sin ^{2}A \right )\\ &=\displaystyle \frac{\cos ^{4}A}{\sin ^{2}A}\\ \sqrt{xy}&=\displaystyle \frac{\cos A}{\sin A}\times \cos A\\ \color{black}\textrm{Se}&\color{black}\textrm{lanjutnya}\\ x^{2}-y^{2}&=\left (\cot A+\cos A \right )^{2}-\left ( \cot A-\cos A \right )^{2}\\ (x+y)&(x-y)=(\cot A+\cos A+\cot A-\cos A)\\ &\qquad\times (\cot A+\cos A-(\cot A-\cos A))\\ x^{2}-y^{2}&=2\cot A\times 2\cos A\\ &=4\times \displaystyle \frac{\cos A}{\sin A}\times \cos A\\ &=\color{magenta}4\sqrt{xy} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Jika nilai}\: \: \cos A+\sin A=\sqrt{2}\cos A\\ &\textrm{maka nilai}\: \: \left ( \cos A-\sin A \right )=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sqrt{2}\cos A\\ \textrm{b}.&-\sqrt{2}\sin A\\ \color{red}\textrm{c}.&\sqrt{2}\sin A\\ \textrm{d}.&\displaystyle \frac{1}{\sqrt{2}}\sec A\\ \textrm{e}.&\displaystyle \frac{1}{\sqrt{2}}\csc A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\cos A+\sin A&=\sqrt{2}\cos A\\ \left (\cos A+\sin A \right )^{2}&=\left (\sqrt{2}\cos A \right )^{2}\\ 1+2\sin A\cos A&=2\cos ^{2}A\\ 2\sin A\cos A&=2\cos ^{2}A-1\\ \textrm{maka}&\\ \left (\cos A-\sin A \right )^{2}&=1-2\sin A\cos A\\ &=1-\left ( 2\cos ^{2}A-1 \right )\\ &=2-2\cos ^{2}A\\ &=2\left ( 1-\cos ^{2}A \right )\\ &=2\sin ^{2}A\\ \cos A-\sin A&=\sqrt{2\sin ^{2}A}\\ &=\sin A\sqrt{2}\\ &=\color{magenta}\sqrt{2}.\sin A \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Sembiring, S., Zulkifli, M., Marsito, dan Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU.
2. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

Contoh Soal 5 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{ll}\\ 21.&\textrm{Nilai dari}\: \: \displaystyle \frac{1}{\sec ^{2}A}+\frac{1}{\csc ^{2}A}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \color{red}\textrm{d}.&\displaystyle 1\\ {e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{1}{\sec ^{2}A}+\frac{1}{\csc ^{2}A}\\ &=\cos ^{2}A+\sin ^{2}=1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Nilai dari}\: \: \displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\cot B\times \cot C \\ \color{red}\textrm{b}.&\displaystyle \tan B\times \tan C\\ \textrm{c}.&\displaystyle \sec B\times \csc C\\ \textrm{d}.&\displaystyle \tan B\times \cot C\\ {e}.&\displaystyle \tan B\times \csc C \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}\\ &=\displaystyle \frac{\tan B+\tan C}{\displaystyle \frac{1}{\tan B}+\frac{1}{\tan C}}\\ &=\displaystyle \frac{\tan B+\tan C}{\left ( \displaystyle \frac{\tan B+\tan C}{\tan B\times \tan C} \right )}\\ &=\tan B\times \tan C \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Nilai dari}\\ &\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=....\\ &\begin{array}{llll}\\ \textrm{a}.&2\tan A \\ \textrm{b}.&2\cot A\\ \textrm{c}.&\displaystyle 2\sec A\\ \color{red}\textrm{d}.&\displaystyle 2\csc A\\ {e}.&\displaystyle 2\tan A.\sec A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}\\ &=\tan A\left (\displaystyle \frac{1}{\displaystyle \frac{1}{\cos A}-1}+\frac{1}{\displaystyle \frac{1}{\cos A}+1} \right )\\ &=\displaystyle \frac{\sin A}{\cos A}\left ( \displaystyle \frac{\cos A}{1-\cos A}+\frac{\cos A}{1+\cos A} \right )\\ &=\displaystyle \frac{\sin A}{1-\cos A}+\frac{\sin A}{1+\cos A}\\ &=\displaystyle \frac{\sin A(1+\cos A)+\sin A(1-\cos A)}{(1-\cos A)(1+\cos A)}\\ &=\displaystyle \frac{2\sin A}{1-\cos ^{2}}\\ &=\displaystyle \frac{2\sin A}{\sin ^{2}A}\\ &=\displaystyle \frac{2}{\sin A}\\ &=2\csc A \end{aligned}\\\\ &\textrm{Sebagai catatanya}\\ &\textrm{Anda bisa gunakan cara yang lain} \end{array}$

$\begin{array}{ll}\\ 24.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\sin \left ( 2x-20^{\circ} \right )=-\cos \left ( 3x+50^{\circ} \right )\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-30^{\circ}\\ \textrm{b}.&-25^{\circ}\\ \color{red}\textrm{c}.&20^{\circ}\\ \textrm{d}.&25^{\circ}\\ {e}.&30^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\sin \left ( 2x-20^{\circ} \right )&=-\cos \left ( 3x+50^{\circ} \right )\\ \sin \left ( 20^{\circ}-2x \right )&=\cos \left ( 3x+50^{\circ} \right )\\ \sin A&=\cos B,\: \: \color{black}\textrm{artinya}\\ A+B&=90^{\circ},\: \: \color{magenta}\textrm{maka}\\ \left ( 20^{\circ}-2x \right )+\left ( 3x+50^{\circ} \right )&=90^{\circ}\\ x+70^{\circ}&=90^{\circ}\\ x&=90^{\circ}-70^{\circ}\\ &=20^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\tan \left ( 2x+60^{\circ} \right )=\cot \left ( 90^{\circ}-3x \right )\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&20^{\circ}\\ \textrm{b}.&30^{\circ}\\ \textrm{c}.&40^{\circ}\\ \textrm{d}.&50^{\circ}\\ \color{red}\textrm{e}.&60^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\tan \left ( 2x+60^{\circ} \right )&=\cot \left ( 90^{\circ}-3x \right )\\ \tan (2x+60^{\circ})&=\tan 3x\\ 2x+60^{\circ}&=3x\\ 2x-3x&=-60^{\circ}\\ -x&=-60^{\circ}\\ x&=60^{\circ} \end{aligned} \end{array}$

Contoh Soal 4 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{ll}\\ 16.&\textrm{Jika}\: \: \tan^{2} x +\sec x =5 \: \: \textrm{untuk}\\ &0\leq x\leq \displaystyle \frac{\pi }{2}\: \: \textrm{maka nilai} \: \: \cos x=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \color{red}\textrm{b}.&\displaystyle \frac{1}{2}\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&\displaystyle \frac{1}{\sqrt{2}}\\ \textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Ingat bahwa}&\: \: 0\leq x\leq \displaystyle \frac{\pi }{2}\\ \textrm{berarti sudu}&\textrm{t}\: \: x\: \: \textrm{berada di kuadran I}\\ \textrm{sehingga ak}&\textrm{an menyebabkan nilai}\\ & \color{magenta}\cos x=+\\ \color{black}\textrm{Selanjutnya}&\\ \tan^{2} x +\sec x &=5\\ \sec ^{2}x-1+\sec x&=5\\ \sec ^{2}x+\sec x-6&=0\\ (\sec x+3)(\sec x-2)&=0\\ \sec x=-3\: \: \textrm{atau}&\sec x=2\\ \textrm{untuk}\: \: \sec x&=-3\: \: (\color{red}\textrm{tidak memenuhi})\\ \textrm{untuk}\: \: \sec x&=2\: \: (\color{magenta}\textrm{memenuhi})\\ \color{black}\textrm{Selanjutnya}&\: \color{black}\textrm{lagi}\\ \sec x&=2\\ \displaystyle \frac{1}{\cos x}&=2\\ \cos x&=\color{magenta}\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Nilai}\\ &\left ( \sin A+\cos A \right )^{2}+\left ( \sin A-\cos A \right )^{2}=....\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \color{red}\textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 3\\ \textrm{d}.&\displaystyle 3\cos A\\ {e}.&\displaystyle 4\sin A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\left ( \sin A+\cos A \right )^{2}+\left ( \sin A-\cos A \right )^{2}\\ &=\sin ^{2}A+2\sin A\cos A+\cos ^{2}A\\ &\quad +\sin ^{2}A-2\sin A\cos A+\cos ^{2}A\\ &=1+1\\ &=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Nilai}\: \: \sqrt{\displaystyle \frac{1+\sin A}{1-\sin A}}=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\sec A+\tan A\\ \textrm{b}.&\sec ^{2}A+\tan ^{2}A\\ \textrm{c}.&\sec ^{2}A-\tan ^{2}A\\ \textrm{d}.&\tan ^{2}A-\sec ^{2}A\\ {e}.&\sec A\times \tan A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\sqrt{\displaystyle \frac{1+\sin A}{1-\sin A}}\\ &=\sqrt{\displaystyle \frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}}\\ &=\sqrt{\displaystyle \frac{\left ( 1+\sin A \right )^{2}}{1-\sin ^{2}A}}\\ &=\sqrt{\displaystyle \frac{\left ( 1+\sin A \right )^{2}}{\cos ^{2}A}}\\ &=\displaystyle \frac{1+\sin A}{\cos A}\\ &=\displaystyle \frac{1}{\cos A}+\frac{\sin A}{\cos A}\\ &=\sec A+\tan A \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: 0^{\circ}\leq \theta \leqslant 90^{\circ},\: \textrm{maka nilai}\\ &\left ( \displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta } \right )=....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \color{red}\textrm{b}.&\displaystyle 0\\ \textrm{c}.&\displaystyle \frac{1}{4}\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ {e}.&\displaystyle 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\left ( \displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta } \right )\\ &=\left ( \displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }\times \frac{4+5\cos \theta }{4+5\cos \theta } \right )\\ &\qquad-\left ( \displaystyle \frac{3+5\sin \theta }{4+5\cos \theta }\times \frac{3-5\sin \theta }{3-5\sin \theta } \right )\\ &=\displaystyle \frac{25\cos ^{2}-16-\left ( 9-25\sin ^{2}\theta \right )}{(3-5\sin \theta )(4+5\cos \theta )}\\ &=\displaystyle \frac{25\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )-25}{(3-5\sin \theta )(4+5\cos \theta )}\\ &=\displaystyle \frac{0}{(3-5\sin \theta )(4+5\cos \theta )}\\ &=\color{magenta}0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Nilai}\\ &\left ( 1+\cot \theta -\csc \theta \right )\left ( 1+\tan \theta +\sec \theta \right )=....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \color{red}\textrm{e}.&\displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\left ( 1+\cot \theta -\csc \theta \right )\left ( 1+\tan \theta +\sec \theta \right )\\ &=\left ( 1+\displaystyle \frac{\cos \theta }{\sin \theta } -\frac{1}{\sin \theta } \right )\left ( 1+\frac{\sin \theta }{\cos \theta } +\frac{1}{\cos \theta } \right )\\ &=\left ( \displaystyle \frac{\sin \theta +\cos \theta -1}{\sin \theta } \right )\left ( \displaystyle \frac{\cos \theta +\sin \theta +1}{\cos \theta } \right )\\ &=\displaystyle \frac{\left (\sin \theta +\cos \theta \right )^{2}-1}{\sin \theta \cos \theta }\\ &=\displaystyle \frac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }\\ &=2 \end{aligned} \end{array}$

Contoh Soal 3 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{ll}\\ 11.&\textrm{Himpunan penyelesaian persamaan}\\ &3\cos 2x+5\sin x+1=0\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 2\pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ \displaystyle \frac{7}{6}\pi ,\frac{11}{6}\pi \right \}\\ \textrm{b}.&\left \{ \displaystyle \frac{5}{6}\pi ,\frac{11}{6}\pi \right \}\\ \textrm{c}.&\left \{ \displaystyle \frac{1}{6}\pi ,\frac{7}{6}\pi \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{1}{5}\pi ,\frac{5}{6}\pi \right \}\\ \textrm{e}.&\left \{ \displaystyle \frac{5}{6}\pi ,\frac{7}{6}\pi \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}3\cos 2x+5\sin x+1&=0\\ 3\left ( 1-2\sin ^{2}x \right )+5\sin x+1&=0\\ -6\sin ^{2}+5\sin x+4&=0\\ 6\sin ^{2}x-5\sin x-4&=0\\ \left ( 3\sin x-4 \right )\left ( 2\sin x+1 \right )&=0\\ \sin x=\displaystyle \frac{4}{3}\: \: \textrm{atau}\: \: \sin x&=-\frac{1}{2}\\ \sin x&=\sin 150^{\circ}=\frac{5}{6}\pi \\ x&=\begin{cases} \displaystyle \frac{7}{6}\pi &+k.2\pi \\ \pi -\displaystyle \frac{7}{6}\pi & +k.2\pi \end{cases}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=\displaystyle \frac{7}{6}\pi \\ x_{2}&=-\displaystyle \frac{1}{6}\pi \\ \textrm{saat}\: \: k&=1\\ x_{1}&=\color{red}\displaystyle \frac{7}{6}\pi +2\pi \\ x_{2}&=-\displaystyle \frac{1}{6}\pi +2\pi =\frac{11}{6}\pi \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Himpunan penyelesaian dari}\\ &\sqrt{3}\sin 2x+2\cos ^{2}x=-1\: \: \textrm{untuk}\\ &0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 240^{\circ},300^{\circ} \right \}\\ \textrm{b}.&\left \{ 30^{\circ},60^{\circ} \right \}\\ \textrm{c}.&\left \{ 150^{\circ},315^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 120^{\circ},300^{\circ} \right \}\\ \textrm{e}.&\left \{ 60^{\circ},150^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\sqrt{3}\sin 2x+2\cos ^{2}x&=-1\\ \sqrt{3}\sin 2x+1+\cos 2x&=-1\\ \sqrt{3}\sin 2x+\cos 2x&=-2\\ \sqrt{\sqrt{3}^{2}+1^{2}}\cos \left ( 2x-\alpha \right )&=-2\\ a=x=1,\: \: b&=y=\sqrt{3}\\ \alpha &=\arctan \displaystyle \frac{b}{a}=\arctan \frac{\sqrt{3}}{1}\\ \alpha &=60^{\circ}\\ \textrm{maka persamaan akan}&\: \textrm{menjadi}\\ 2\cos \left ( 2x-60^{\circ} \right )&=-2\\ \cos \left ( 2x-60^{\circ} \right )&=-1\\ \cos \left ( 2x-60^{\circ} \right )&=\cos 180^{\circ}\\ \left ( 2x-60^{\circ} \right )&=\pm 180^{\circ}+k.360^{\circ}\\ 2x&=60^{\circ}\pm 180^{\circ}+k.360^{\circ}\\ x&=30^{\circ}\pm 90^{\circ}+k.180^{\circ}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=120^{\circ}\\ x_{2}&=-60^{\circ}\\ \textrm{saat}\: \: k&=1\\ x_{3}&=120^{\circ}+360^{\circ}=....\\ x_{2}&=-60^{\circ}+360^{\circ}=300^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: 3\sin \theta +4\cos \theta =5 \: \: \textrm{maka}\\ &\textrm{nilai dari} \: \: \sin \theta \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0,3\\ \color{red}\textrm{b}.&0,60\\ \textrm{c}.&0,75\\ \textrm{d}.&0,80\\ \textrm{e}.&1,20 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}3\sin \theta +4\cos \theta &=5\\ k\cos \left ( \theta -\alpha \right )&=5\\ a=x=4,\: &b=y=3\\ \theta &=\arctan \displaystyle \frac{b}{a}\\ \theta &=\arctan \displaystyle \frac{3}{4}\\ \color{black}\textrm{atau}\: \: &\\ \tan \theta &=\displaystyle \frac{3}{4}\\ \color{black}\textrm{maka}\: \: &\\ \sin \theta &=\displaystyle \frac{3}{\sqrt{3^{2}+4^{2}}}\\ &=\displaystyle \frac{3}{5}\\ &=\color{magenta}0,6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Jika}\: \: \tan \theta +\sec \theta =x \: \: \textrm{maka}\\ &\textrm{nilai dari} \: \: \tan \theta \: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2x}{x^{2}-1}\\ \textrm{b}.&\displaystyle \frac{2x}{x^{2}+1}\\ \textrm{c}.&\displaystyle \frac{x^{2}+1}{2x}\\ \color{red}\textrm{d}.&\displaystyle \frac{x^{2}-1}{2x}\\ \textrm{e}.&\displaystyle \frac{x^{2}-1}{x^{2}+1} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\color{black}\textrm{Langkah}&\: \: 1\\ \tan \theta +\sec \theta &=x\\ \displaystyle \frac{\sin \theta }{\cos \theta } +\frac{1}{\cos \theta }&=x\\ \displaystyle \frac{\sin \theta +1}{\cos \theta }&=x\\ \sin \theta +1&=x\cos \theta ...........1\\ \color{black}\textrm{Langkah}&\: \: 2\\ \left (\tan \theta +\sec \theta \right )^{2}&=x^{2}\\ \tan ^{2}\theta +2\tan \theta \sec \theta +\sec ^{2}\theta &=x^{2}\\ \sec ^{2}\theta -1+2\tan \theta \sec \theta &+\sec ^{2}\theta =x^{2}\\ 2\sec ^{2}\theta +2\tan \theta \sec \theta &=x^{2}+1\\ \displaystyle \frac{2}{\cos ^{2}\theta }+\frac{2\sin \theta }{\cos ^{2}\theta }&=x^{2}+1\\ 1+\sin \theta &=\left (\displaystyle \frac{x^{2}+1}{2} \right )\cos ^{2}\theta.....2\\ \color{black}\textrm{Langkah}&\: \: 3\\ \left (\displaystyle \frac{x^{2}+1}{2} \right )\cos ^{2}&=x\cos \theta \\ \cos \theta &=\displaystyle \frac{2x}{x^{2}+1}\\ \textrm{maka}\: \: &(\color{magenta}\textbf{dengan sisi segitiga})\\ \tan \theta =\displaystyle \frac{\sqrt{\left ( x^{2}+1 \right )^{2}-(2x)^{2}}}{2x}&=\displaystyle \frac{\sqrt{x^{4}+2x^{2}+1-4x^{2}}}{2x}\\ \tan \theta &=\displaystyle \frac{\sqrt{x^{4}-2x^{2}+1}}{2x}\\ &=\displaystyle \frac{\sqrt{\left ( x^{2}-1 \right )^{2}}}{2x}\\ &=\displaystyle \frac{x^{2}-1}{2x} \end{aligned} \end{array}$

$\begin{aligned}.\: \: \qquad \textbf{Sehingga}&\: \textbf{dari kasus di atas didapatkan}\\ \color{magenta}\sin \theta &=\displaystyle \frac{x^{2}-1}{x^{2}+1}\\ \color{magenta}\cos \theta &=\displaystyle \frac{2x}{x^{2}+1}\\ \color{magenta}\tan \theta &=\displaystyle \frac{x^{2}-1}{2x} \end{aligned}$

$\begin{array}{ll}\\ 15.&\textrm{Jika}\: \: \sec x +\tan x =\displaystyle \frac{3}{2} \: \: \textrm{untuk}\\ &0\leq x\leq \displaystyle \frac{\pi }{2}\: \: \textrm{maka nilai} \: \: \sin x=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{5}{13}\\ \textrm{b}.&\displaystyle \frac{12}{13}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle \frac{2}{13}\\ \textrm{e}.&\displaystyle \frac{5}{12} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\sec x +\tan x &=\displaystyle \frac{3}{2}\\ \textrm{ingat saat}&\: \color{black}\textrm{mengerjakan soal no}.14\\ \sec \theta +\tan \theta &=x\\ \sin \theta &=\displaystyle \frac{x^{2}-1}{x^{2}+1},\: \: \color{black}\textrm{maka}\\ \sin x&=\color{magenta}\displaystyle \frac{\left ( \displaystyle \frac{3}{2} \right )^{2}-1}{\left (\displaystyle \frac{3}{2} \right )^{2}+1}\\ &=\displaystyle \frac{\displaystyle \frac{9}{4}-1}{\displaystyle \frac{9}{4}+1}\\ &=\frac{\displaystyle \frac{5}{4}}{\displaystyle \frac{13}{4}}\\ &=\color{magenta}\displaystyle \frac{5}{13} \end{aligned} \end{array}$

Contoh Soal 2 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{ll}\\ 6.&\textrm{Bentuk}\: \: \sqrt{3}\cos x-\sin x\: \: \textrm{untuk}\: \: 0\leq x\leq 2\pi ,\\ &\textrm{dapat dinyatakan sebagai}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2\cos \left ( x+\displaystyle \frac{\pi }{6} \right )\\ \textrm{b}.&2\cos \left ( x+\displaystyle \frac{7\pi }{6} \right ) \\ \color{red}\textrm{c}.&2\cos \left ( x-\displaystyle \frac{11\pi }{6} \right )\\ \textrm{d}.&2\cos \left ( x-\displaystyle \frac{7\pi }{6} \right )\\ \textrm{e}.&2\cos \left ( x-\displaystyle \frac{\pi }{6} \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{c}\\ &\color{blue}\begin{aligned}\sqrt{3}\cos x-\sin x&=k\cos (x-\alpha )\\ (1)\qquad & (a,b)=\begin{cases} a &=\sqrt{3} \\ b &=-1 \end{cases}\\ \textrm{maka}\: &\textrm{titik ada dikadran IV}\\ (2)\: \quad k&=\sqrt{a^{2}+b^{2}}\\ &=\sqrt{\sqrt{3}^{2}+(-1)^{2}}\\ &=\sqrt{4}=2\\ (3)\quad \alpha &=\arctan \frac{b}{a}=\arctan \left (\frac{-1}{\sqrt{3}} \right )=-30^{\circ}\\ &=\left ( 360^{\circ}-30^{\circ} \right )=330^{\circ}=\displaystyle \frac{11}{6}\pi \\ \textrm{sehingga}&\\ \sqrt{3}\cos x-\sin x&=2\cos \left ( x-\displaystyle \frac{11}{6}\pi \right )\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Nilai-nilai}\: \: x\: \: \textrm{yang terletak pada}\: \: 0\leq x\leq 2\pi ,\\ &\textrm{yang memenuhi persamaan}\: \: \sqrt{3}\cos x+\sin x=\sqrt{2}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&75^{\circ}\: \: \textrm{atau}\: \: 285^{\circ}\\ \color{red}\textrm{b}.&75^{\circ}\: \: \textrm{atau}\: \: 345^{\circ}\\ \textrm{c}.&15^{\circ}\: \: \textrm{atau}\: \: 285^{\circ}\\ \textrm{d}.&15^{\circ}\: \: \textrm{atau}\: \: 345^{\circ}\\ \textrm{e}.&15^{\circ}\: \: \textrm{atau}\: \: 75^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{b}\\ &\color{blue}\begin{aligned}\sqrt{3}\cos x+\sin x&=\sqrt{2}\\ \sqrt{\sqrt{3}^{2}+1^{2}}&\left ( \cos \left ( \alpha -\arctan \displaystyle \frac{1}{\sqrt{3}} \right ) \right )=\sqrt{2}\\ 2\cos \left ( x-30^{\circ} \right )&=\sqrt{2}\\ \cos \left ( x-30^{\circ} \right )&=\displaystyle \frac{\sqrt{2}}{2}=\frac{1}{2}\sqrt{2}\\ \cos \left ( x-30^{\circ} \right )&=\cos 45^{\circ}\\ \left ( x-30^{\circ} \right )&=\pm 45^{\circ}+k.360^{\circ}\\ x&=30^{\circ}\pm 45^{\circ}+k.360^{\circ}\\ \textrm{saat}\: \: k&=0\\ x_{1}&=75^{\circ}\\ x_{2}&=\color{red}-15^{\circ}\\ \textrm{saat}\: \: k&=1\\ x_{3}&=75^{\circ}+360^{\circ}=\color{red}435^{\circ}\\ x_{4}&=-15^{\circ}+360^{\circ}=345^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Diketahui fungsi trigonometri}\: \: f(x)=\displaystyle \frac{1}{2}\sin 3x\\ &\textrm{perhatikanlah pernyataan-pernyataan berikut}\\ &(1)\quad \textrm{hasil dari}\: \: f(0)+f\left ( \displaystyle \frac{\pi }{6} \right )=1\\ &(2)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{6} \right )+f\left ( \displaystyle \frac{\pi }{3} \right )=\displaystyle \frac{1}{2}\\ &(3)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{6} \right )-f\left ( \displaystyle \frac{\pi }{3} \right )=\displaystyle \frac{1}{2}\\ &(4)\quad \textrm{hasil dari}\: \: f\left ( \displaystyle \frac{\pi }{3} \right )-f\left ( \displaystyle \frac{\pi }{6} \right )=1\\ &\textrm{Pernyataan yang tepat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&(1)\: \: \textrm{dan}\: \: (2)\\ \textrm{b}.&(1)\: \: \textrm{dan}\: \: (3)\\ \textrm{c}.&(1)\: \: \textrm{dan}\: \: (4)\\ \color{red}\textrm{d}.&(2)\: \: \textrm{dan}\: \: (3)\\ \textrm{e}.&(3)\: \: \textrm{dan}\: \: (4) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{d}\\ &\color{blue}\begin{aligned}\textrm{Diketahui}&\\ f(x)&=\displaystyle \frac{1}{2}\sin 3x\\ \textrm{maka}\qquad&\\ f(0)&=\displaystyle \frac{1}{2}\sin 3(0^{\circ})=0\\ f\left ( \displaystyle \frac{\pi }{3} \right )&=\displaystyle \frac{1}{2}\sin 3\left ( \displaystyle \frac{\pi }{3} \right )=0\\ f\left ( \displaystyle \frac{\pi }{6} \right )&=\displaystyle \frac{1}{2}\sin 3\left ( \displaystyle \frac{\pi }{6} \right )=\displaystyle \frac{1}{2}\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Himpunan penyelesaian dari}\: \: \sin x-\sqrt{3}\cos x=-1\\ &\textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 0^{\circ},120^{\circ} \right \}\\ \textrm{b}.&\left \{ 90^{\circ},330^{\circ} \right \}\\ \textrm{c}.&\left \{ 60^{\circ},180^{\circ} \right \}\\ \textrm{d}.&\left \{ 90^{\circ},120^{\circ} \right \}\\ \color{red}\textrm{e}.&\left \{ 30^{\circ},270^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\sin x-\sqrt{3}\cos x&=-1\\ \sqrt{1^{2}+\left ( -\sqrt{3} \right )^{2}}\cos \left ( x-\alpha \right )&=-1\\ a&=x=-\sqrt{3},\: \: b=y=1\\ &\textbf{kuadran II}\\ \alpha &=\arctan \left ( \displaystyle \frac{1}{-\sqrt{3}} \right )=-30^{\circ}\\ &=180^{\circ}-30^{\circ}=150^{\circ}\\ \textrm{maka persamaan akan}&\: \textrm{menjadi}\\ 2\cos \left ( x-150^{\circ} \right )&=-1\\ \cos \left ( x-150^{\circ} \right )&=\displaystyle \frac{-1}{2}\\ \cos \left ( x-150^{\circ} \right )&=\cos 120^{\circ}\\ \left ( x-150^{\circ} \right )&=\pm 120^{\circ}+k.360^{\circ}\\ x&=150^{\circ}\pm 120^{\circ}+k.360^{\circ}\\ \textrm{saat}\: \: &k=0\\ x_{1}&=270^{\circ}\\ x_{2}&=30^{\circ}\\ \textrm{saat}\: \: &k=1\\ x_{3}&=270^{\circ}+360^{\circ}=....\\ x_{4}&=30^{\circ}+360^{\circ}=.... \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Nilai}\: \: \tan x\: \: \textrm{yang memenuhi persamaan}\\ &\cos 2x+7\cos x-3=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\sqrt{3}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{c}.&\displaystyle \frac{1}{3}\sqrt{3}\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{e}.&\displaystyle \frac{1}{5}\sqrt{5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\cos 2x+7\cos x-3&=0\\ 2\cos ^{2}x-1+7\cos x-3&=0\\ 2\cos ^{2}x+7\cos x-4&=0\\ \left ( \cos x+4 \right )\left (2 \cos -1 \right )&=0\\ \color{red}\cos x=-4\: \: \color{black}\textrm{atau}\: \: \color{magenta}\cos x&=\displaystyle \frac{1}{2}=\cos 60^{\circ}\\ x&=60^{\circ},\\ \textrm{maka}&\\ \tan 60^{\circ}&=\sqrt{3}\\\\ \textrm{ dan ingat bahwa}&\: \: \cos x=-4\: \: \color{red}\textrm{tidak memenuhi} \end{aligned} \end{array}$

Contoh Soal 1 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{ll}\\ 1.&\textrm{Himpunan penyelesaian dari}\\ &\sin 2x=\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 30^{\circ},210^{\circ} \right \}\\ \textrm{b}.&\left \{ 60^{\circ},240^{\circ} \right \}\\ \textrm{c}.&\left \{ 30^{\circ},60^{\circ},210^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 30^{\circ},60^{\circ},210^{\circ},240^{\circ} \right \}\\ \textrm{e}.&\left \{ 30^{\circ},60^{\circ},210^{\circ},240^{\circ},270^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\sin 2x&=\displaystyle \frac{1}{2}\sqrt{3}\\ \sin 2x&=\sin 60^{\circ}\\ 2x&=\begin{cases} 60^{\circ} & +k.360^{\circ} \\ \left ( 180^{\circ}-60^{\circ} \right ) & +k.360^{\circ} \end{cases}\\ x&=\begin{cases} 30^{\circ} & +k.180^{\circ}\\ 60^{\circ} & +k.180^{\circ} \end{cases}\\ \textrm{saat}&\: \: k=0\\ x&=\begin{cases} 30^{\circ} & \\ 60^{\circ} & \end{cases}\\ \textrm{saat}&\: \: k=1\\ x&=\begin{cases} 30^{\circ} & +1.180^{\circ}=210^{\circ}\\ 60^{\circ} & +1.180^{\circ}=240^{\circ} \end{cases}\\ \textrm{saat}&\: \: k=2\\ x&=\begin{cases} 30^{\circ} & +2.180^{\circ}=\color{red}390^{\circ}\\ 60^{\circ} & +2.180^{\circ}=\color{red}420^{\circ} \end{cases}\\ \color{red}\textrm{kedua}&\color{red}\textrm{nnya tidak memenuhi} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Himpunan penyelesaian dari}\\ &\tan 2x-\sqrt{3}=0\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 15^{\circ},105^{\circ},195^{\circ},285^{\circ} \right \}\\ \color{red}\textrm{b}.&\left \{ 30^{\circ},120^{\circ},210^{\circ},300^{\circ} \right \}\\ \textrm{c}.&\left \{ 45^{\circ},135^{\circ},225^{\circ},315^{\circ} \right \}\\ \textrm{d}.&\left \{ 15^{\circ},105^{\circ},195^{\circ},285^{\circ} \right \}\\ \textrm{e}.&\left \{ 15^{\circ},30^{\circ},45^{\circ},60^{\circ},75^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\tan 2x&-\sqrt{3}=0\\ \tan 2x&=\sqrt{3}\\ \tan 2x&=\tan 60^{\circ}\\ 2x&=60+k.180^{\circ}\\ x&=30^{\circ}+k.90^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=30^{\circ}\\ \textrm{saat}&\: \: k=1\\ x&=30^{\circ}+90^{\circ}=120^{\circ}\\ \textrm{saat}&\: \: k=2\\ x&=30^{\circ}+180^{\circ}=210^{\circ}\\ \textrm{saat}&\: \: k=3\\ x&=30^{\circ}+270^{\circ}=300^{\circ}\\ \textrm{saat}&\: \: k=4\\ x&=30^{\circ}+360^{\circ}=\color{red}390^{\circ}\\ \color{red}\textrm{tidak}&\: \color{red}\textrm{memenuhi} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Himpunan penyelesaian dari}\\ &\cos 3x=-\displaystyle \frac{1}{2}\sqrt{3}\: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 180^{\circ}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 40^{\circ},80^{\circ} \right \}\\ \textrm{b}.&\left \{ 50^{\circ},70^{\circ} \right \}\\ \textrm{c}.&\left \{ 40^{\circ},70^{\circ},80^{\circ} \right \}\\ \color{red}\textrm{d}.&\left \{ 50^{\circ},70^{\circ},170^{\circ} \right \}\\ \textrm{e}.&\left \{ 50^{\circ},80^{\circ},170^{\circ} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\cos 3x&=-\displaystyle \frac{1}{2}\sqrt{3}\\ \cos 3x&=-\cos 30^{\circ}\\ \cos 3x&=\cos \left (180^{\circ}-30^{\circ} \right )=\cos 150^{\circ}\\ 3x&=\pm 150^{\circ}+k.360^{\circ}\\ x&=\pm 50^{\circ}+k.120^{\circ}\\ \textrm{saat}&\: \: k=0\\ x&=\pm 50^{\circ}\: \rightarrow x=50^{\circ}\: \: (\textrm{mm})\\ \textrm{saat}&\: \: k=1\\ x&=\pm 50^{\circ}+120^{\circ}=\begin{cases} 170^{\circ} & (\textrm{mm}) \\ 70^{\circ} & (\textrm{mm}) \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &2\cos ^{2}x+\cos x-1=0\: \: \textrm{untuk}\: \: 0\leq x\leq \pi \\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \pi \\ \textrm{b}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \frac{2}{3}\pi \\ \textrm{c}.&\displaystyle \frac{1}{3}\pi \: \: \textrm{dan}\: \: \frac{3}{4}\pi \\ \textrm{d}.&\displaystyle \frac{1}{4}\pi \: \: \textrm{dan}\: \: \frac{3}{4}\pi \\ \textrm{e}.&\displaystyle \frac{1}{4}\pi \: \: \textrm{dan}\: \: \frac{2}{3}\pi \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}2\cos ^{2}x+\cos x-1&=0\\ \left (2\cos x-1 \right )\left (\cos x+1 \right )&=0\\ \cos x=\displaystyle \frac{1}{2}\: \: \color{magenta}\textrm{atau}\: \: &\cos x=-1\\ \cos x=\cos 60^{\circ}=\cos \frac{1}{3}\pi \: \: &\\ \color{magenta}\textrm{atau}\: \: \cos x&=\cos 180^{\circ}=\cos \pi \\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Untuk}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\tan ^{2}x-\tan x-6=0\: \: \textrm{pada}\: \: 0\leq x\leq \pi ,\\ &\textrm{maka himpunan nilai}\: \: \sin x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\left \{ \displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \}\\ \textrm{b}.&\left \{ \displaystyle \frac{3\sqrt{10}}{10},-\frac{2\sqrt{5}}{5} \right \} \\ \textrm{c}.&\left \{ -\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \} \\ \textrm{d}.&\left \{ \displaystyle \frac{\sqrt{10}}{10},\frac{\sqrt{5}}{5} \right \} \\ \textrm{e}.&\left \{ \displaystyle \frac{\sqrt{10}}{10},\frac{2\sqrt{5}}{5} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\tan ^{2}x-\tan x-6&=0\\ \left (\tan x-3 \right )\left (\tan x+2 \right )&=0\\ \tan x=3\: \: \textrm{atau}\: \: &\tan x=-2\\ \tan x=\displaystyle \frac{3}{1}\: \: \textrm{atau}\: \: &\tan x=\frac{-2}{1}\\ \sin x=\displaystyle \frac{3}{\sqrt{1^{2}+3^{2}}}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{\sqrt{1^{2}+2^{2}}}\\ \sin x=\displaystyle \frac{3}{\sqrt{10}}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{\sqrt{5}}\\ \sin x=\displaystyle \frac{3}{10}\sqrt{10}\: \: \textrm{atau}\: \: &\sin x=\frac{2}{5}\sqrt{5} \end{aligned} \end{array}$

Lanjutan 3 : Grafik Fungsi Trigonometri (Matematika Peminatan Kelas XI)

$\color{blue}\textrm{3. Grafik Fungsi Tangen}$

$\color{blue}\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \color{black}x&0&\frac{\pi }{6}&\frac{\pi }{4}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{3\pi }{4}&\frac{5\pi }{6}&\pi \\\hline \color{red}f(x)&0&\frac{1}{3}\sqrt{3}&1&\sqrt{3}&\infty &-\sqrt{3}&-1&-\frac{1}{3}\sqrt{3}&0\\\hline \color{black}x&\frac{7\pi }{6}&\frac{5\pi }{4}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{7\pi }{4}&\frac{11\pi }{6}&2\pi&\\\hline \color{red}f(x)&\frac{1}{3}\sqrt{3}&1&\sqrt{3}&\infty &-\sqrt{3}&-1&-\frac{1}{3}\sqrt{3}&0&\\\hline \end{array}$

Pada fungsi Tangen demikian juga nanti Cotangen ada beberapa nilai yang tida terdefinisi. Dalam fungsi Tangen fungsi, nilai fungsi yang tidak terdefini terdapat saat nilai  $x=\displaystyle \frac{\pi }{2}=90^{\circ}$ dan $x=\displaystyle \frac{3\pi }{2}=270^{\circ}$. Sehingga pada saat posisi nilai itu, maka dibuatlah garis putus-putus pada grafik yang dan ditampakkan berupa grais vertikal yang selanjutnya garis vertikal itu disebut sebagai asimtot.

LIMIT FUNGSI ALJABAR

 $\color{blue}\textrm{A. Pendahuluan}$

Mengingat kembali definisi limit yang telah dipelajari sebelumnya di kelas XI, yaitu limit fungsi aljabar $f(x)$ yang didefinisikan dengan:

$\begin{aligned}\underset{x\rightarrow a }{\textrm{lim}}\: f(x)=L&\: \: \textbf{adalah}\, \: \textrm{Jika}\: \: x\: \: \textrm{mendekati}\: \: a\\ &\textrm{dengan tidak sama dengan}\: \: a,\\ &\textrm{maka nilai}\: \: f(x)\: \: \textrm{mendekati}\: \: L \end{aligned}$.

Perhatikan definisi di atas istilah  $x\: \: \textrm{mendekati}\: \: a$ dituliskan dengan simbol  $(x\rightarrow a)$. Suatu nilai limit dianggap ada jika nilai $f(x)$ mendekati  $a$ dari arah kiri sama dengan nilai $f(x)$ mendekati  $a$ dari arah kanan dengan nilai yang sama misalnya $L$. Jika disimbolkan pernyataan ini menjadi berikut

$\begin{aligned}\underset{x\rightarrow \color{blue}a^{-} }{\textrm{lim}}\: f(x)=\underset{x\rightarrow \color{blue}a^{+} }{\textrm{lim}}\: f(x)=\underset{x\rightarrow \color{blue}a }{\textrm{lim}}\: f(x)=L&\: \: \ \end{aligned}$.

$\begin{aligned}\textrm{Perlu di}&\textrm{perhatikan bahwa didekati dari}\\ \bullet \: \: \textbf{kiri}\: &\textrm{disimbolkan dengan}\: \: \underset{x\rightarrow \color{blue}a^{-} }{\textrm{lim}}\: f(x),\: \: \textrm{dan}\\ \bullet \: \: \, \textbf{kan}&\textbf{an}\: \: \textrm{disimbolkan dengan}\: \: \underset{x\rightarrow \color{blue}a^{+} }{\textrm{lim}}\: f(x) \end{aligned}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai limit dari}\: \: f(x)=\displaystyle \frac{x^{2}-4}{x-2}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikanlah ketika fungsi} \: \: \displaystyle \frac{x^{2}-4}{x-2}\\ &\textrm{di sekitar}\: \: x=2\: \: \textrm{sebagaimana dalam tabel}\\ &\textrm{berikut} \end{aligned}\end{array}$.

$\begin{aligned}.\qquad&\textrm{Jadi, nilai}\: \: \underset{x\rightarrow \color{blue}2}{\textrm{lim}} \: \displaystyle \frac{x^{2}-4}{x-2}=2 \: \: \textrm{atau dapat dikatakan}\\ &\textrm{nilai} \: \: \underset{x\rightarrow \color{blue}2}{\textrm{lim}} \: \displaystyle \frac{x^{2}-4}{x-2}\: \: \textbf{ada}\\ &\textrm{meskipun nilai substitusi langsung}\: \: x=2\: \: \textrm{yaitu}\\ &f(0)=\displaystyle \frac{0^{2}-0}{0-0}=\frac{0}{0}\: \: \textrm{berupa bentuk tak}\\ &\textrm{tentu. Berikut ilustrasinya} \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Selidikilah limit fungsi berikut apakah}\\ &\textrm{memiliki harga limit}\\ &\underset{x\rightarrow \color{red}5}{\textrm{lim}} \:f(x),\: \: \textrm{untuk}\: \: f(x)=\begin{cases} x &\textrm{saat}\: \: x<5 \\ 5-x &\textrm{saat}\: \: x\geq 5 \end{cases}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikanlah ketika didekati dari kiri}\\\ &\textrm{yaitu}\: \: x\rightarrow \color{red}5^{-},\: \: \color{black}\textrm{maka}\: \: \underset{x\rightarrow \color{red}5^{-}}{\textrm{lim}} \:f(x)=x\\ & \textrm{atau}\: \: \underset{x\rightarrow \color{red}5^{-}}{\textrm{lim}} \:f(x)=5\\ &\textrm{boleh juga dituliskan dengan}\\ &\underset{x\rightarrow \color{red}5^{-}}{\textrm{lim}} \:f(x)=x=5.\: \: \textrm{Sedangkan ketika}\\ &\textrm{didekati dari arah kanan yaitu}\: \: x\rightarrow \color{red}5^{+},\\ &\textrm{maka}\: \: \underset{x\rightarrow \color{red}5^{+}}{\textrm{lim}} \:f(x)=\underset{x\rightarrow \color{red}5^{+}}{\textrm{lim}} \: (5-x)=5-5=0.\\ &\textrm{Karena nilai}\: \: \underset{x\rightarrow \color{red}5^{-}}{\textrm{lim}} \:f(x)\neq \underset{x\rightarrow \color{red}5^{+}}{\textrm{lim}} \:f(x),\: \: \textrm{maka}\\ &\textrm{nilai atau harga}\: \: \underset{x\rightarrow \color{red}5}{\textrm{lim}} \:f(x)\: \: \textbf{tidak ada}\\ &\textrm{Berikut ilustrasi gambarnya} \end{aligned}\end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Selidikilah limit fungsi berikut apakah}\\ &\textrm{memiliki harga limit}\\ &\underset{x\rightarrow \color{red}0}{\textrm{lim}} \:f(x),\: \: \textrm{untuk}\: \: f(x)=\cos x\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikanlah ketika didekati dari kiri}\\\ &\textrm{yaitu}\: \: x\rightarrow \color{red}0^{-},\: \: \color{black}\textrm{maka}\: \: \underset{x\rightarrow \color{red}0^{-}}{\textrm{lim}} \: \cos x\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline x&-0,5&-0,4&-0,3&-0,2&-0,1&0\\\hline \cos x&...&...&0,999986&0,999994&0,9999985&1\\\hline \end{array}\\ &\textrm{Sedangkan ketika}\\ &\textrm{didekati dari arah kanan yaitu}\: \: x\rightarrow \color{red}0^{+},\\ &\textrm{maka}\: \: \underset{x\rightarrow \color{red}0^{+}}{\textrm{lim}} \:\cos x\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline x&0&0,1&0,2&0,3&0,4&0,5\\\hline \cos x&1&0,9999985&0,999994&0,999986&...&...\\\hline \end{array}\\ &\textrm{Karena nilai}\: \: \underset{x\rightarrow \color{red}0^{-}}{\textrm{lim}} \:f(x)= \underset{x\rightarrow \color{red}0^{+}}{\textrm{lim}} \:f(x)=1,\: \: \textrm{maka}\\ &\textrm{nilai}\: \: \underset{x\rightarrow \color{red}0}{\textrm{lim}} \:\cos x\: \: \textbf{ ada}\\ &\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\color{blue}\textrm{B. Sifat-Sifat Limit Fungsi}$

$\begin{aligned}&\textrm{Misalkan}\: \: f\: \: \textrm{dan}\: \: g\: \: \textrm{adalah fungsi-fungsi yang}\\ &\textrm{mempunyai nilai limit di titik sekitar}\: \: x=a\\ &\textrm{atau}\: \: (x\rightarrow a)\: \: \textrm{dan}\: \: c\: \: \textrm{adalah suatu konstanta}\\ &\textrm{serta}\: \: n\: \: \textrm{adalah suatu bilangan bulat positif},\\ &\textrm{maka berlaku sifat-sifat berikut}:\\ &\begin{array}{ll}\\ 1.&\underset{x\rightarrow a }{\textrm{lim}}\: \displaystyle c=c\\ 2.&\underset{x\rightarrow a }{\textrm{lim}}\: \displaystyle x^{n}=a^{n}\\ 3.&\underset{x\rightarrow a }{\textrm{lim}}\: c.f(x)=c.\underset{x\rightarrow a }{\textrm{lim}}\: f(x)\\ 4.&\underset{x\rightarrow a}{\textrm{lim}}\: \left ( f(x)\pm g(x) \right )=\underset{x\rightarrow a }{\textrm{lim}}\: f(x)\pm \underset{x\rightarrow a }{\textrm{lim}}\: g(x)\\ 5.&\underset{x\rightarrow a }{\textrm{lim}}\: \left ( f(x)\times g(x) \right )=\underset{x\rightarrow a }{\textrm{lim}}\: f(x)\times \underset{x\rightarrow a }{\textrm{lim}}\: g(x)\\ 6.&\underset{x\rightarrow a }{\textrm{lim}}\: \displaystyle \frac{f(x)}{g(x)}=\displaystyle \frac{\underset{x\rightarrow a }{\textrm{lim}}\: f(x)}{\underset{x\rightarrow a }{\textrm{lim}}\: g(x)}\\ 7.&\underset{x\rightarrow a }{\textrm{lim}}\: \left (f(x) \right )^{n}= \left [\underset{x\rightarrow a }{\textrm{lim}}\: f(x)) \right ]^{n}\\ 8.&\underset{x\rightarrow a }{\textrm{lim}}\: \sqrt[n]{f(x)}=\sqrt[n]{\underset{x\rightarrow \infty}{\textrm{lim}}\: f(x)},\quad \textrm{dengan}\: \: \underset{x\rightarrow a }{\textrm{lim}}\: f(x)\geq 0\\ &\qquad\qquad\qquad\qquad\qquad\qquad\quad \textrm{dan}\: \: n\: \: \textrm{genap} \end{array} \end{aligned}$

Lanjutan 2 : Grafik Fungsi Trigonometri (Matematika Peminatan Kelas XI)

$\color{blue}\textrm{2. Grafik Fungsi Cosinus}$

$\color{blue}\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \color{black}x&0&\frac{\pi }{6}&\frac{\pi }{4}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{3\pi }{4}&\frac{5\pi }{6}&\pi \\\hline \color{red}f(x)&1&\frac{1}{2}\sqrt{3}&\frac{1}{2}\sqrt{2}&\frac{1}{2}&0&-\frac{1}{2}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{3}&-1\\\hline \color{black}x&\frac{7\pi }{6}&\frac{5\pi }{4}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{7\pi }{4}&\frac{11\pi }{6}&2\pi&\\\hline \color{red}f(x)&-\frac{1}{2}\sqrt{3}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}&0&\frac{1}{2}&\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{3}&1&\\\hline \end{array}$

Notasi Sigma Lanjutan Induksi Matematika (Matematika Wajib Kelas XI)

$\LARGE\color{blue}\textrm{A. Pendahuluan}$

Notasi sigma dari asalnya dari yaitu dari huruf yunani yang memiliki makna jumlah. Dalam matematika lambang notasi sigma $"\sum"$  selanjutnya akan menunjukkan penjumlahan yang teratur sehingga penulisan sebuah deret dari suatu bilangan yang berpola tertentu dapat disederhanakan lebih ringkas.

Sebagai ilustrasinya untuk deretny adalah sebagai berikut

$\begin{array}{ll}\\ \textrm{a}.\quad 1+2+3+4+5+\cdots +100\\ \textrm{b}.\quad 1+3+5+7+9+\cdots +199\\ \textrm{c}.\quad 1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+\cdots +100^{2}\end{array}$

Dari bentuk deret di atas jika dimodelkan dengan notasi sigma maka bentuknya akan menjadi lebih sederhana, yaitu:

$\color{blue}\begin{aligned}&\sum_{i=1}^{n}a_{i}=\color{magenta}a_{1}+a_{2}+a_{3}+\cdots +a_{n}\\ &\textrm{Dibaca}:\: \: "\textrm{Jumlah}\: \textrm{dari}\: a_{i}\: \textrm{untuk}\: \: i\\ &\textrm{dari 1 sampai dengan}\: \: n" \: \: \textrm{dan}\: \: a_{i}\: \textrm{adalah suku ke}-i \end{aligned}$

Sehingga contoh ilustrasi deret di atas jika dinotasikan dengan notasi sigma menjadi

$\begin{array}{ll}\\ \textrm{a}.\quad 1+2+3+4+5+\cdots +100=\sum_{i=1}^{100}i\\\\ \textrm{b}.\quad 1+3+5+7+9+\cdots +199=\sum_{i=1}^{100}(2i-1)\\\\ \textrm{c}.\quad 1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+\cdots +100^{2}=\sum_{i=1}^{100}i^{2}\end{array}$

$\LARGE\color{blue}\textrm{B. Sifat-Sifat Notasi Sigma}$

Misalkan diketahui $a_{k}$  dan $b_{k}$  adalah suku ke-k dan C adalah sebuah konstanta, maka

$\color{blue}\begin{array}{l}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}C=nC\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}C.a_{k}=C\sum_{k=1}^{n}a_{k}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}\left ( a_{k}+b_{k} \right )=\displaystyle \sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}\left ( a_{k}+b_{k} \right )^{2}=\displaystyle \sum_{k=1}^{n}a_{k}^{2}+2\sum_{k=1}^{n}a_{k}b_{k}+\sum_{k=1}^{n}b_{k}^{2}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n-1}a_{k}+a_{n}\\ &\blacklozenge \quad \displaystyle \sum_{k=1}^{n}a_{k}=\sum_{k=1}^{m}a_{k}+\sum_{k=m+1}^{n}a_{k},\quad 1<m<n \end{array}$

$\LARGE\color{yellow}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Uraikan jumlah berikut dengan lengkap}\\ &\begin{array}{lll}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{4}k&\textrm{f}.\quad \displaystyle \sum_{k=1}^{3}2^{k}\\ &\textrm{b}.\quad \displaystyle \sum_{k=1}^{4}(k-3)&\textrm{g}.\quad \displaystyle \sum_{k=1}^{3}\frac{1}{3^{k}}\\ &\textrm{c}.\quad \displaystyle \sum_{k=1}^{4}5k&\textrm{h}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+1 \right )\\ &\textrm{d}.\quad \displaystyle \sum_{k=1}^{3}(4k+2)&\textrm{i}.\quad \displaystyle \sum_{k=1}^{4}5\left ( \frac{2}{3} \right )^{k}\\ &\textrm{e}.\quad \displaystyle \sum_{k=1}^{3}(2k+3)&\textrm{j}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+2k-3 \right ) \end{array} \end{array}$

$\begin{aligned}&\textrm{Jawab}\\ &\color{blue}\begin{array}{ll}\\ 1.&\textrm{Perhatikanlah,}\\ &\begin{array}{lll}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{4}k=1+2+3+4=10\\ &\textrm{b}.\quad \displaystyle \sum_{k=1}^{4}(k-3)=(1-3)+(2-3)+(3-3)+(4-3)=-2.\\ &\textrm{c}.\quad \displaystyle \sum_{k=1}^{4}5k=5.1+5.2+5.3+5.4=50\\ &\textrm{d}.\quad \displaystyle \sum_{k=1}^{3}(4k+2)=(4.1+2)+(4.2+2)+(4.3+2)=30\\ &\textrm{e}.\quad \displaystyle \sum_{k=1}^{3}(2k+3)=(2.1 +3)+(2.2 +3)+(2.3 +3)=21 \\ &\textrm{f}.\quad \displaystyle \sum_{k=1}^{3}2^{k}=2^{1}+2^{2}+2^{3}+2^{4}=2+4+8+16=30\\ &\textrm{g}.\quad \displaystyle \sum_{k=1}^{3}\frac{1}{3^{k}}=\displaystyle \frac{1}{3^{1}}+\frac{1}{3^{2}}+\frac{1}{3^{3}}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}=\displaystyle \frac{9+3+1}{27}=\frac{13}{27}\\ &\textrm{h}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+1 \right )=\cdots +\cdots +\cdots +\cdots +\cdots \\ &\textrm{i}.\quad \displaystyle \sum_{k=1}^{4}5\left ( \frac{2}{3} \right )^{k}=\cdots +\cdots +\cdots +\cdots \\ &\textrm{j}.\quad \displaystyle \sum_{k=1}^{5}\left ( k^{2}+2k-3 \right )=\cdots +\cdots +\cdots +\cdots +\cdots \\ \end{array} \end{array} \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Nyatakanlah penjumlahan berikut dengan notasi sigma}\\ &\textrm{a}.\quad 2+4+8+16+32+64\\ &\textrm{b}.\quad 2+6+18+54+162\\ &\textrm{c}.\quad 15+24+35+48\\ &\textrm{d}.\quad \displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}\\ &\textrm{e}.\quad ab+a^{2}b^{2}+a^{3}b^{3}+a^{4}b^{4} \end{array}$

$\color{blue}\begin{aligned}&\color{black}\textrm{Jawab}\\ &(\textrm{a})\quad 2+4+8+16+32+64=\displaystyle \sum_{k=1}^{6}2^{k}\\ &(\textrm{b})\quad 2+6+18+54+162=\displaystyle \sum_{k=1}^{5}2.3^{k-1}\\ &(\textrm{c})\quad 15+24+35+48=\displaystyle \sum_{k=1}^{4}\left ( k^{2}+6k+8 \right )\\ &(\textrm{d})\quad \displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}=\displaystyle \sum_{k=1}^{5}\frac{2^{k}}{(2k+1)}\\ &(\textrm{e})\quad ab+a^{2}b^{2}+a^{3}b^{3}+a^{4}b^{4}=\displaystyle \sum_{k=1}^{4}(ab)^{k}\\ \end{aligned}$

$\begin{array}{ll}\\ 3.&\textrm{Dengan menggunakan kaidah notasi sigma},\\ &\textrm{tunjukkan bahwa}\\ &\textrm{a}.\quad \displaystyle \sum_{k=1}^{6}(2k+3)=2\sum_{k=1}^{6}k+18\\ &\textrm{b}.\quad \displaystyle \sum_{k=3}^{8}(k+3)=\sum_{k=1}^{6}k+30\\ &\textrm{c}.\quad \displaystyle \sum_{k=2}^{5}\left ( 2k^{2}+3k+3 \right )=2\sum_{k=1}^{4}k^{2}+7\sum_{k=1}^{4}k+32\\ &\textrm{d}.\quad \displaystyle \sum_{k=0}^{5}k^{2}=\sum_{k=1}^{6}k^{2}-2\sum_{k=1}^{6}k+6\\ &\textrm{e}.\quad \displaystyle \sum_{k=3}^{6}\left ( k^{2}+2k-3 \right )=\sum_{k=1}^{6}k^{2}+6\sum_{k=1}^{4}+20 \end{array}$

$\color{blue}\begin{aligned}&\color{black}\textrm{Jawab}\\ &(\textrm{a})\quad \displaystyle \sum_{k=1}^{6}(2k+3)=\sum_{k=1}^{6}2k+\sum_{k=1}^{6}3\\ &\: \qquad =\sum_{k=1}^{6}2k+6.3=2\sum_{k=1}^{6}k+18\\ &(\textrm{b})\quad \displaystyle \sum_{k=3}^{8}(k+3)=\sum_{k=3-2}^{8-2}\left ( (k+2)+3 \right )\\ &\: \qquad =\sum_{k=1}^{6}(k+5)=\sum_{k=1}^{6}k+\sum_{k=1}^{6}5=\sum_{k=1}^{6}k+6.5\\ &\: \qquad=\sum_{k=1}^{6}k+30\\ &(\textrm{c})\quad \displaystyle \sum_{k=2}^{5}\left ( 2k^{2}+3k+3 \right )\\ &\: \qquad=\sum_{k=2-1}^{5-1}\left ( 2(k+1)^{2}+3(k+1)+3 \right )=\cdots\\ &(\textrm{d})\quad \displaystyle\sum_{k=0}^{5}k^{2}=\sum_{k=0+1}^{5+1}(k-1)^{2}=\cdots\\ &(\textrm{e})\quad \displaystyle \sum_{k=3}^{6}\left ( k^{2}+2k-3 \right )=\cdots\\ \end{aligned}$

$\LARGE\color{green}\fbox{LATIHAN SOAL}$

$\begin{array}{ll}\\ .&\textrm{Buktikanlah bahwa}\\ &\textrm{a}.\quad \displaystyle \sum_{k=6}^{12}k^{2}=\sum_{k=1}^{7}k^{2}+10\sum_{k=1}^{7}k+175\\ &\textrm{b}.\quad \displaystyle \sum_{k=1}^{n}(3k-1)^{2}=9\sum_{k=1}^{n}k^{2}-6\sum_{k=1}^{n}k+n\\ &\textrm{c}.\quad \displaystyle \sum_{k=m}^{n}a_{k}=\sum_{k=m+p}^{n+p}a_{k-p}\\ &\textrm{d}.\quad \displaystyle \sum_{i=m}^{n}a_{i}=\sum_{i=1}^{n}a_{i}-\sum_{i=1}^{m-1}a_{i}\\ &\textrm{e}.\quad \displaystyle \sum_{k=1}^{n}a_{k}=\sum_{k=0}^{n-1}a_{k+1}=\sum_{k=2}^{n+2}a_{k-1}\\ &\textrm{f}.\quad \displaystyle \sum_{k=1}^{n-5}a_{k}=\sum_{k=1}^{n}a_{k}-\sum_{k=(n-5)+1}^{n}a_{k} \end{array}$

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono, Kurnianingsih, S. 2005. Matematika untuk SMA dan MA Kelas XII Program Ilmu Alam. Jakarta: Gelora Aksara Pratama.
2. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: ESIS.

Lanjutan 1 : Grafik Fungsi Trigonometri (Matematika Peminatan Kelas XI)

$\LARGE\textrm{B. Grafik Fungsi Trigonometri}$

Dalam melukis grafik fungsi trigonometri nantinya yang ditampilkan adalah nilai-nilai untuk sudut istimewa saja. Dan selanjutnya besar sudutnya disajikan dalam derajat dan atau radian.

Selanjutnya perhatikanlah uraian berikut

$\color{blue}\textrm{1. Grafik Fungsi Sinus}$

$\color{blue}\begin{array}{|c|c|c|c|c|c|c|c|c|c| }\hline \color{magenta}x&0&\frac{\pi }{6}&\frac{\pi }{4}&\frac{\pi }{3}&\frac{\pi }{2}&\frac{2\pi }{3}&\frac{3\pi }{4}&\frac{5\pi }{6}&\pi \\\hline \color{red}f(x)&0&\frac{1}{2}&\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{3}&1&\frac{1}{2}\sqrt{3}&\frac{1}{2}\sqrt{2}&\frac{1}{2}&0\\\hline \color{magenta}x&\frac{7\pi }{6}&\frac{5\pi }{4}&\frac{4\pi }{3}&\frac{3\pi }{2}&\frac{5\pi }{3}&\frac{7\pi }{4}&\frac{11\pi }{6}&2\pi & \\\hline \color{red}f(x)&-\frac{1}{2}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{3}&-1&-\frac{1}{2}\sqrt{3}&-\frac{1}{2}\sqrt{2}&-\frac{1}{2}&0&\\\hline \end{array}$