Fungsi Trigonometri dan Grafiknya

F. Fungsi Trigonometri dan Grafiknya

F. 1 Fungsi Trigonometri

Perhatikan ilustrasi berikut ini

Dengan

$\begin{array}{c|c}\\ \begin{aligned}&\textrm{Dalil/rumus Pythagoras}\\ &a^{2}+b^{2} =c^{2}\\ &\color{red}\textrm{atau}\\ &c=\sqrt{a^{2}+b^{2}} \end{aligned}&\begin{aligned}&\sin \angle ACB=\displaystyle \frac{a}{c}\\ &\cos \angle ACB=\displaystyle \frac{b}{c}\\ &\tan \angle ACB=\displaystyle \frac{a}{b}=\displaystyle \frac{\sin \angle ACB}{\cos \angle ACB} \end{aligned} \end{array}$.

Adapun gambar dari fungsi atau pemetaan trigonometrinya dari setiap sudut $\alpha $ ke salah satu nilai dari $\sin \alpha$ , $\cos \alpha$, maupun $\tan \alpha$  dalam wilayah bilangan real adalah  sebagaimana ilustrasi berikut:

$\begin{aligned}&\textrm{Misalkan}\: \: A\: \: \textrm{dan}\: \: B\: \: \textrm{dua himpunan}\\ &\textrm{Suatu relasi}\: \: F\subseteq A\times B\: \: \textrm{disebut fungsi jika}\\ &\textrm{setiap}\: a\in A,\: \textrm{maka hanya ada tepat satu}\: \: b\in B\\ &\textrm{dengan}\: \: (a,b)\in F.\\ &\textrm{Fungsi}\: \: F\: \: \textrm{disebut dengan fungsi dari}\: \: A\: \: \textrm{ke}\: \: B\\ &\textrm{Selanjutnya}\: \: A\: \: \textrm{dinamakan}\: \: \textbf{Domain}\: \: \textrm{atau}\\ &\textrm{daerah asal atau juga daerah definisi fungsi}\\ &\textrm{dan}\: \: B\: \: \textrm{disebut}\: \: \textbf{Kodomain}\\ &\textrm{Himpunan}\: \: \left \{ b\in B|(a,b)\in F \right \}\: \textrm{selanjutnya disebut}\\ &\textrm{sebagai}\: \: \textbf{nilai fungsi}\\ &\textrm{Jika}\: \: (a,b)\in F,\: \: \textrm{maka dapat tuliskan dengan}\\ &b=F(a),\: \: \textrm{yaitu nilai fungsi}\: \: F\: \: \textrm{di titik}\: \: a\\\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|c|l|c|}\hline \textrm{No}&\: \: \quad\color{red}\textrm{Gambar}&\color{red}\textrm{Fungsi}\: \left ( f:\mathbb{R}\Rightarrow \mathbb{R} \right )\\\hline 1&\textrm{Fungsi Sinus}&\begin{aligned}&f:\alpha \Rightarrow \sin \alpha  \end{aligned}\\\hline 2&\textrm{Fungsi Cosinus}&\begin{aligned}&f:\alpha \Rightarrow \cos \alpha  \end{aligned}\\\hline 3&\textrm{Fungsi Tangen}&\begin{aligned}&f:\alpha \Rightarrow \tan \alpha  \end{aligned}\\\hline \end{array}  \end{aligned}$.

$\begin{aligned}&\textrm{Jangan lupa, sebagai pengingat kita untuk}\\ &\textrm{nilai sudut istimewanya adalah sebagai berikut}:\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline \alpha &0^{\circ}&30&45^{\circ}&60^{\circ}&90^{\circ}&180^{\circ}\\\hline \sin \alpha &0&\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0\\\hline \cos \alpha &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1\\\hline \tan \alpha &0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&\color{red}\textrm{TD}&0\\\hline \end{array} \end{aligned}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Jika diketahui}\: \: f(x)=\sin 2x,\: \: \textrm{tentukan nilai}\\ &\textrm{a}.\quad f(60^{\circ})\\ &\textrm{b}.\quad f\left ( \displaystyle \frac{1}{3}\pi  \right )\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&f(60^{\circ})=\sin 2\left ( 60^{\circ} \right )=\sin 120^{\circ}\\ &\: \: \: \qquad =\sin \left ( 180^{\circ}-60^{\circ} \right )=\sin 60^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.\quad&f\left ( \displaystyle \frac{1}{3}\pi  \right )=\sin 2\left ( \displaystyle \frac{1}{3}\pi  \right )=\sin \left ( \displaystyle \frac{2}{3}\pi  \right )\\ &\: \: \qquad\quad =\sin \left ( \displaystyle \frac{2}{3}(180^{\circ}) \right )=\sin 120^{\circ}=\displaystyle \frac{1}{2}\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika diketahui}\: \: f(x)=\sin x,\: \: \textrm{tentukan harga}\\ &x\: \: \textrm{jika diketahui}\: (\: x\: \: \textrm{sudut lancip})\\ &\textrm{a}.\quad f(x)=\displaystyle \frac{1}{2}\\ &\textrm{b}.\quad f(x)=\displaystyle \frac{1}{4}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&f(x)=\displaystyle \frac{1}{2}=\sin x\Rightarrow x=30^{\circ}\\ \textrm{b}.\quad&f(x)=\displaystyle \frac{1}{4}=\sin x\Rightarrow x=\color{red}\sin ^{-1}\left ( \displaystyle \frac{1}{4} \right )\\ &\textrm{hal ini dikarenakan}\: \: \displaystyle \frac{1}{4}\: \: \textrm{bukanlah}\\ &\textrm{nilai dari salah satu sudut istimewa}\\ &\textrm{untuk fungsi}\: \: \textbf{sinus} \end{aligned} \end{array}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Jika diketahui}\: \: f(x)=\cos 2x,\: \: \textrm{tentukan nilai}\\  &\textrm{a}.\quad f(60^{\circ})\\ &\textrm{b}.\quad f\left ( \displaystyle \frac{1}{3}\pi  \right )\end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika diketahui}\: \: f(x)=\tan x,\: \: \textrm{tentukan nilai}\\  &\textrm{a}.\quad f(60^{\circ})\\ &\textrm{b}.\quad f\left ( \displaystyle \frac{1}{3}\pi  \right )\end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika diketahui}\: \: f(x)=\cos x,\: \: \textrm{tentukan harga}\\ &x\: \: \textrm{jika diketahui}\: (\: x\: \: \textrm{sudut lancip})\\ &\textrm{a}.\quad f(x)=\displaystyle \frac{1}{2}\\ &\textrm{b}.\quad f(x)=\displaystyle \frac{1}{4} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika diketahui}\: \: f(x)=\tan x,\: \: \textrm{tentukan harga}\\ &x\: \: \textrm{jika diketahui}\: (\: x\: \: \textrm{sudut lancip})\\ &\textrm{a}.\quad f(x)=\displaystyle \frac{1}{3}\sqrt{3}\\ &\textrm{b}.\quad f(x)=\displaystyle \frac{1}{6}\sqrt{3} \end{array}$

DAFTAR PUSTAKA

1. Budhi, W.S. 2014. Bupena Matematika SMA/MA Kelas X Kelompok Wajib. Jakarta: ERLANGGA.
2. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Lanjutan Identitas Trigonometri

 $\begin{aligned}&\textrm{E. 3 Menentukan Nilai Perbandingan Trigonometri}\\ &\quad\textrm{pada Segitiga Siku-Siku} \end{aligned}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \tan \theta =\displaystyle \frac{a}{x} \\ &\textrm{Tentukanlah nilai}\: \: \displaystyle \frac{x}{\sqrt{a^{2}+x^{2}}}\\\\ &\color{blue}\textrm{Jawab}:\\ &\textrm{Perhatikanlah gambar segitiga AOX berikut} \end{array}$

$.\: \qquad\begin{aligned}&\textrm{Dengan rumus Pythagoras dapatr ditentukan}\\ &\textrm{panjang ruas}\: \: \textrm{AX, yaitu}:\\ &AO^{2}+OX^{2} =AX^{2}\\ &\textrm{atau}\\ &AX^{2}=AO^{2}+OX^{2} \\ &AX=\sqrt{AO^{2}+OX^{2}}\\ &\qquad =\sqrt{x^{2}+a^{2}},\\ &\textrm{maka}\\ &\bullet \quad \sin \theta =\displaystyle \frac{a}{\sqrt{x^{2}+a^{2}}}\\ &\bullet \quad \cos \theta =\displaystyle \frac{x}{\sqrt{x^{2}+a^{2}}} \\ &\textrm{Jadi, nilai}\: \: \displaystyle \frac{x}{\sqrt{x^{2}+a^{2}}}=\color{red}\cos \theta \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: \sin \beta +\cos \beta =\displaystyle \frac{6}{5},\: \textrm{tentukanlah}\\ &\textrm{a}.\quad \sin \beta \cos \beta \\ &\textrm{b}.\quad \sin ^{3}\beta +\cos ^{3}\beta \\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\sin \beta +\cos \beta=\displaystyle \frac{6}{5}\\ &\color{red}\textrm{saat masing-masing ruas dikuadratkan,}\\ &\textrm{maka}\\ &\left (\sin \beta +\cos \beta \right )^{2}=\left (\displaystyle \frac{6}{5} \right )^{2}\\ &\sin ^{2}\beta +2\sin \beta \cos \beta +\cos ^{2}\beta =\displaystyle \frac{36}{25}\\ &\sin ^{2}\beta +\cos ^{2}\beta +2\sin \beta \cos \beta=\displaystyle \frac{36}{25}\\ &1+2\sin \beta \cos \beta=\displaystyle \frac{36}{25}\\ &2\sin \beta \cos \beta=\displaystyle \frac{36}{25}-1\\ &2\sin \beta \cos \beta=\displaystyle \frac{36-25}{25}=\frac{11}{25}\\ &\sin \beta \cos \beta=\color{blue}\displaystyle \frac{11}{50} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\sin ^{3}\beta +\cos ^{3}\beta \\ &=\left ( \sin \beta +\cos \beta \right )\left ( \sin ^{2}\beta +\cos ^{2}\beta -\sin \beta \cos \beta \right )\\ &=\left ( \sin \beta +\cos \beta \right )\left ( 1 -\sin \beta \cos \beta \right )\\ &=\left ( \displaystyle \frac{6}{5} \right ).\left ( 1-\displaystyle \frac{11}{50} \right )\\ &=\left ( \displaystyle \frac{6}{5} \right ).\left ( \displaystyle \frac{50-11}{50} \right )\\ &=\left ( \displaystyle \frac{6}{5} \right ).\left ( \displaystyle \frac{39}{50} \right )\\ &=\displaystyle \color{blue} \frac{3\times 39}{5\times 25}=\frac{117}{125} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: \tan \alpha =\displaystyle \frac{1}{\sqrt{7}},\: \textrm{tentukanlah}\\ &\left ( \displaystyle \frac{\csc ^{2}\alpha -\sec ^{2}\alpha }{\csc ^{2}\alpha +\sec ^{2}\alpha} \right ) \\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui bahwa}:\\ \tan \alpha &=\displaystyle \frac{1}{\sqrt{7}},\: \: \color{red}\textrm{dan ingat juga bahwa}\\ \sec ^{2}\alpha &=\tan ^{2}\alpha +1=\left ( \displaystyle \frac{1}{\sqrt{7}} \right )^{2}+1=\frac{1}{7}+1=\frac{8}{7}\\ \color{red}\textrm{Demik}&\color{red}\textrm{ian juga},\: \color{black}\cot \alpha =\displaystyle \frac{1}{\tan \alpha } =\displaystyle \frac{1}{\left ( \frac{1}{\sqrt{7}} \right )}=\sqrt{7},\\ \textrm{maka},&\: \: \csc ^{2}\alpha =\cot ^{2}\alpha +1=\left ( \sqrt{7} \right )^{2}+1=7+1=8\\ \textrm{Selanj}&\textrm{utnya}\\ &\left ( \displaystyle \frac{\csc ^{2}\alpha -\sec ^{2}\alpha }{\csc ^{2}\alpha +\sec ^{2}\alpha} \right )=\left ( \displaystyle \frac{8-\displaystyle \frac{8}{7}}{8+\displaystyle \frac{8}{7}} \right )\\ &=\displaystyle \frac{\displaystyle \frac{56-8}{7}}{\displaystyle \frac{56+8}{7}} \\ &=\displaystyle \frac{48}{64}\\ &=\color{blue}\displaystyle \frac{3}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: \beta\: \: \textrm{sudut lancip dan}\: \: \cos \beta =\displaystyle \frac{3}{5},\\ &\textrm{tentukan nilai dari}\: \: \displaystyle \frac{\sin \beta \tan \beta -1}{2\tan ^{2}\beta }\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\\ \cos \beta &=\displaystyle \frac{3}{5}\Rightarrow \sin ^{2}\beta +\cos ^{2}\beta =1\\ \sin ^{2}\beta &+\cos ^{2}\beta =1\\ \sin \beta &=\sqrt{1-\cos ^{2}\beta}=\sqrt{1-\left ( \displaystyle \frac{3}{5} \right )^{2}}\\ &=\sqrt{1-\displaystyle \frac{9}{25}}=\sqrt{\displaystyle \frac{16}{25}}=\displaystyle \frac{4}{5}\\ \textrm{Sehingga}\: &\tan \beta =\displaystyle \frac{\sin \beta }{\cos \beta }=\frac{\displaystyle \frac{4}{5}}{\displaystyle \frac{3}{5}}=\frac{4}{3}\\ &\color{red}\displaystyle \frac{\sin \beta \tan \beta -1}{2\tan ^{2}\beta }\color{black}=\displaystyle \frac{\displaystyle \frac{4}{5}\times \frac{4}{3}-1}{2\left ( \displaystyle \frac{4}{3} \right )^{2}}\\ &\: \, \quad\quad\quad\quad\quad\quad =\displaystyle \frac{\displaystyle \frac{16}{15}-1}{\displaystyle \frac{32}{9}}=\displaystyle \frac{\displaystyle \frac{1}{15}}{\displaystyle \frac{32}{9}}=\displaystyle \frac{9}{32\times 15}\\ &\: \, \quad\quad\quad\quad\quad\quad =\displaystyle \frac{3}{32\times 5}\\ &\: \, \quad\quad\quad\quad\quad\quad =\color{blue}\displaystyle \frac{3}{160} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

Identitas Trigonometri

 $\Large\textrm{E  Identitas Trigonometri}$.

E. 1  Nilai Trigonometri Sudut

$\textrm{a.  Perbandingan Trigonometri dalam Segitiga Siku-Siku}$.

Perhatikanlah ilustrasi sebuah segitiga siku-siku sama kaki berikut

Diketahui pula bahwa :

$\begin{matrix} \bullet \quad \sin 45^{\circ}=\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}\\ \bullet \quad \cos 45^{\circ}=\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}\\ \bullet \quad \tan 45^{\circ}=1 \qquad\qquad\: \: \end{matrix}$.

$\begin{matrix} \bullet \quad \csc 45^{\circ}=\displaystyle \sqrt{2}\\ \bullet \quad \sec 45^{\circ}=\displaystyle \sqrt{2}\\ \bullet \quad \cot 45^{\circ}=1 \: \: \, \end{matrix}$.

Berikut ilustrasi segitiga dengan sudut istimewa yang lain yaitu $30^{\circ}$ dan  $60^{\circ}$.

$\begin{array}{|c|c|}\hline \begin{matrix} \bullet \quad \color{purple}\sin 30^{\circ}=\displaystyle \frac{1}{2}\\ \bullet \quad \cos 30^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \bullet \quad \tan 30^{\circ}=\displaystyle \frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3} \: \: \,\\ \bullet \quad \color{blue}\sin 60^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \bullet \quad \cos 60^{\circ}=\displaystyle \frac{1}{2}\\ \bullet \quad \tan 30^{\circ}=\displaystyle \sqrt{3}\\ \end{matrix} &\begin{matrix} \bullet \quad \csc 30^{\circ}=\displaystyle 2\\ \bullet \quad \sec 30^{\circ}=\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}\\ \bullet \quad \cot 30^{\circ}=\displaystyle \sqrt{3} \: \: \,\\ \bullet \quad \color{red}\csc 60^{\circ}=\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}\\ \bullet \quad \sec 60^{\circ}=\displaystyle 2\\ \bullet \quad \color{purple}\cot 30^{\circ}=\displaystyle \frac{1}{3}\sqrt{3}\\ \end{matrix} \\\hline \end{array}$

Perhatikan segitiga ABC siku-siku di C berikut

Perhatikanlah segitiga OAB berikut

$\begin{aligned}\textrm{a}.\quad&\color{purple}\sin \alpha =\displaystyle \frac{y}{r}\\ \textrm{b}.\quad&\cos \alpha =\displaystyle \frac{x}{r}\\ \textrm{c}.\quad&\color{blue}\tan \alpha =\displaystyle \frac{y}{x}\\ \textrm{d}.\quad&\csc \alpha =\displaystyle \frac{r}{y}\\ \textrm{e}.\quad&\sec \alpha =\displaystyle \frac{r}{x}\\ \textrm{f}.\quad&\color{red}\cot \alpha =\displaystyle \frac{x}{y}\\ \end{aligned}$.

E. 2  Identitas Trigonometri Dasar

$\textrm{a.  Dalil Pythagoras Segitiga Siku-Siku}$.

$\begin{array}{c|c}\\ \begin{aligned}&\textrm{Dalil/rumus Pythagoras}\\ &a^{2}+b^{2} =c^{2}\\ &\color{red}\textrm{atau}\\ &c=\sqrt{a^{2}+b^{2}} \end{aligned}&\begin{aligned}&\sin \angle ACB=\displaystyle \frac{a}{c}\\ &\cos \angle ACB=\displaystyle \frac{b}{c}\\ &\tan \angle ACB=\displaystyle \frac{a}{b}=\displaystyle \frac{\sin \angle ACB}{\cos \angle ACB}\\ &\csc \angle ACB=\displaystyle \frac{c}{a}\\ &\sec \angle ACB=\displaystyle \frac{c}{b}\\ &\cot \angle ACB=\displaystyle \frac{b}{a}=\displaystyle \frac{\cos \angle ACB}{\sin \angle ACB} \end{aligned} \end{array}$

$\color{purple}\textrm{b. Identitas trigonometri pada segitiga siku-siku}$.

$\begin{aligned}&\textrm{Dalil/rumus Pythagoras}\\ &a^{2}+b^{2} =c^{2}\\ &\textrm{Perhatikan lagi gambar di poin c di atas}\\ &\begin{array}{|c|l|}\hline 1.&\textrm{Rumus saat dibagi dengan}\: \: c^{2}\\ &\displaystyle \frac{a^{2}}{c^{2}}+\displaystyle \frac{b^{2}}{c^{2}}=\displaystyle \frac{c^{2}}{c^{2}}\Leftrightarrow \color{red}\displaystyle \frac{a^{2}}{c^{2}}+\displaystyle \frac{b^{2}}{c^{2}}=1\\&\\ &\textrm{menjadi}\: \: \: \left ( \displaystyle \frac{a}{c} \right )^{2}+\left ( \displaystyle \frac{b}{c} \right )^{2}=1\\ &\color{blue}\sin ^{2}\angle ACB+\cos ^{2}\angle ACB=1\\\hline 2&\textrm{Rumus saat dibagi dengan}\: \: b^{2}\\ &\displaystyle \frac{a^{2}}{b^{2}}+\displaystyle \frac{b^{2}}{b^{2}}=\displaystyle \frac{c^{2}}{b^{2}}\Leftrightarrow \color{red}\displaystyle \frac{a^{2}}{b^{2}}+1=\displaystyle \frac{c^{2}}{b^{2}}\\&\\ &\textrm{menjadi}\: \: \: \left ( \displaystyle \frac{a}{b} \right )^{2}+1=\left ( \displaystyle \frac{c}{b} \right )^{2}\\ &\color{blue}\tan ^{2}\angle ACB+1=\sec ^{2}\angle ACB\\\hline 3&\textrm{Rumus saat dibagi dengan}\: \: a^{2}\\ &\displaystyle \frac{a^{2}}{a^{2}}+\displaystyle \frac{b^{2}}{a^{2}}=\displaystyle \frac{c^{2}}{a^{2}}\Leftrightarrow \color{red}1+\displaystyle \frac{b^{2}}{a^{2}}=\displaystyle \frac{c^{2}}{a^{2}}\\&\\ &\textrm{menjadi}\: \: \: 1+\left ( \displaystyle \frac{b}{a} \right )^{2}=\left ( \displaystyle \frac{c}{a} \right )^{2}\\ &\color{blue}1+\cot ^{2}\angle ACB=\csc ^{2}\angle ACB\\\hline \end{array} \end{aligned}$

$\color{purple}\textrm{c. Tabel trigonometri nilai sudut istimewa}$.

$\begin{array}{|c|c|c|c|c|c|c|}\hline \alpha &0^{\circ}&30&45^{\circ}&60^{\circ}&90^{\circ}&180^{\circ}\\\hline \sin \alpha &0&\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0\\\hline \cos \alpha &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1\\\hline \tan \alpha &0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&\color{red}\textrm{TD}&0\\\hline \end{array}$.

$\begin{aligned}&\color{blue}\textrm{d. Macam-Macam Identitas Trigonometri Dasar}\\ &1.\quad \csc \alpha =\displaystyle \frac{1}{\sin \alpha }\qquad\qquad 5.\quad \tan \alpha =\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ &2.\quad \sec \alpha =\displaystyle \frac{1}{\cos \alpha }\qquad\qquad 6.\quad \tan^{2} \alpha +1=\sec ^{2}\alpha \\ &3.\quad \cot \alpha =\displaystyle \frac{1}{\tan \alpha }\qquad\qquad 7.\quad \cot^{2} \alpha +1=\csc ^{2}\alpha \\ &4.\quad \cot \alpha =\displaystyle \frac{\cos \alpha }{\sin \alpha }\qquad\qquad 8.\quad \sin^{2} \alpha +\cos ^{2}=1\\ \end{aligned}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad\tan \alpha =\displaystyle \frac{\sin \alpha \cos \alpha }{1-\sin ^{2}\alpha }\\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\tan \alpha &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\times \frac{\cos \alpha }{\cos \alpha }\\ &=\displaystyle \frac{\sin \alpha \cos \alpha }{\cos^{2} \alpha }\\ &=\displaystyle \frac{\sin \alpha \cos \alpha }{1-\sin^{2} \alpha }\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{1}{\sqrt{\tan ^{2}\beta }}\times \sin \beta =\cos \beta \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{1}{\sqrt{\tan ^{2}\beta }}\times \sin \beta &=\displaystyle \frac{1}{\tan \beta }\times \sin \beta \\ &=\displaystyle \frac{\cos \beta }{\sin \beta }\times \sin \beta \\ &=\cos \beta \qquad\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{\cos ^{2}\gamma }{1-\sin \gamma } =1+\sin \gamma \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{\cos ^{2}\gamma }{1-\sin \gamma } &=\displaystyle \frac{1-\sin^{2} \gamma }{1-\sin \gamma }\\ &=\displaystyle \frac{(1-\sin \gamma )(1+\sin \gamma )}{1-\sin \gamma }\\ &=1+\sin \gamma \qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta } =\cos ^{2}\theta -\sin ^{2}\theta \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta } &=\displaystyle \frac{1-\tan ^{2}\theta }{\sec ^{2}\theta }=\displaystyle \frac{1-\displaystyle \frac{\sin ^{2}\theta }{\cos ^{2}\theta }}{\displaystyle \frac{1}{\cos ^{2}\theta }}\\ &=\displaystyle \frac{\displaystyle \frac{\cos ^{2}\theta -\sin ^{2}\theta }{\cos ^{2}\theta }}{\displaystyle \frac{1}{\cos ^{2}\theta }} \\ &=\cos ^{2}\theta -\sin ^{2}\theta\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \cos ^{4}\alpha -\sin ^{4}\alpha =1-2\sin ^{2}\alpha \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\cos ^{4}\alpha -\sin ^{4}\alpha &=\left ( \cos ^{2}\alpha \right )^{2} -\left (\sin ^{2}\alpha \right )^{2}\\ &=\left (\cos ^{2}\alpha -\sin ^{2}\alpha \right )\left ( \cos ^{2}\alpha +\sin ^{2}\alpha \right )\\ &=\left (\cos ^{2}\alpha -\sin ^{2}\alpha \right )\times 1\\ &=\cos ^{2}\alpha -\sin ^{2}\alpha \\ &=\left ( 1-\sin ^{2}\alpha \right )-\sin ^{2}\alpha \\ &=1-2\sin ^{2}\alpha \qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{\sin \beta \sec \beta }{\sin ^{2}\beta -\tan ^{2}\beta } =-\displaystyle \frac{\cos \beta }{\sin ^{3}\beta } \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{\sin \beta \sec \beta }{\sin ^{2}\beta -\tan ^{2}\beta }&=\displaystyle \frac{\sin \beta \left ( \displaystyle \frac{1}{\cos \beta } \right ) }{\sin ^{2}\beta -\displaystyle \frac{\sin ^{2}\beta }{\cos ^{2}\beta } } \\ &=\displaystyle \frac{\left ( \displaystyle \frac{\sin \beta }{\cos \beta } \right )}{\sin ^{2}\beta \left ( 1-\displaystyle \frac{1}{\cos ^{2}\beta } \right )}\times \frac{\cos ^{2}\beta }{\cos ^{2}\beta }\\ &=\displaystyle \frac{\sin \beta \cos \beta }{\sin ^{2}\beta \left ( \cos ^{2}\beta -1 \right )}\\ &=\displaystyle \frac{\cos \beta }{\sin \beta \left ( -\sin ^{2}\beta \right )}\\ &=-\displaystyle \frac{\cos \beta }{\sin ^{3}\beta } \qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tunjukkan bahwa} \\ &\displaystyle \frac{1}{1-\displaystyle \frac{1}{1+\displaystyle \frac{1}{\tan ^{2}x}}}=\sec ^{2}x \\\\&\color{blue}\textrm{Bukti}:\\&\begin{aligned}&\displaystyle \frac{1}{1-\displaystyle \frac{1}{1+\displaystyle \frac{1}{\tan ^{2}x}}}=\displaystyle \frac{1}{1-\displaystyle \frac{1}{1+\cot ^{2}x}}\\ &=\displaystyle \frac{1}{1-\displaystyle \frac{1}{\csc ^{2}x}}=\displaystyle \frac{1}{1-\sin ^{2}x}=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x\qquad \blacksquare  \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Noormandiri, B. K. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
2. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
3. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Contoh 6 Soal dan Pembahasan Materi Hubungan Dua Lingkaran

$\begin{array}{ll}\\ 26.&\textrm{Diketahui lingkaran-lingkaran}\\ & x^{2}+y^{2}-2x+3y+k=0\: \: \textrm{dan}\: \\  &x^{2}+y^{2}+8x-6y-7=0\: \: \textrm{saling}\\ &\textrm{berpotongan ortogonal saat}\: \: k=\: ....\\ &\textrm{a}.\quad \color{red}-10\\ &\textrm{b}.\quad -3\\ &\textrm{c}.\quad 1\\ &\textrm{d}.\quad 5\\ &\textrm{e}.\quad 8\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}-2x+3y+k=0&\begin{cases} P_{1} &=\left ( 1,-\displaystyle \frac{3}{2} \right ) \\  r_{1} & = \sqrt{\displaystyle \frac{13-4k}{4}} \end{cases}\\\hline \begin{aligned}L_{2}&\equiv x^{2}+y^{2}+8x-6y-7=0  \end{aligned}&\begin{cases} P_{2} &=\left ( -4,3 \right ) \\  r_{2} & = \sqrt{32} \end{cases}\\\hline \end{array} \\ &\textrm{Syarat dua lingkaran berpotongan ortogonal}\\ &\begin{aligned}&\left (P_{1}P_{2}  \right )^{2}=r_{1}^{2}+r_{2}^{2}\\ &\Leftrightarrow \left ( 1+4 \right )^{2}+\left ( -\displaystyle \frac{3}{2}-3 \right )^{2}=\left ( \sqrt{\displaystyle \frac{13-4k}{4}} \right )^{2}+\sqrt{32}^{2}\\ &\Leftrightarrow \: 25+\displaystyle \frac{81}{4}=\displaystyle \frac{13-4k}{4}+32\\ &\Leftrightarrow \: 100+81=13-4k+128\\ &\Leftrightarrow \: k=-10 \end{aligned} \\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut}  \end{array}$.

$\begin{array}{ll}\\ 27.&\textrm{Persamaan lingkaran yang berpotongan}\\ &\textrm{lingkaran lain}\: \:  x^{2}+y^{2}+2x+y-11=0\\ &\textrm{secara tegak lurus dan melalui}\: \: (4,3)\: \: \textrm{serta}\\ &\textrm{pusatnya pada}\: \: 9x+4y=37\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}x^{2}+y^{2}-10x+4y+3=0\\ &\textrm{b}.\quad x^{2}+y^{2}-8x+10y+6=0\\ &\textrm{c}.\quad x^{2}+y^{2}+4x-8y+7=0\\ &\textrm{d}.\quad x^{2}+y^{2}+6x+y+5=0\\ &\textrm{e}.\quad x^{2}+y^{2}+12x+6y+5=0\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}+2x+y-11=0&\begin{cases} P_{1} &=\left ( -1,-\displaystyle \frac{1}{2} \right ) \\  r_{1} & = \sqrt{\displaystyle \frac{49}{4}}=\displaystyle \frac{7}{2} \end{cases}\\\hline \begin{aligned}L_{2}&\equiv (x-a)^{2}+(y-b)^{2}=r^{2}  \end{aligned}&\begin{cases} P_{2} &=\left ( a,b \right ) \\  r_{2} & = r \end{cases}\\\hline \end{array}\\ &\textrm{Karena berpotongan tegak lurus, maka}\\ &\begin{aligned}&\left (P_{1}P_{2}  \right )^{2}=r_{1}^{2}+r_{2}^{2}\\ &\Leftrightarrow \left ( -1-a \right )^{2}+\left ( -\displaystyle \frac{1}{2}-b \right )^{2}=\displaystyle \frac{49}{4}+r^{2}\\ &\Leftrightarrow a^{2}+2a+1+b^{2}+b+\displaystyle \frac{1}{4}=\displaystyle \frac{49}{4}+r^{2}\\ &\Leftrightarrow \color{blue}a^{2}+b^{2}+2a+b+\displaystyle \frac{5}{4}=\displaystyle \frac{49}{4}+r^{2}\\ &\Leftrightarrow a^{2}+b^{2}+2a+b-11=r^{2}\: .......(1)\\ \end{aligned} \\ &\textrm{Selanjutnya}\\ &\begin{aligned}&\textrm{Lingkaran}\: \: L_{2}\: \: \textrm{melalui titik}\: \: (4,3), \textrm{artinya}\\ &\textrm{bahwa}\: :\: (4-a)^{2}+(3-b)^{2}=r^{2}\\ &\Leftrightarrow a^{2}-8a+16+b^{2}-6b+9=r^{2}\\ &\Leftrightarrow a^{2}+b^{2}-8a-6b+25=r^{2}\: .......(2)\\ &\textrm{Pusat lingkaran}\: \: L_{2}\: \: \textrm{melalui garis}\: \: 9x+4y=37\\ &\textrm{artinya}:\: 9a+4b=37\: ...............(3)\\  \end{aligned}\\ &\begin{aligned}&\textrm{Dengan eliminasi}\: 1\: \&\: 2\: \: \textrm{dapat diperoleh}:\\ &\begin{array}{rll} a^{2}+b^{2}-8a-6b+25&=r^{2}&\\ a^{2}+b^{2}+2a+b-11&=r^{2}&-\\\hline -10a-7b+36&=0&\textrm{atau}\\ 10a+7b&=36&......(4) \end{array}\\ &\textrm{Dari persamaan}\: 3\: \&\: 4\: \: \textrm{dapat diperoleh}:\\ & \end{aligned}\\ &\begin{array}{rll} 10a+7b&=36&(\times 4)\\ 9a+4b&=37&(\times 7)\\\hline 40a+28b&=144&\\ 63a+28b&=259&\\\hline -23a\: \quad\quad&=-115&\\ a&=\displaystyle \frac{-115}{-23}&=5\\ 10(5)+7b&=36&\\ 7b&=-14\\ b&=-2 \end{array}\\ &\textrm{Adapun langkah berikutnya}\\ &\begin{aligned}&L_{2}\equiv (4-a)^{2}+(3-b)^{2}=r^{2}\\ &L_{2}\equiv (4-5)^{2}+(3+2)^{2}=r^{2}\\ &L_{2}\equiv r^{2}=25+1=26\\ &\textrm{Sehingga},\: L_{2}\equiv (x-5)^{2}+(y+2)^{2}=26\\ &\Leftrightarrow x^{2}+y^{2}-10x+4y+25+4-26=0\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-10x+4y+3=0 \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnya} \end{array}$.

Jika diperjelas dengan tambahan garis 9x+4y=37

$\begin{array}{ll}\\ 28.&\textrm{Diketahui lingkaran pertama berpusat di}\: \:  (1,2)\\ &\textrm{dan menyinggung garis}\: \: 3x-4y+10=0.\\ &\textrm{Jika ada lingkaran kedua dengan pusat}\: \: (4,6)\\ &\textrm{dan menyinggung lingkaran yang pertama},\\ &\textrm{maka persamaan lingkaran yang kedua}\\ &\textrm{tersebut adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-8x-12y+48=0\\ &\textrm{b}.\quad x^{2}+y^{2}-8x-12y+43=0\\ &\textrm{c}.\quad \color{red}x^{2}+y^{2}-8x-12y+36=0\\ &\textrm{d}.\quad x^{2}+y^{2}-8x-12y+27=0\\ &\textrm{e}.\quad x^{2}+y^{2}-8x-12y+16=0\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa kedua lingkaran saling}\\ &\color{blue}\textrm{bersinggungan di luar},\: \color{black}\textrm{maka}\\ &\begin{aligned}r_{1}+r_{2}&=P_{1}P_{2}\\ &=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}\\ &=\sqrt{(1-4)^{2}+(2-6)^{2}}\\ &=\sqrt{3^{2}+4^{2}}=\sqrt{5^{2}}=5 \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned}r_{\textrm{pertama}}&=\left |\displaystyle \frac{3(1)-4(2)+10}{\sqrt{3^{2}+4^{2}}}   \right |\\ &=\left | \displaystyle \frac{3-8+10}{\sqrt{5^{2}}} \right |=\left | \displaystyle \frac{5}{5} \right |=\left | 1 \right |=1\\ \textrm{sehingga} &\\ r_{\textrm{kedua}}&=5-r_{\textrm{pertama}}=5-1=4 \end{aligned}\\ &\textrm{maka persamaan lingkaran keduanya adalah}:\\ &\begin{aligned}&(x-4)^{2}+(y-6)^{2}=4^{2}\\ &\Leftrightarrow x^{2}-8x+16+y^{2}-12y+36=16\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-8x-12y+36=0 \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnya} \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Garis kuasa (tali busur sekutu)}\\ &\textrm{dari lingkaran}\\ &L_{1}\equiv x^{2}+y^{2}+6x-4y-12=0\\ &\textrm{dan}\: \: L_{2}\equiv x^{2}+y^{2}-12y=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 3x+4y+9=0\\ &\textrm{b}.\quad 3x-4y-8=0\\ &\textrm{c}.\quad 3x-4y+7=0\\ &\textrm{d}.\quad 3x+4y-7=0\\ &\textrm{e}.\quad \color{red}3x+4y-6=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &L_{1}\equiv x^{2}+y^{2}+6x-4y-12=0,\\ &\textrm{dan}\: \: L_{2}\equiv x^{2}+y^{2}-12y=0\\ &\textrm{Persamaan}\: \: \color{red}\textrm{garis kuasa}\: \color{black}\textrm{dari kedua}\\ &\textrm{lingkaran tersebut adalah}:\\ &\color{blue}L_{1}(x,y)- L_{2}(x,y)=0\\ &\Leftrightarrow x^{2}+y^{2}+6x-4y-12\\ &-(x^{2}+y^{2}-12y)=0\\ &\Leftrightarrow 6x+8y-12=0\\ &\Leftrightarrow \color{red}3x+4y-6=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Jika dua lingkaran}\\ & x^{2}+y^{2}=9\: \: \textrm{dan}\\ &x^{2}+y^{2}-4y+2y+3=0\: \: \textrm{yang}\\ &\textrm{berpotongan di}\: \: (x_{1},y_{1})\: \: \textrm{dan}\: \: (x_{2},y_{2}),\\ &\textrm{maka nilai}\: \: 5(x_{1}+x_{2})\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}24\\ &\textrm{b}.\quad 26\\ &\textrm{c}.\quad 28\\ &\textrm{d}.\quad 30\\ &\textrm{e}.\quad 32\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &L_{1}\equiv x^{2}+y^{2}-9=0\: \: \textrm{dan}\\ &L_{2}\equiv x^{2}+y^{2}-4x+2y+3\\ &\textrm{Persamaan}\: \: \color{red}\textrm{garis kuasa}\: \color{black}\textrm{dari kedua}\\ &\textrm{lingkaran tersebut adalah}:\\ &\color{blue}L_{1}(x,y)- L_{2}(x,y)=0\\ &\Leftrightarrow x^{2}+y^{2}-9\\ &-(x^{2}+y^{2}-4y+2y+3)=0\\ &\Leftrightarrow 4x-2y-12=0\\ &\Leftrightarrow 2x-y-6=0\\ &\Leftrightarrow y=6-2x \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned}&x^{2}+y^{2}-9=0\\ &\Leftrightarrow x^{2}+(6-2x)^{2}-9=0\\ &\Leftrightarrow x^{2}+36-24x+4x^{2}-9=0\\ &\Leftrightarrow 5x^{2}-24x+27=0\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{24\pm \sqrt{576-540}}{10}\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{24\pm \sqrt{36}}{10}=\frac{24\pm 6}{10}\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{24\pm \sqrt{36}}{10}=\frac{24\pm 6}{10}\\ &\Leftrightarrow \quad x_{1}=3\: \: \textrm{atau}\: \: x_{2}=1,8\\ &\textrm{maka}\: \: 5(x_{1}+x_{2})=5\left ( 3+1,8 \right )=\color{red}24 \end{aligned} \end{array}$.

Lanjutan Ketaksamaan Schur

(Bagian Kedua)

Bagian Pertama silahkan klik di sini

1. Penyederhanaan dengan pola siklik dan simetri

$.\quad\begin{aligned}&\color{red}\textrm{Mengenal penulisan pola}\: \textbf{Siklik dan Simetri}\\ &\textrm{Misal untuk}\: \: n=3,\: \: \textrm{pada penulisan unsur}\\ &x,y,\: \: \textrm{dan}\: \: z,\: \textrm{maka}\\ &\begin{array}{|l|l|}\hline \textbf{Pola Siklik}&\textbf{Pola Simetri}\\\hline\begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}=x^{2}+y^{2}+z^{2}\\ &\\ &\\ &\\ &\\ & \end{aligned} &\begin{aligned}\displaystyle\sum_{\textrm{sym}}^{.}x^{2}&=x^{2}+x^{2}\\ &+y^{2}+y^{2}\\ \\ &+z^{2}+z^{2}\\ \\ &=2\left (x^{2}+y^{2}+z^{2}  \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{3}=x^{3}+y^{3}+z^{3}\end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{3}=2\left (x^{3}+y^{3}+z^{3}  \right ) \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y=x^{2}y+y^{2}z+z^{2}x\\ &\\ &\\ & \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}x^{2}y&=x^{2}y+x^{2}z\\ &+y^{2}x+y^{2}z\\\\ &+z^{2}x+z^{2}y \end{aligned}\\ \begin{aligned}\displaystyle \sum_{\textrm{siklik}}^{.}xyz&=xyz+yzx+zxy\\ &=3xyz \end{aligned}&\begin{aligned}\displaystyle \sum_{\textrm{sym}}^{.}xyz&=xyz+xzy+\cdots \\ &=6xyz \end{aligned}  \\\hline \end{array} \end{aligned}$

2. Bentuk ketaksamaan berdasar nilai r

Masih ingat kita pada ketaksamaan Schur saat $r=1$, yaitu,:

$\begin{aligned}&a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\\ &=a^{3}+b^{3}+c^{3}-\left( a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b \right)+3abc\\ &=\displaystyle \sum_{\textrm{sym}}^{.}\left( a^3-2a^{2}b+abc \right)\ge 0 \end{aligned}$.

Selanjutnya saat  $\color{red}r=1$, kita bisa mendaptkan

$\begin{aligned}1.\quad &a^{3}+b^{3}+c^{3}+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)\\ 2.\quad &abc\geq (a+b-c)(b+c-a)(c+a-b)\\ 3.\quad &(a+b+c)^{3}+9abc\geq 4(a+b+c)(ab+bc+ca) \end{aligned}$.

Dan saat  $\color{red}r=2$, kita bisa mendaptkan

$\begin{aligned}a^{4}+&b^{4}+c^{4}+abc(a+b+c)\geq ab(a^{2}+b^{2})+bc(b^{2}+c^{2})+ca(c^{2}+a^{2}) \end{aligned}$.

3. Beberapa formulasi bantu

$\begin{aligned}1.\quad&\color{red}(a+b+c)(a^{2}+b^{2}+c^{2})\\&=a^{3}+b^{3}+c^{3}+a(b^{2}+c^{2})+b(a^{2}+c^{2})+c(a^{2}+b^{2})\\ 2.\quad&\color{red}(a+b+c)(a^{2}+b^{2}+c^{2})\\ &=a^{3}+b^{3}+c^{3}+a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)\\ 3.\quad&\color{red}(a+b+c)(ab+ac+bc)\\ &=3abc+ab(a+b)+ac(a+c)+bc(b+c)\\ 4.\quad&\color{red}(a+b+c)(ab+ac+bc)\\ &=3abc+a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)\\ 5.\quad&\color{red}(a+b+c)^{3}+3abc\\ &=a^{3}+b^{3}+c^{3}+3(a+b+c)(ab+ac+bc)\\ 6.\quad&\color{red}(a+b)(a+c)(b+c)\\ &=2abc++a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)\\ 7.\quad&\color{red}(a+b-c)(a+c-b)(b+c-a)\\ &=-2abc-(a^{3}+b^{3}+c^{3})+a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)\\ 8.\quad&\color{red}(a-b)(b-c)(c-a)\\ &=a^{2}(c-b)+b^{2}(a-c)+c^{2}(b-a)\\  \end{aligned}$.

4. Penyederhanaan ketaksamaan metode pqr

$\begin{array}{|l|l|}\hline .\qquad\qquad\textbf{Kesamaan}&\quad \qquad\textbf{Ketaksamaan}\\\hline\begin{aligned}1.\quad&\displaystyle \sum_{sik}^{.}x^{2}=p^{2}-2q\\ 2.\quad&\displaystyle \sum_{sik}^{.}x^{3}=p(p^{2}-3q)+3r\\ 3.\quad&\sum_{sik}^{.}x^{2}y^{2}=q^{2}-2pr\\ 4.\quad&\prod (x+y)=pq-r\\ 5.\quad&\sum_{sik}^{.}xy(x+y)=pq-3r\\ 6.\quad&\sum_{sik}^{.}x^{2}(y+z)=pq-3r\\ 7.\quad&\prod (1+x)=1+p+q+r\\ &\\ &\\ &\\ &\\ & \end{aligned} &\begin{aligned}1.\quad&pq\geq 9r\\ 2.\quad&p^{2}\geq 3q\\ 3.\quad&q^{2}\geq 3pr\\ 4.\quad&p^{3}\geq 27r\\ 5.\quad&q^{3}\geq 27r^{2}\\ 6.\quad&p^{3}r\geq q^{3}\\ 7.\quad&p^{3}+9r\geq 4pq\\ 8.\quad&2p^{3}+9r\geq 7pq\\ 9.\quad&2p^{3}+27r\geq 9pq\\ 10.\quad&2p^{3}+9r^{2}\geq 7pqr\\ 11.\quad&q^{3}+9r^{2}\geq 4pqr\\ 12.\quad&2q^{3}+27r^{2}\geq 9pqr\\ 13.\quad&p^{4}+3q^{2}\geq 4p^{2}q\\ 14.\quad&p^{4}+4q^{2}+6pr\geq 5p^{2}q\\ 15.\quad&p^{2}q+3pr\geq 4q^{2}\\ 16.\quad&pq^{2}\geq 2p^{2}r+3qr\\ 17.\quad&p^{2}q^{2}+12r^{2}\geq 4p^{3}r+pqr \end{aligned}  \\\hline \end{array}$.

5. Ketaksamaan Schur bentuk pqr

Perhatikan poin 2 di atas saat $r=1$, kitaa akan medapatkan bentuk berikut ini:

$\begin{aligned}&(a+b+c)^{3}+9abc\geq 4(a+b+c)(ab+bc+ca)\\ &\Leftrightarrow \color{red}p^{3}+9r\geq 4pq \end{aligned}$.

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

$\begin{array}{ll}\\ 1&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan bahwa}\\ &(a+b+c)(ab+ac+bc)\geq 9abc\\\\ &\textbf{Bukti}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Untuk pembuktian pernyataan di atas}\\ &\textrm{dengan AM-GM kita memiliki}\\ &\bullet\quad a+b+c\geq 3\sqrt[3]{abc}\\ &\bullet\quad ab+ac+bc\geq 3\sqrt[3]{(abc)^{2}}\\ &\textrm{maka hasil dari}\\ &(a+b+c)(ab+ac+bc)\geq 3\sqrt{abc}.3\sqrt[3]{(abc)^{2}}\\ &\Leftrightarrow pq\geq 3\sqrt[3]{r}.3\sqrt[3]{r^{2}}\\ &\Leftrightarrow pq\geq 9\sqrt[3]{r^{3}}\\ &\Leftrightarrow \color{red}pq\geq 9r\qquad \color{black}\blacksquare   \end{array}$.

$\begin{array}{ll}\\ 2&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan bahwa}\\ &(a+b+c)^{2}\geq 3(ab+ac+bc)\\\\ &\textbf{Bukti}\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Untuk pembuktian pernyataan di atas}\\ &\textrm{dengan AM-GM kita memiliki}\\ &\bullet\quad a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\ &\textrm{Dan juga sebuah kesamaan}\\ &\bullet \quad a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+ac+bc)\\ &\textrm{maka dari kedua bentuk di atas kita akan}\\ &\textrm{dapatkan bentuk}\\ &a^{2}+b^{2}+c^{2}\geq ab+ac+bc\\ &\Leftrightarrow (a+b+c)^{2}-2(ab+ac+bc)\geq ab+ac+bc\\ &\Leftrightarrow (a+b+c)^{2}\geq 3(ab+ac+bc)\\ &\Leftrightarrow  \color{red}p^{2}\geq 3q\qquad \color{black}\blacksquare\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Telah kita ketahui bahwa}\\ &(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+ac+bc)\\ &\textrm{Dengan ketaksamaan}\: \textbf{renata}\: \textrm{kita akan}\\ &\textrm{dapatkan bentuk}\\ &(a+b+c)^{2}\geq ab+bc+ca+2(ab+ac+bc)\\ &\Leftrightarrow (a+b+c)^{2}\geq 3(ab+ac+bc)\\ &\Leftrightarrow  \color{red}p^{2}\geq 3q\qquad \color{black}\blacksquare  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan}\\ &\textrm{bahwa}\: \: (ab+ac+bc)^{2}\geq 3abc(a+b+c)\\\\ &\textbf{Bukti}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Perhatikan kesamaan berikut}\\ &\begin{aligned}&(ab+ac+bc)^{2}=a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+2abc(a+b+c)\\  &\textrm{Dengan AM-GM dan GM-AM kita dapatkan}\\ &\Leftrightarrow (ab+ac+bc)^{2}\geq 3\sqrt[3]{a^{4}b^{4}c^{4}}+2abc(a+b+c)\\ &\Leftrightarrow (ab+ac+bc)^{2}\geq 3abc\sqrt[3]{abc}+2abc(a+b+c)\\ &\Leftrightarrow (ab+ac+bc)^{2}\geq abc(a+b+c)+2abc(a+b+c)\\ &\Leftrightarrow (ab+ac+bc)^{2}\geq 3abc(a+b+c)\\ &\Leftrightarrow \color{red}q^{2}\geq 3pr\qquad \color{black}\blacksquare  \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan bahwa}\\ &(a+b+c)^{3}\geq 27abc\\\\ &\textbf{Bukti}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Untuk pembuktian pernyataan di atas}\\ &\textrm{dengan AM-GM kita memiliki}\\ &\bullet\quad a+b+c\geq 3\sqrt[3]{abc}\\ &\textrm{Masing-masing ruas pangkatkan 3, maka}\\ &(a+b+c)^{3}\geq 27abc\Leftrightarrow \color{red}p^{3}\geq 27r\qquad \color{black}\blacksquare  \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Untuk}\: \: a,b,c\: \: \textrm{tak negatif, tunjukkan bahwa}\\ &(a+b+c)^{3}\geq 27abc\\\\ &\textbf{Bukti}\\ &\textrm{Misalkan}\\ &p=a+b+c,\: q=ab+ac+bc,\\ &\textrm{dan}\: \: r=abc\\ &\textrm{Untuk pembuktian pernyataan di atas}\\ &\textrm{dengan AM-GM kita memiliki}\\ &\bullet\quad ab+ac+bc\geq 3\sqrt[3]{(abc)^{2}}\\ &\textrm{Masing-masing ruas pangkatkan 3, maka}\\ &(ab+ac+bc)^{3}\geq 27(abc)^{2}\Leftrightarrow \color{red}q^{3}\geq 27r^{2}\qquad \color{black}\blacksquare  \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}+b^{2}+c^{2}+2abc+1\geq 2(ab+acb+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Darij Grinberg})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{dilanjutkan dengan}\: \: \textbf{ketaksamaan Schur}\\ &\textrm{serta menggesernya ke ruas kiri, maka}\\ &a^{2}+b^{2}+c^{2}+2abc+1- 2(ab+ac+bc)\\ &\geq a^{2}+b^{2}+c^{2}+3(abc)^{^{\frac{2}{3}}}+1\geq 2(ab+ac+bc)\\ &\geq \left ((a)^{^{\frac{2}{3}}}  \right )^{3}+\left ((b)^{^{\frac{2}{3}}}  \right )^{3}+\left ((c)^{^{\frac{2}{3}}}  \right )^{3}+3(abc)^{^{\frac{2}{3}}}-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &= a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -b^{.^{\frac{2}{3}}} \right )^{2}+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\geq 0\qquad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 7.&(\textbf{APMO 2004})\\ &\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad (x^{2}+2)(y^{2}+2)(z^{2}+2)\geq 9(xy+yz+zx)\\\\ &\textbf{Bukti}\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Dengan menjabarkan akan didapatkan}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{Perhatikan bahwa}\\ &\bullet \quad\color{red}2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}-4\displaystyle \sum_{\textrm{siklik}}^{.}xy+6=2\displaystyle \sum_{\textrm{siklik}}^{.}(xy-1)^{2}\geq 0\\ &\bullet  \quad \color{red}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\: \: \color{black}\textrm{atau}\: \: \color{red}x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\textrm{Kita cukup membuktikan bahwa}\\ & x^{2}y^{2}z^{2}+\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+2\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow  x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\begin{aligned}&\textrm{Untuk}\: \: a,b,c\: \: \textrm{bilangan real positif},\\ &\textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=1\\ &\textrm{memberikan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq \displaystyle \sum_{\textrm{siklik}}^{.}a^{2}b+\displaystyle \sum_{\textrm{siklik}}^{.}ab^{2}\\ &\qquad\qquad\qquad =ab(a+b)+bc(b+c)+ca(c+a)\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan AM-GM}\: \: \textrm{didapakan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}(ab)^{\frac{3}{2}}\\ &\textrm{Pilih}\: \: a=x^{\frac{2}{3}},\: b=y^{\frac{2}{3}},\: c=z^{\frac{2}{3}},\: \textrm{maka didapatkan}\\ &(x^{\frac{2}{3}})^{3}+(y^{\frac{2}{3}})^{3}+(z^{\frac{2}{3}})^{3}+3(xyz)^{\frac{2}{3}}\geq 2(xy+yz+zx)\\ &\textrm{Selanjutnya kita selesaikan ini}\: ,\: x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow  x^{2}y^{2}z^{2}+2 \geq 3(xyz)^{\frac{2}{3}}\\ &\textrm{Misalkan}\: \: (xyz)^{\frac{2}{3}}=t,\: \textrm{maka}\\ &t^{3}+2\geq 3t\Leftrightarrow t^{3}-3t+2\geq 0\\ &(t-1)^{2}(t+2)\geq 0\: \: \textrm{adalah hal benar} \end{aligned}   \end{array}$.

$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{atau dalam bentuk utuhnya, yaitu}\\ &x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\geq 9(xy+xz+yz)\\ &\textrm{Sekarang kira uraikan satu persatu bagian}\\ &\bullet\quad x^{2}y^{2}z^{2}+1+1\geq 3\sqrt[3]{(xyz)^{2}}\geq \displaystyle \frac{9abc}{a+b+c}=\frac{9r}{p}\\ &\qquad \textrm{ingat bahwa}\: \: \textbf{jika ada}\: \: \displaystyle \frac{9r}{p}\geq 4q-p^{2}\\ &\qquad =4(xy+xz+yz)-(x+y+z)^{2}\\ &\qquad \textrm{adalah}\: \: \textbf{ketaksamaan Schur saat}\: \: \color{red}r=1\\ &\bullet \quad x^{2}y^{2}+1+x^{2}z^{2}+1+y^{2}z^{2}+1\geq 2(xy+xz+yz)\\ &\bullet\quad  x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\qquad \textrm{keduanya didapat dengan ketaksamaan}\: \: \textbf{AM-GM}\\&x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\\ &=x^{2}y^{2}z^{2}+2+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}+3)+4(x^{2}+y^{2}+z^{2})\\ &\geq 4(xy+xz+yz)-(x+y+z)^{2}+4(xy+xz+yz)+4(xy+xz+yz)\\ &\geq 12(xy+xz+yz)-(x+y+z)^{2}\\ &\geq 12(xy+xz+yz)-3(xy+xz+yz)\\ &=9(xy+xz+yz)\qquad \blacksquare    \end{aligned}$.

$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 3}\\ &(x^{2}+2)(y^{2}+2)(z^{2}+2)- 9(xy+yz+zx)\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\\ &=x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})\\ &\quad +4(x^{2}+y^{2}+z^{2})+8- 9(xy+xz+yz)\\ &=4(x^{2}+y^{2}+z^{2})+2\left ((x^{2}y^{2}+1) +(x^{2}z^{2}+1)+(y^{2}z^{2}+1) \right )\\ &\quad +(x^{2}y^{2}z^{2}+1)+1-9(xy+xz+yz)\\ &\geq 4(x^{2}+y^{2}+z^{2})+4(xy+xz+yz)\\ &\quad +2xyz+1-9(xy+xz+yz)\\ &=(x^{2}+y^{2}+z^{2})+3(x^{2}+y^{2}+z^{2})\\ &\quad +2xyz+1-5(x^{2}+y^{2}+z^{2})\\ &\geq x^{2}+y^{2}+z^{2}+3(xy+xz+yz)\\ &+2xyz-5(xy+xz+yz)\\ &=x^{2}+y^{2}+z^{2}+2xyz+1-2(xy+xz+yz)\geq 0\\ &\textrm{adalah benar dengan bukti ada pada}\\ &\textrm{nomor soal sebelumnya}  \end{aligned}$.

$\begin{array}{ll}\\ 8.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad 2(a^{2}+b^{2}+c^{2})+abc+8\geq 5(a+b+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Tran Nam Dung})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{menggeser ke ruas kiri dan masing-masing}\\ &\textrm{serta mengalikan semunya dengan 6, maka}\\ &12(a^{2}+b^{2}+c^{2})+6abc+48- 30(a+b+c)\\ &= 12(a^{2}+b^{2}+c^{2})+3(2abc+1)+45- 5.2.3(a+b+c)\\ &\geq 2(a^{2}+b^{2}+c^{2})+9\sqrt[3]{(abc)^{2}}+45- 5\left ((a+b+c)^{2}+9  \right )\\ &=12(a^{2}+b^{2}+c^{2})+\displaystyle \frac{9abc}{\sqrt[3]{abc}}-5\left ((a^{2}+b^{2}+c^{2})+2(ab+ac+bc)  \right )\\ &=7(a^{2}+b^{2}+c^{2})+ \displaystyle \frac{9abc}{\sqrt[3]{abc}}-10(ab+ac+bc)\\ &\geq  7(a^{2}+b^{2}+c^{2})+\displaystyle \frac{27abc}{a+b+c}-10(ab+ac+bc)\\ &\begin{aligned}&\textrm{dengan}\: \: \textbf{ketaksamaan Schur},\: \: \textrm{yaitu}:\\ &p^{3}+9r\geq 4pq\Leftrightarrow \displaystyle \color{red}\frac{9r}{p}\geq 4q-p^{2}\\ &\textrm{maka ketaksamaan akan menjadi}\\ &\geq 7(a^{2}+b^{2}+c^{2})+\color{blue}3(4q-p^{2})-10q\\ &\geq 7(a^{2}+b^{2}+c^{2})+2q-3p^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2(ab+ac+bc)-3(a+b+c)^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2q-3\left ((a^{2}+b^{2}+c^{2})+2q  \right )\\ &=4(a^{2}+b^{2}+c^{2})+2q-6q\\ &=4(a^{2}+b^{2}+c^{2})-4q\\ &=4(a^{2}+b^{2}+c^{2})-4(ab+ac+bc)\\ &=4(a^{2}+b^{2}+c^{2}-ab-ac-bc)\geq 0\quad \blacksquare  \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Venkatachala, B.J. 2009. Inequalities An Approach Through Problems (2nd). India: SPRINGER.
2. Vo Tranh Van..... Bat Dang Thuc Schur Va Phuong Phap Doi Bien P, Q, R.
3. Vo Quoc Ba Can. 2007. Bai Viet Ve Bat Dang Thuc Schur Va Vornicu Schur.
4. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Contoh 5 Soal dan Pembahasan Materi Hubungan Dua Lingkaran

$\begin{array}{ll}\\ 21.&\textrm{Titik Kuasa dari lingkaran-lingkaran}\\ &\textrm{berikut}\\ &L_{1}\equiv x^{2}+y^{2}+x+y-14=0\\ &L_{2}\equiv x^{2}+y^{2}=13\\ &L_{3}\equiv x^{2}+y^{2}+3x-2y-26=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}(3,-2)\\ &\textrm{b}.\quad (2,-3)\\ &\textrm{c}.\quad (-3,2)\\ &\textrm{d}.\quad (-2,3)\\ &\textrm{e}.\quad (3,2)\\\\ &\textbf{Jawab}:\\ &\textrm{Dengan eliminasi, kita mendapatkan}\\ &\begin{aligned}&\begin{array}{lrlll} (L_{1})&x^{2}+y^{2}+x+y&=&14\\ (L_{2})&x^{2}+y^{2}&=&13&-\\\hline &x+y&=&1&....(1) \end{array}\\ &\textrm{dan}\\ &\begin{array}{lrlll} (L_{3})&x^{2}+y^{2}+3x-2y&=&26\\ (L_{2})&x^{2}+y^{2}&=&13&-\\\hline &3x-2y&=&13&....(2) \end{array}\\ &\textrm{Selanjutnya kita eliminasi}\: (1)\& (2)\\ &\textrm{dan hasilnya adalah}:\\ &\begin{array}{rrlrl} \color{blue}(2)&3x-2y&=&13\\ \color{blue}(1)&3x+3y&=&3&-\qquad (\times 3)\\\hline &-5y&=&10&\\ &y&=&\color{red}-2&\Rightarrow x=\color{red}3 \end{array}\\ &\textrm{Jadi, titik kuasa ketiganya}: (3,-2) \end{aligned}\\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut}  \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Titik-titik potong dari persekutuan dua}\\ &\textrm{lingkaran}\: \: L_{1}\equiv (x-2)^{2}+y^{2}=10\: \: \: \textrm{dan}\\ &L_{2}\equiv x^{2}+(y-2)^{2}=10\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad (3,3)\: \: \textrm{dan}\: \: (1,1)\\ &\textrm{b}.\quad \color{red}(3,3)\: \: \textrm{dan}\: \: (-1,-1)\\ &\textrm{c}.\quad (3,-3)\: \: \textrm{dan}\: \: (1,1)\\ &\textrm{d}.\quad (-3,3)\: \: \textrm{dan}\: \: (1,1)\\ &\textrm{e}.\quad (-3,-3)\: \: \textrm{dan}\: \: (-1,-1)\\\\ &\textbf{Jawab}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\textrm{Dengan substitusi opsi pilihan jawaban}\\ &\textrm{maka akan ketemu jawabannya langsung}\\ &\color{blue}\textrm{Alternatif 2}\\ &\textrm{Dengan eliminasi dan ilustrasi gambar}\\ &\begin{array}{lrlll} (L_{1})&(x-2)^{2}+y^{2}&=&10\\ (L_{2})&x^{2}+(y-2)^{2}&=&10&\\&\color{blue}\textrm{menjadi}\\ (L_{1})&x^2+y^2-4x&=&6\\ (L_{2})&x^2+y^2-4y&=&6&-\\\hline &-4x+4y&=&0&\\ &\color{blue}\textrm{maka hasilnya}\\ &y&=&x \end{array} \\ &\textrm{Jelas opsi jawaban c, d salah}\\ &\textrm{karena}\: \: y=x,\\ &\textbf{Dengan bantuan ilustrasi, pilihan jawaban}\\ &\textbf{akan tampak dengan jelas}  \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Persamaan tali busur persekutuan dua}\\ &\textrm{lingkaran}\: \: L_{1}\equiv (x-3)^{2}+y^{2}=16\: \: \: \textrm{dan}\\ &L_{2}\equiv x^{2}+(y-3)^{2}=16\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad y=-2x\\ &\textrm{b}.\quad y=-x\\ &\textrm{c}.\quad \color{red}y=x\\ &\textrm{d}.\quad y=2x\\ &\textrm{e}.\quad y=\displaystyle \frac{1}{2}x\\\\ &\textbf{Jawab}:\\ &\textrm{Dengan eliminasi, kita mendapatkan}\\ &\begin{array}{lrlll} (L_{1})&(x-3)^{2}+y^{2}&=&16\\ (L_{2})&x^{2}+(y-3)^{2}&=&16&\\&\color{blue}\textrm{menjadi}\\ (L_{1})&x^2+y^2-6x&=&9\\ (L_{2})&x^2+y^2-6y&=&9&-\\\hline &-6x+6y&=&0&\\ &\color{blue}\textrm{maka hasilnya}\\ &y&=&x \end{array}\\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut}  \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Banyaknya garis singgung persekutuan}\\ &\textrm{lingkaran-lingkaran}\: x^{2}+y^{2}+2x-6y+9=0\\ &\textrm{dan}\: \:  x^{2}+y^{2}+8x-6y+9=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 0\\ &\textrm{b}.\quad \color{red}1\\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 3\\ &\textrm{e}.\quad 4\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}+2x-6y+9=0&\begin{cases} P_{1} &=(-1,3) \\  r_{1} & = 1 \end{cases}\\\hline L_{2}\equiv x^{2}+y^{2}+8x-6y+9=0&\begin{cases} P_{2} &=(-4,3) \\  r_{2} & = 4 \end{cases}\\\hline \end{array} \\ &\textrm{Perhatikan pula bahwa}\: \: r_{2}-r_{1}=4-1=3\\ &\begin{aligned}&\textrm{Karena}\: \: P_{1}P_{2}=r_{2}-r_{1},\: \textrm{hal ini berarti lingkaran}\\ &L_{1}\: \: \textrm{bersinggungan di dalam dengan lingkaran}\: L_{2}\\ &\textrm{Sehingga kedua lingkaran ini hanya akan }\\ &\textrm{memiliki}\: \: \color{red}\textrm{satu}\: \: \color{black}\textrm{garis singgung persekutuan} \end{aligned}\\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut}  \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Persamaan lingkaran dengan jari-jari}\: \: 5\\ &\textrm{dan menyinggung lingkaran lain}\\ & x^{2}+y^{2}-2x-4y-20=0\: \: \: \textrm{di titik}\\ &(5,5)\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-2x-4y-120=0\\ &\textrm{b}.\quad x^{2}+y^{2}-2x-4y-120=0\\ &\textrm{c}.\quad x^{2}+y^{2}-2x-4y-120=0\\ &\textrm{d}.\quad \color{red}x^{2}+y^{2}-2x-4y-120=0\\ &\textrm{e}.\quad x^{2}+y^{2}-2x-4y-120=0\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahi bahwa}\\ &\begin{aligned}&\begin{array}{rrlll} (L_{1})&(x-a)^2+(y-b)^{2}&=&5^{2}\\ (L_{2})&x^{2}+y^{2}-2x-4y&=&20& \end{array} \\ &\textrm{Titik singgung dua lingkaran}\\ &\textrm{di titik}\: \: (5,5),\: \textrm{artinya}\\ &\begin{pmatrix} 5\\  5 \end{pmatrix}=\displaystyle \frac{\begin{pmatrix} a\\  b \end{pmatrix}+\begin{pmatrix} 1\\  2 \end{pmatrix}}{2}\\ &\Leftrightarrow \begin{pmatrix} 10\\  10 \end{pmatrix}=\begin{pmatrix} a\\  b \end{pmatrix}+\begin{pmatrix} 1\\  2 \end{pmatrix}\\ &\Leftrightarrow \begin{pmatrix} a\\  b \end{pmatrix}=\begin{pmatrix} 10-1\\  10-2 \end{pmatrix}=\begin{pmatrix} 9\\  8 \end{pmatrix} \end{aligned}\\ &\begin{aligned}&\textrm{maka persamaan lingkarannya adalah}:\\ &\Leftrightarrow (x-9)^{2}+(y-8)^{2}=5^{2}\\ &\Leftrightarrow x^{2}+y^{2}-18x-16y+120=0 \end{aligned}\\  &\textbf{Berikut ilustrasi gambarnya}  \end{array}$.

Operasi Vektor di Ruang (Lanjutan Materi Vektor di Ruang)

Sebelumnya silahkan lihat di sini

 $\color{blue}\textrm{A. Operasi Vektor Dalam Ruang}$

Operasi vektor pada dimensi tiga kurang lebih sama dengan operasi pada vektor berdimensi dua.

$\color{blue}\textrm{A.  Penjumlahan dan Pengurangan}$.

$\begin{aligned}\textrm{Jika}\: & \textrm{diketahui sebagai misal}\\ \bar{u}&=a\bar{i}+b\bar{j}+c\bar{k}\: \: \: \color{red}\textrm{dan}\\ \bar{v}&=p\bar{i}+q\bar{j}+r\bar{k}\\ \textrm{mak}&\textrm{a}\\ \textbf{Pen}&\textbf{jumlahan dua vektor di atas adalah}\\ \bar{u}+\bar{v}&=(a+p)\bar{i}+(b+q)\bar{j}+(c+r)\bar{k}\\ \textbf{dem}&\textbf{ikian juga untuk pengurangan}\\ \bar{u}-\bar{v}&=(a-p)\bar{i}+(b-q)\bar{j}+(c-r)\bar{k}\\ \end{aligned}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Jika diketahui}\: \: \bar{a}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \bar{b}=\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\textrm{Tentukanlah hasil dari}\\ &\textrm{a}.\quad \bar{a}+\bar{b}\\ &\textrm{b}.\quad \bar{a}-\bar{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bar{a}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \bar{b}=\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix},\\ &\textrm{maka}\\ &\bar{a}+\bar{b}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}+\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\qquad =\begin{pmatrix} 1+8\\ 3+(-2)\\ 7+0 \end{pmatrix}=\begin{pmatrix} 9\\ 1\\ 7 \end{pmatrix}\\ &\textrm{Dan untuk}\: \: \bar{a}-\bar{b}\: \: \textrm{adalah}:\\ &\bar{a}-\bar{b}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}-\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\qquad =\begin{pmatrix} 1-8\\ 3-(-2)\\ 7-0 \end{pmatrix}=\begin{pmatrix} -7\\ 5\\ 7 \end{pmatrix} \end{aligned} \end{array}$.

$\color{blue}\textrm{B. 1. Perkalian Skalar dengan Vektor}$.

Misalkan suatu skalar   $m$  dan suatu vektor  $\bar{u}=a\bar{i}+b\bar{j}+c\bar{k}$, maka perkalian $m$  dengan vektor  $\bar{u}$ tersebut adalah  $\bar{u}=ma\bar{i}+mb\bar{j}+mc\bar{k}$.

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

$\begin{array}{ll} 2.&\textrm{Jika}\: \: \bar{a}=\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix},\: \: \textrm{tentukanlah nilai}\\ &\textrm{dari}\: \: 2\bar{a}\: \: \: \textrm{dan}\: \: -3\bar{a}\\\\ &\textbf{Jawab}\\ &\begin{aligned}2\bar{a}&=2\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix}=\begin{pmatrix} 4044\\ 4042\\ 4040 \end{pmatrix},\: \: \textrm{dan}\\ -3\bar{a}&=-3\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix}=\begin{pmatrix} -6066\\ -6063\\ -6060 \end{pmatrix} \end{aligned} \end{array}$

$\color{blue}\textrm{F. 2. Perkalian Skalar Dua Vektor}$.

Hasil dari perkalian skalar dua vektor $\bar{a}$  dan  $\bar{b}$ adalah :  $\bar{a}\: \: \bullet\: \: \bar{b}$.

Dengan

$\bar{a}\: \: \bullet\: \: \bar{b}=\left | \bar{a} \right |\left | \bar{b} \right |\cos \theta$.  sehingga

$\begin{aligned}&\textrm{Tanda dari hasil skalar ini adalah}\\ &\begin{array}{|l|l|l|}\hline \textbf{Besar sudut}\: \: \: \theta &\textbf{Tanda}&\textrm{Bentuk}\\\hline 0^{\circ}\leq \theta < 90^{\circ}&\textrm{Positif}&\color{red}\textrm{Lancip}\\\hline \theta =90^{\circ}&\textrm{Nol}&\textrm{Siku-siku}\\\hline 90^{\circ}< \theta \leq 180^{\circ}&\textrm{Negatif}&\color{blue}\textrm{Tumpul}\\\hline \end{array}\\ &\textrm{Untuk}\: \: \theta \: \: \textrm{berupa sudut istimewa}:\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline \theta &0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}&180^{\circ}\\\hline \cos \theta &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1\\\hline \end{array} \end{aligned}$

Adapun secara rumus untuk menentukan besar sudutnya adalah:

$\cos \theta =\displaystyle \frac{\bar{a}\: \: \bullet\: \: \bar{b} }{\left | \bar{a} \right |\left | \bar{b} \right |}$.

Sebagai ilustrasinya perhatikanlah gambar berikut

Selain hasil di atas ada cara lain menyelesaikan perkalian skalar dua vektor, yaitu:

$\begin{aligned}\textrm{Jika}\: & \textrm{diketahui sebagai misal}\\ \bar{u}&=a\bar{i}+b\bar{j}+c\bar{k}\: \: \: \color{red}\textrm{dan}\\ \bar{v}&=p\bar{i}+q\bar{j}+r\bar{k}\\ \textrm{mak}&\textrm{a}\\ \textbf{Per}&\textbf{kalian skalar dua vektor adalah}:\\ \bar{u}\: \bullet \: &\bar{v}=\left ( a\bar{i}+b\bar{j}+c\bar{k} \right )\left ( p\bar{i}+q\bar{j}+r\bar{k} \right )\\ &\: \: =ap.\bar{i}\: \bullet \bar{i}+aq.\bar{i}\: \bullet \: \bar{j}+ar.\bar{i}\: \bullet \: \bar{k}\\ &\: \: \: \: \: \: \: +bp.\bar{j}\: \bullet \: \bar{i}+bq.\bar{j}\: \bullet \: \bar{j}+br.\bar{j}\: \bullet \: \bar{k}\\ &\: \: \: \: \: \: \: +cp.\bar{k}\: \bullet \: \bar{i}+cq.\bar{k}\: \bullet \: \bar{j}+cr.\bar{k}\: \bullet \: \bar{k}\\ &\: \: =ap+0+0+0+bq+0+0+0+cr\\ &\: \: =\color{red}a p+b q+c r \end{aligned}$

$\begin{aligned}&\textrm{Sebagai penjelasannya adalah}:\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{i}=\left | \bar{i} \right |\left | \bar{i} \right |\cos 0^{\circ}=1.1.1=1\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{j}=\left | \bar{i} \right |\left | \bar{j} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{k}=\left | \bar{i} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{i}=\bar{i}\: \: \bullet \: \: \bar{j}=0\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{j}=\left | \bar{j} \right |\left | \bar{j} \right |\cos 0^{\circ}=1.1.1=1\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{k}=\left | \bar{j} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{i}=\bar{i}\: \: \bullet \: \: \bar{k}=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{j}=\bar{j}\: \: \bullet \: \: \bar{k}=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{k}=\left | \bar{k} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.1=1 \end{aligned}$

$\begin{aligned}&\textrm{Atau jika ditabelkan nilainya}\\ &\begin{array}{|c|c|c|c|}\hline \bar{u}\: \bullet \: \bar{v}&p\bar{i}&q\bar{j}&r\bar{k}\\\hline a\bar{k}&ap&0&0\\ b\bar{j}&0&bq&0\\ c\bar{k}&0&0&cr\\\hline \end{array} \end{aligned}$

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

$\begin{array}{ll} 3.&\textrm{Jika}\: \: \vec{a}=\begin{pmatrix} 1\\ 2\\ 4 \end{pmatrix},\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 5\\ 4\\ 0 \end{pmatrix}\\\ & \textrm{tentukanlah nilai}\: \: \textrm{dari}\: \: \vec{a}\bullet \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=1.5+2.4+4.0=5+8+0=13 \end{aligned} \end{array}$

$\begin{array}{ll} 4.&\textrm{Jika diketahui}\: \: \vec{a}=\vec{i}-2\vec{j}+3\vec{k},\\ & \textrm{dan}\: \: \vec{b}=3\vec{i}-4\vec{j}+m\vec{k}\: \: \textrm{serta}\\\ & \textrm{nilai}\: \: \vec{a}\bullet \vec{b}=-4,\: \: \textrm{maka tentukan}\\ &\textrm{nilai}\: \: m\\\\ &\textbf{Jawab}\\ &\textrm{Diketahui bahwa}\\ &\color{red}\triangleright \quad \color{black}\vec{a}=\vec{i}-2\vec{j}+3\vec{k}=\begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix},\: \: \textrm{dan}\\ &\color{red}\triangleright \quad \color{black}\vec{b}=3\vec{i}-4\vec{j}+m\vec{k}=\begin{pmatrix} 3\\ -4\\ m \end{pmatrix}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=1.3+3.(-4)+3.m\\ -4&=3+8+3m\\ -3m&=11+4\\ m&=-\displaystyle \frac{15}{3}\\ &=\color{blue}-5 \end{aligned} \end{array}$

$\begin{array}{ll} 5.&\textrm{Diketahui}\: \: \left |\vec{a} \right |=10,\: \left | \vec{b} \right |=6.\\ & \textrm{Jika}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{membentuk sudut}\\ &60^{\circ}.\: \textrm{Tentukanlah nilai}\: \: \vec{a}\bullet \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ &=10.6.\cos 60^{\circ}\\ &=60.\left ( \displaystyle \frac{1}{2} \right )\\ &=\color{blue}30\\ \textrm{Jadi}&\: \textrm{hasil kali skalarnya adalah 30} \end{aligned} \end{array}$.

Contoh Soal 3 Distribusi Normal

 $\begin{array}{ll}\\ 11.&\textrm{Tersiar kabar bahwa harga beras di pasar di}\\ &\textrm{wilayah B adalah}\: \: \textrm{Rp}8.000,00/\textrm{Kg dengan}\\ &\textrm{simpangan baku}\: \: \textrm{Rp}1.500,00.\: \textrm{Berdasar kabar}\\ & \textrm{tersebut dilakukan penelitian dengan mengambil}\\ &\textrm{sampel secara acak sebanyak}\: \: 60\: \: \textrm{kios yang dan}\\ &\textrm{diperoleh rata-rata harga beras}\: \: \textrm{Rp}7.800,00/\textrm{Kg}\\ &\textrm{Jika penghitungan menggunakan tingkat}\\ &\textrm{signifikansi}\: \: 5\%,\: \textrm{maka kesimpulan berikut }\\ &\textrm{yang tepat adalah}\:  ....\\ &\textrm{a}.\quad \textrm{harga beras di pasar lebih dari}\: \: \textrm{Rp}7.800,00/\textrm{Kg}\\ &\textrm{b}.\quad \textrm{harga beras di pasar lebih dari}\: \: \textrm{Rp}8.000,00/\textrm{Kg}\\ &\textrm{c}.\quad \textrm{harga beras di pasar kurang dari}\: \: \textrm{Rp}0.000,00/\textrm{Kg}\\ &\textrm{d}.\quad \textrm{harga beras di pasar}\: \: \textrm{Rp}7.800,00/\textrm{Kg}\\ &\textrm{e}.\quad \color{red}\textrm{harga beras di pasar}\: \: \textrm{Rp}8.000,00/\textrm{Kg}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\&\begin{aligned}&\underline{\textrm{Hipotesis}}\\ &\textrm{Rata-rata harga beras dipasar}\: \: \textrm{Rp}8.000,00/\textrm{Kg}\\ &\textrm{H}_{0}:\mu =8.000\\ &\textrm{H}_{0}:\mu \neq 8.000\\ &\underline{\textrm{Daerah Kritis}}\\ &\textrm{Taraf nyata yang dipilih adalah}=\alpha =0,05=5\%\\ &\displaystyle \frac{\alpha }{2}=2,5\%=0,025\\ &\textrm{z}_{0,025}=1,96\\ &\textrm{maka daerah kritis/penolakannya adalah}\\ &\textrm{z}<-1,96\: \: \textrm{atau}\: \: \textrm{z}>1,96 \end{aligned}  \end{array}$.

$.\: \qquad\begin{aligned}&\underline{\textrm{Nilai Satistik Uji}}\\ &\textrm{Dihitung dengan rumus}:\: \textrm{z}=\displaystyle \frac{\overline{\textrm{x}}-\mu _{0}}{\displaystyle \frac{\sigma }{\sqrt{n}}}\\ &\textrm{z}=\displaystyle \frac{7.800-8.000}{\displaystyle \frac{1.500}{\sqrt{60}}}=-\displaystyle \frac{200\sqrt{60}}{1.500}=-1,03\\ &\underline{\textrm{Keputusan Uji}}\\ &\textrm{Karena nilai}\: \: -1,96<\textrm{z}<1,96,\\ &\textrm{maka}\: \: \textrm{H}_{0}\: \: \color{red}\textrm{diterima}\\ &\underline{\textrm{Kesimpulan}}\\ &\textrm{Rata-rata harga beras dipasar}\: \: \color{red}\textrm{Rp}8.000,00/\textrm{Kg} \end{aligned}$ .

Contoh Soal 2 Distribusi Normal

$\begin{array}{ll}\\ 6.&\textrm{Luas daerah yang diarsir di bawah}\\ &\textrm{kurva normal baku berikut adalah}\:  ....\\ &\textrm{a}.\quad 0,4861\\ &\textrm{b}.\quad \color{red}0,4878\\&\textrm{c}.\quad 0,4881\\ &\textrm{d}.\quad 0,4938\\ &\textrm{e}.\quad 0,4946\\\\   \end{array}$.

$.\qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Perhatikan tabel}\: \color{blue}\textrm{distribusi normal}\:  \color{black}\textrm{berikut}\\&\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\hline \textrm{z} &0&1&2&3&4&5&6&7&8&9\\\hline \vdots &&&&&&\downarrow &&&&\\ \vdots &&&&&&&&&&\\ 2,2 &&&&&&\color{red}0,4878&&&&\\ \vdots &&&&&&&&&&\\\hline \end{array}\\ &\textrm{Sehingga nilai}\: \: \textrm{z}=2,25\: \: \textrm{luasnya}=0,4878 \end{aligned}$.

$\begin{array}{ll}\\ 7.&\textrm{Luas daerah yang diarsir di bawah}\\ &\textrm{kurva normal baku berikut adalah}\:  ....\\ &\textrm{a}.\quad \color{red}0,1138\\ &\textrm{b}.\quad 0,3810\\&\textrm{c}.\quad 0,3862\\ &\textrm{d}.\quad 0,4948\\ &\textrm{e}.\quad 0,5000\\\\   \end{array}$.

$.\qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Perhatikan tabel}\: \color{blue}\textrm{distribusi normal}\:  \color{black}\textrm{berikut}\\&\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\hline \textrm{z} &0&1&2&3&4&5&6&7&8&9\\\hline \vdots &&&&&&&\downarrow &&\downarrow&\\ \vdots &&&&&&&&&&\\ 1,1 &&&&&&&\downarrow&&\color{red}0,3810&\\ \vdots &&&&&&&&&&\\ 2,5&&&&&&&\color{red}0,4948&&&\\ \vdots &&&&&&&&&&\\\hline \end{array}\\ &\textrm{Sehingga nilai}\: \: \textrm{z}=1,18\: \: \textrm{luasnya}=0,3810\\ &\textrm{Dan nilai}\: \: \textrm{z}=2,56\: \: \textrm{luasnya}=0,4948\\ &\textrm{maka luas arsiran}\\ &=P(1,18<Z<2,56)\\ &=P(0<Z<2,56)-P(0<Z<1,18)\\ &=0,4948-0,3810\\ &=\color{red}0,1138 \end{aligned}$.

$\begin{array}{ll}\\ 8.&\textrm{Luas daerah yang diarsir pada gambar}\\ &\textrm{di bawah adalah 0,9332, maka nilai}\: \: \textrm{z}=\:  ....\\    \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{a}.\quad 1,05\\ &\textrm{b}.\quad 1,06\\&\textrm{c}.\quad 1,16\\ &\textrm{d}.\quad \color{red}1,50\\ &\textrm{e}.\quad 1,60\\\\ &\textbf{Jawab}:\\ &\textrm{Luas arsiran adalah}\\ &=P(Z<z)=0,9332\\ &=\color{blue}0,5\color{black}+0,4332\\ &=0,5+P(0<Z<z)\\ &\textrm{lihat/konfirmasi ke tabel}\\ &\textrm{z}=\color{red}1,50\\ &\begin{aligned}&\textrm{Perhatikan tabel}\: \color{blue}\textrm{distribusi normal}\:  \color{black}\textrm{berikut}\\&\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\hline \textrm{z} &0&1&2&3&4&5&6&7&8&9\\\hline \vdots &\downarrow &&&&&&&&&\\ \vdots &&&&&&&&&&\\ 1,5 &\color{red}0,4332&&&&&&&&&\\ \vdots &&&&&&&&&&\\\hline \end{array}\\ &\textrm{Sehingga luasnya}=0,4878,\: \: \textrm{batas z}=1,50  \end{aligned}    \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Pada suatu kelas seorang guru matematika}\\ &\textrm{menyatakan bahwa nilai ulangan mapel }\\ & \textrm{yang diampunya tidak kurang dari}\: \:68\\ &\textrm{Untuk menentukan uji tersebut, maka guru}\\ &\textrm{yang bersangkutan memilih 10 anak secara}\\ &\textrm{acak untuk ditanyai nilai hasil ulangannya}\\ &\textrm{Hipotesis}\: \: \textrm{H}_{0}\: \: \textrm{dan}\: \: \textrm{H}_{1}\: \: \textrm{yang tepat dari kondisi}\\ &\textrm{kondisi di atas adalah}\:  ....\\ &\textrm{a}.\quad \begin{matrix} \textrm{H}_{0}:\mu =68\\  \textrm{H}_{1}:\mu \neq 68 \end{matrix}\\ &\textrm{b}.\quad \begin{matrix} \textrm{H}_{0}:\mu =68\\  \textrm{H}_{1}:\mu > 68 \end{matrix}\\ &\textrm{c}.\quad \color{red}\begin{matrix} \textrm{H}_{0}:\mu \geq 68\\  \textrm{H}_{1}:\mu < 68 \end{matrix}\\ &\textrm{d}.\quad \begin{matrix} \textrm{H}_{0}:\mu \leq 68\\  \textrm{H}_{1}:\mu >q 68 \end{matrix}\\ &\textrm{e}.\quad \begin{matrix} \textrm{H}_{0}:\mu >68\\  \textrm{H}_{1}:\mu \neq 68 \end{matrix}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas}\\ &\textrm{Dan ini contoh uji satu pihak, yaitu kiri}  \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Seorang petugas}\: \: customer\: service\: \: \textrm{menyatakan}\\ &\textrm{bahwa di dealer A dapat mensevis rata-rata }\\ &\textrm{sebanyak}\: \: 75\: \: \textrm{unit sepeda motor perhari}.\: \textrm{Untuk}\\ & \textrm{mengecek kebenaran pernyataan di atas diambil}\\ &\textrm{sampel beberapa hari secara random sebanyak }\\ &\textrm{20 hari. Dari penelitian ini diperoleh rata-rata}\\ &\textrm{78 unit dengan simpangan bakunya 5. Hasil}\\ &\textrm{perhitungan}\: \: \textrm{z}_{0}-\: \textrm{nya adalah}\:  ....\\ &\textrm{a}.\quad 2,35\\ &\textrm{b}.\quad 2,43\\ &\textrm{c}.\quad 2,55\\ &\textrm{d}.\quad \color{red}2,68\\ &\textrm{e}.\quad 2,75\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Diketahui data dianggap berdistribusi}\\ &\textrm{normal baku}\: N(0,1)\: \: \textrm{dengan}\\ &\textrm{Rata-rata sampel}=\overline{\textrm{x}}=\color{blue}78\: \: \color{black}\textrm{unit sepeda motor}\\ &\textrm{Rata-rata populasinya yang diuji}=\mu _{0}=75\\ &\textrm{Simpangan bakunya}=\sigma =\color{blue}5\: \: \color{black}\textrm{unit}\\ &\textrm{dengan banyak data sampel}=n=\: \color{blue}20\: \: \color{black}\textrm{hari}\\ &\textrm{Penghitungan nilainya}\: \: z-\textrm{nya}\\ &=\displaystyle \frac{\overline{\textrm{x}}-\mu _{0}}{\displaystyle \frac{\sigma }{\sqrt{n}}}\\ &=\displaystyle \frac{78-75}{\displaystyle \frac{5}{\sqrt{20}}}=\displaystyle \frac{3}{\displaystyle \frac{5}{2\sqrt{5}}}\\ &=\displaystyle \frac{6}{\sqrt{5}}=\frac{6}{2,236}=\color{red}2,683 \end{aligned}  \end{array}$.

Contoh Soal 1 Distribusi Normal

 $\begin{array}{ll}\\ 1.&\textrm{Fungsi distribusi normal variabel acak X}\\ &\textrm{dengan}\: \: \mu =8\: \: \textrm{dan}\: \: \sigma =2\: \: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \displaystyle f(x)=\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\frac{(x-8)^{2}}{2}}}\\ &\textrm{b}.\quad \displaystyle f(x)=\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\frac{(x-8)^{2}}{4}}}\\&\textrm{c}.\quad \displaystyle f(x)=\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\frac{(x-8)^{2}}{2}}}\\&\textrm{d}.\quad \displaystyle f(x)=\displaystyle \frac{1}{\sqrt{8\pi }}e^{.^{-\frac{(x-8)^{2}}{4}}}\\&\textrm{e}.\quad \color{red}\displaystyle f(x)=\displaystyle \frac{1}{\sqrt{8\pi }}e^{.^{-\frac{(x-8)^{2}}{8}}}\\\\&\textbf{Jawab}:\quad \textbf{e}\\&\begin{aligned}\displaystyle f(x)&=\displaystyle \frac{1}{\sigma \sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\left (\frac{x-\mu}{\sigma } \right )^{2}}},\: \:  \textrm{dengan}\: \: \left\{\begin{matrix} \mu =8\\ \sigma =2 \end{matrix}\right.\\&=\displaystyle \frac{1}{2 \sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\left (\frac{x-8}{2 } \right )^{2}}}\\ &=\color{red}\displaystyle \frac{1}{\sqrt{8\pi }}e^{.^{-\displaystyle \frac{(x-8)^{2}}{8}}} \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika variabel acak}\: \: Z\: \: \textrm{berdistribusi normal}\\ &\textrm{N}(0,1),\:  \textrm{nilai}\: \: \textrm{P}(Z< 2)\: \: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \displaystyle \int_{0}^{2}\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\displaystyle z^{2}}}\: dz\\ &\textrm{b}.\quad \displaystyle \int_{2}^{\infty }\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\displaystyle z^{2}}}\: dz\\ &\textrm{c}.\quad \color{red}\displaystyle \int_{-\infty }^{2}\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\displaystyle z^{2}}}\: dz\\ &\textrm{d}.\quad \displaystyle \int_{0}^{2}\displaystyle \frac{1}{\sigma \sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\left ( \displaystyle \frac{\textrm{x}-\mu }{\sigma } \right )^{2}}}\: dz\\ &\textrm{e}.\quad \displaystyle \int_{0}^{2}\displaystyle \frac{1}{\sigma \sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\left ( \displaystyle \frac{\textrm{x}-\mu }{\sigma } \right )^{2}}}\: dz\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&P(\textrm{Z}<2)\: ,\qquad \textrm{Z}\sim \textrm{N}(0,1)\\ &=P(-\infty <\textrm{Z}<0)+P(0<\textrm{Z}<2)\\ &=\color{red}\displaystyle \int_{-\infty }^{2}\displaystyle \frac{1}{\sqrt{2\pi }}e^{.^{-\displaystyle \frac{1}{2}\displaystyle z^{2}}}\: dz \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Jika luas daerah di bawah kurva}\\ &\textrm{berdistribusi normal pada interval}\: \: \textrm{Z}>z\\ &\textrm{adalah}\: \: L,\: \: \textrm{nilai}\: \: \textrm{P}(-z<Z< z)\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \displaystyle 0,5+L\\ &\textrm{b}.\quad 0,5-L\\ &\textrm{c}.\quad \displaystyle 1-L\\ &\textrm{d}.\quad \color{red}\displaystyle 1-2L\\ &\textrm{e}.\quad \displaystyle 2L\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}P&(-z<Z<z)\\ &=0,5-L+0,5-L\\ &=\color{red}1-2L\\ &\textrm{Berikut ilustrasi kurva beserta luasnya} \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui}\: \: \textrm{X}\: \: \sim \textrm{N}(20,4)\: \: \textrm{dan}\: \:  Z\sim N(0,1)\\ &\textrm{Jika}\: \: P(0<Z<1)=0,3413,\: \: \textrm{maka nilai}\\ &P(X<24)\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \displaystyle 0,1587\\ &\textrm{b}.\quad \displaystyle 0,3174\\ &\textrm{c}.\quad \displaystyle 0,3413\\ &\textrm{d}.\quad \displaystyle 0,6826\\ &\textrm{e}.\quad \color{red}\displaystyle 0,8413\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: X\sim N(20,4)\begin{cases} \mu  & =20 \\  \sigma  & =4  \end{cases}\\ &\textrm{Dan diketahui pula}\: \: P(0<Z<1)=0,3413\\ &\textrm{Jika}\: \: Z\sim N(0,1),\: \: \textrm{maka untuk}\: P(X<24)\\ &\textrm{transformasi}\: \: \textrm{x}=24\: \: \textrm{menjadi}\\ &\textrm{z}=\displaystyle \frac{\textrm{x}-\mu }{\sigma }=\frac{24-20}{4}=\frac{4}{4}=1\\ &\textrm{Selanjutnya}\\ &\begin{aligned}P(X<24)&=P(Z<1)\\ &=0,5+P(0<Z<1)\\ &=0,5+0,3413\\ &=\color{red}0,8413 \end{aligned} \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Nilai kuartil atas dari data}\\ &\textrm{berdistribusi normal baku}=q\\ & \textrm{Pernyataan berikut yang tepat adalah}\: ....\\ &\textrm{a}.\quad \color{red}\textrm{Luas daerah pada}\: (Z<q)=0,25\\ &\textrm{b}.\quad \textrm{Luas daerah pada}\: (Z>q)=0,25\\ &\textrm{c}.\quad \textrm{Luas daerah pada}\: (0<Z<q)=0,25\\ &\textrm{d}.\quad \textrm{Luas daerah pada}\: (Z<-0,25)=q\\ &\textrm{e}.\quad \textrm{Luas daerah pada}\: (0<Z<0,25)=q\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Pembahasan diserahkan kepada pembaca}\\ &\textrm{yang budiman}  \end{array}$.