Lanjutan Materi Fungsi Trigonometri dan Grafiknya

F. 2 Garfik Fungsi Trigonometri

F. 2. 1 Grafik Fungsi Sinus

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

F. 2. 2 Grafik Fungsi Cosinus

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

F. 2. 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 Cotangennya ada beberapa nilai fungsinya yang tidak terdefinisi. Dalam fungsi Tangen fungsi, nilai fungsi yang tidak terdefini terdapat pada 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 bantu berupa garis putus-putus pada grafik yang dan ditampakkan berupa garis vertikal yang selanjutnya garis vertikal itu disebut sebagai asimtot.

F. 2. 4 Menggambar Grafik Fungsi Trigonometri

$\begin{aligned}&\textrm{untuk bentuk}\\ &f(x)=\begin{cases} y &=a\sin bx+c \\  y &=a\cos bx+c \\  y & =a\tan bx+c  \end{cases}\\ &\begin{array}{|c|l|l|}\hline 1.&a&\textrm{Amplitudo}\\\hline 2.&b&\textrm{Periode}\\\hline 3.&c&\textrm{Geseran}\\\hline \end{array}  \end{aligned}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Gambarlah grafik fungsi berikut} \\ &\textrm{jika}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{a}.\quad f(x)=-2\sin x\\ &\textrm{b}.\quad f(x)=3\cos x\\ &\textrm{c}.\quad f(x)=\displaystyle \frac{1}{2}\sin x\\ &\textrm{d}.\quad f(x)=4\cos x\\ &\textrm{e}.\quad f(x)=2\tan x\\\\&\color{blue}\textrm{Jawab}:\\&\begin{aligned}&  \end{aligned} \end{array}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 a}\\ &y=f(x)=-2\sin x=a\sin bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |-2  \right |=2\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array}  \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 b}\\ &y=f(x)=3\cos x=a\cos bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |3  \right |=3\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array}  \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 c}\\ &y=f(x)=\displaystyle \frac{1}{2}\sin x=a\sin bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle \frac{1}{2}  \right |=\displaystyle \frac{1}{2}\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array}  \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 d}\\ &y=f(x)=\displaystyle 4\cos x=a\cos bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle 4  \right |=\displaystyle 4\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array}  \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.1 e}\\ &y=f(x)=\displaystyle 2\tan x=a\tan bx+c\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle 2  \right |=\displaystyle 2\\\hline 2.&b&\textrm{Periode}&\displaystyle \frac{2\pi }{b}=2\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array}  \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Gambarlah grafik fungsi berikut} \\ &\textrm{jika}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{a}.\quad f(x)=\left |-2\sin x  \right |\\ &\textrm{b}.\quad f(x)=\left |3\cos x  \right |\\ &\textrm{c}.\quad f(x)=\left |\displaystyle \frac{1}{2}\sin x  \right |\\ &\textrm{d}.\quad f(x)=\left |4\cos x  \right |\\ &\textrm{e}.\quad f(x)=\left |2\tan x  \right |\\\\&\color{blue}\textrm{Jawab}:\\&\begin{aligned}&  \end{aligned} \end{array}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.2 a}\\ &y=f(x)=\left |-\displaystyle 2\sin x  \right |=\left |a\sin bx+c  \right |\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle 2  \right |=\displaystyle 2\\\hline 2.&b&\textrm{Periode}&\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array}  \end{aligned}$.

$.\: \qquad\begin{aligned}&\color{blue}\textrm{No.2 b}\\ &y=f(x)=\left |\displaystyle 3\cos x  \right |=\left |a\cos bx+c  \right |\\ &\begin{array}{|c|l|l|l|}\hline 1.&a&\textrm{Amplitudo}&\left |\displaystyle 3  \right |=\displaystyle 3\\\hline 2.&b&\textrm{Periode}&\pi \Leftrightarrow b=1\\\hline 3.&c&\textrm{Geseran}&0\\\hline \end{array}  \end{aligned}$.

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

Silahkan selesaikan soal yg belum dibahas

DAFTAR PUSTAKA

1. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

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.

Koordinat Kartesius dan Koordinat Kutub serta Kuadran

C. Koordinat Kartesius dan Koordinat Kutub/Polar

Perhatikan ilustrasi berikut

$\begin{array}{|c|c|}\hline  \textrm{Kartesius}\: \rightarrow \: \textrm{Kutub}&\textrm{Kutub}\: \rightarrow \: \textrm{Kartesius}\\\hline P(x,y)\: \rightarrow \: P\left ( r,\alpha ^{0} \right )&P\left ( r,\alpha ^{0} \right )\: \rightarrow \: P(x,y)\\\hline \begin{aligned}&r=\sqrt{x^{2}+y^{2}},\\ &\displaystyle \tan \alpha ^{0}=\frac{y}{x},\quad \displaystyle \alpha ^{0}=\arctan \frac{y}{x} \end{aligned}&\left\{\begin{matrix} x=r.\cos \alpha ^{0}\\ \\ y=r.\sin \alpha ^{0} \end{matrix}\right.\\\hline \end{array}$.

D. Perbandingan Trigonometri di Berbagai Kuadran

D. 1 Untuk Sudut 0 sampai dengan 360 derajat.

$\begin{array}{ccc|cccc} \textrm{Kuadran II}&&&&\textrm{Kuadran I}&\\ \left (180^{\circ}-\alpha \right )&&&&\textrm{Semua nilai trigon}&\color{blue}\textbf{positif}\\ &&&&&\\\hline \textrm{Kuadran III}&&&&\textrm{Kuadran IV}&\\ \left (180^{\circ}+\alpha \right )&&&&\left (360^{\circ}-\alpha \right )& \\ \end{array}$.

$\begin{array}{llll} (1).&\textrm{Perbandingan Trigonometri untuk Sudut}\: \left ( 90^{0}-\alpha \right )\\ &\textrm{a}.\quad \sin \alpha \rightarrow \sin \left ( 90^{0}-\theta \right )=\cos \theta \\ &\textrm{b}.\quad \cos \alpha \rightarrow \cos \left ( 90^{0}-\theta \right )=\sin \theta \\ &\textrm{c}.\quad \tan \alpha \rightarrow \tan \left ( 90^{0}-\theta \right )=\cot \theta \\ &\textrm{d}.\quad \cot \alpha \rightarrow \cot \left ( 90^{0}-\theta \right )=\tan \theta \end{array}$.

$\begin{array}{llll} (2).&\textrm{Perbandingan Trigonometri untuk Sudut}\: \left ( 180^{0}-\alpha \right )\\ &\textrm{a}.\quad \sin \alpha \rightarrow \sin \left ( 180^{0}-\theta \right )=\sin \theta \\ &\textrm{b}.\quad \cos \alpha \rightarrow \cos \left ( 180^{0}-\theta \right )=-\cos \theta \\ &\textrm{c}.\quad \tan \alpha \rightarrow \tan \left ( 180^{0}-\theta \right )=-\tan \theta \\ &\textrm{d}.\quad \cot \alpha \rightarrow \cot \left ( 180^{0}-\theta \right )=-\cot \theta \end{array}$.

$\begin{array}{llll} (3).&\textrm{Perbandingan Trigonometri untuk Sudut}\: \left ( 180^{0}+\alpha \right )\\ &\textrm{a}.\quad \sin \alpha \rightarrow \sin \left ( 180^{0}+\theta \right )=-\sin \theta \\ &\textrm{b}.\quad \cos \alpha \rightarrow \cos \left ( 180^{0}+\theta \right )=-\cos \theta \\ &\textrm{c}.\quad \tan \alpha \rightarrow \tan \left ( 180^{0}+\theta \right )=\tan \theta \\ &\textrm{d}.\quad \cot \alpha \rightarrow \cot \left ( 180^{0}+\theta \right )=\cot \theta \end{array}$.

$\begin{array}{llll} (4).&\textrm{Perbandingan Trigonometri untuk Sudut}\: \left ( 360^{0}-\alpha \right )\\ &\textrm{a}.\quad \sin \alpha \rightarrow \sin \left ( 360^{0}-\theta \right )=-\sin \theta \\ &\textrm{b}.\quad \cos \alpha \rightarrow \cos \left ( 360^{0}-\theta \right )=\cos \theta \\ &\textrm{c}.\quad \tan \alpha \rightarrow \tan \left ( 360^{0}-\theta \right )=-\tan \theta \\ &\textrm{d}.\quad \cot \alpha \rightarrow \cot \left ( 360^{0}-\theta \right )=-\cot \theta \end{array}$.

D. 2 Untuk Sudut yang Lain

$\begin{aligned}\textrm{a}.\quad&\begin{cases} \sin \left ( -A \right ) & =-\sin A \\ \cos \left ( -A \right ) & =\cos A \\ \tan \left ( -A \right ) & = -\tan A \end{cases}\\ \textrm{b}.\quad&\begin{cases} \csc \left ( -A \right ) &=-\csc A \\ \sec \left ( -A \right ) &=\sec A \\ \cot \left ( -A \right ) &=-\cot A \end{cases}\\ \textrm{c}.\quad&\begin{cases} \sin \left ( n.360^{\circ}+A \right ) & =\sin A \\ \cos \left ( n.360^{\circ}+A \right ) & =\cos A \\ \tan \left ( n.360^{\circ}+A \right ) & =\tan A \end{cases},\qquad n\in \mathbb{N} \end{aligned}$.

Dengan catata: $0^{\circ}=360^{\circ}  =720^{\circ}=1080^{\circ}=n.360^{\circ}$

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

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai}\\ &\textrm{a}.\quad\sin 120^{\circ}\\ &\textrm{b}.\quad\cos 240^{\circ}\\ &\textrm{c}.\quad\tan 315^{\circ}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad\sin 120^{\circ}&=\sin \left ( 180^{\circ}-60^{\circ} \right )=\sin 60^{\circ}\\ &=\displaystyle \frac{1}{2}\sqrt{3},\qquad \color{red}\textrm{atau}\\ &=\sin \left ( 90^{\circ}+30^{\circ} \right )=\cos 30^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.\quad\cos 240^{\circ}&=\cos \left ( 180^{\circ}+60^{\circ} \right) =-\cos 60^{\circ}\\ &=-\displaystyle \frac{1}{2},\qquad \color{red}\textrm{atau}\\ &=\cos \left ( 270^{\circ}-30^{\circ} \right )=-\sin 30^{\circ}=-\frac{1}{2}\\ \textrm{c}.\quad\tan 315^{\circ}&=\tan \left ( 360^{\circ}-45^{\circ} \right )=-\tan 45^{\circ}\\ &=-1,\qquad \color{red}\textrm{atau}\\ &=\tan \left ( 270^{\circ}+45^{\circ} \right )=-\cot 45^{\circ}=-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Buktikan bahwa}\\\\ &\textrm{a}.\quad \displaystyle \frac{\cos \left ( 90^{\circ}-B \right )}{\sec B}+\frac{\sin \left ( 90^{\circ}-B \right )}{\csc B}=2\sin B\cos B\\\\ &\textrm{b}.\quad \tan C+\tan \left ( 90^{\circ}-C \right )=\sec C.\sec \left ( 90^{\circ}-C \right )\\\\ &\textbf{Bukti}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{\cos \left ( 90^{\circ}-B \right )}{\sec B}+\frac{\sin \left ( 90^{\circ}-B \right )}{\csc B}\\ &=\displaystyle \frac{\sin B}{\sec B}+\frac{\cos B}{\csc B}\\ &=\displaystyle \frac{\sin B}{\displaystyle \frac{1}{\cos B}}+\frac{\cos B}{\displaystyle \frac{1}{\sin B}}\\ &=\sin B\cos B+\sin B\cos B\\ &=2\sin B\cos B\qquad\quad \blacksquare \end{aligned} \\ &\begin{aligned}\textrm{b}.\quad&\tan C+\tan \left ( 90^{\circ}-C \right )\\ &=\tan C+\cot C\\ &=\displaystyle \frac{\sin C}{\cos C}+\frac{\cos C}{\sin C}\\ &=\displaystyle \frac{\sin ^{2}C+\cos ^{2}C}{\sin C\cos C}=\displaystyle \frac{1}{\sin C\cos C}\\ &=\displaystyle \frac{1}{\cos C}.\frac{1}{\sin C}\\ &=\sec C.\csc C\\ &=\sec C.\sec \left ( 90^{\circ}-C \right )\qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah nilai}\\ &\textrm{a}.\quad \tan \left ( A-90^{\circ} \right )\sin \left ( -A \right )\\ &\textrm{b}.\quad\cos 540^{\circ}+\sin 690^{\circ}\\ &\textrm{c}.\quad \sin 2021^{\circ}+\cos 2021^{\circ}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\tan \left ( A-90^{\circ} \right )\sin \left ( -A \right )\\ &=\tan \left ( -\left (90^{\circ}-A \right ) \right )\left ( -\sin A \right )\\ &=-\tan \left ( 90^{\circ}-A \right )\left ( -\sin A \right )\\ &= \tan \left ( 90^{\circ}-A \right )\left ( \sin A \right )\\ &=\cot A.\sin A\\ &=\displaystyle \frac{\cos A}{\sin A}.\sin A\\ &=\cos A \end{aligned} \\ &\begin{aligned}\textrm{b}.\quad&\cos 540^{\circ}+\sin 690^{\circ}\\ &=\cos \left ( 360^{\circ}+180^{\circ} \right )+\sin \left ( 720^{\circ}-30^{\circ} \right )\\ &=\cos \left ( 0^{\circ}+180^{\circ} \right )+\sin \left ( 0^{\circ}-30^{\circ} \right )\\ &=\cos 180^{\circ}+ \sin \left ( -30^{\circ} \right ) \\ &=\cos 180^{\circ}-\sin 30^{\circ}\\ &=-1-\displaystyle \frac{1}{2}\\ &=-\displaystyle \frac{3}{2} \end{aligned} \\ &\begin{aligned}\textrm{c}.\quad&\sin 2021^{\circ}+\cos 2021^{\circ}\\ &=\sin \left ( 5.360^{\circ}+221^{\circ} \right )+\cos \left ( 5.360^{\circ}+221^{\circ} \right )\\ &=\sin \left (0^{\circ}+221^{\circ} \right )+\cos \left (0^{\circ}+221^{\circ} \right )\\ &=\sin 221^{\circ}+\cos 221^{\circ}\\ &=\sin \left ( 180^{\circ}+41^{\circ} \right )+\cos \left ( 180^{\circ}+41^{\circ} \right )\\ &=-\sin 41^{\circ}-\cos 41^{\circ} \end{aligned} \end{array}$

Perbandingan Trigonometri Sudut dalam Segitiga Siku-Siku

B. Perbandingan Trigonometri Suatu Sudut

B. 1 Perbandingan Trigonometri pada Segitiha Siku-Siku

Perhatikanlah ilustrasi  dari segitiga ABC siku-siku di C berikut

$\begin{matrix} \sin \alpha =\displaystyle \frac{BC}{AB}\qquad\Leftrightarrow\quad \csc \alpha =\displaystyle \frac{AB}{BC}=\color{red}\displaystyle \frac{1}{\sin \alpha }\\\\ \cos \alpha =\displaystyle \frac{AC}{AB}\qquad\Leftrightarrow \quad \sec \alpha =\displaystyle \frac{AB}{AC}=\color{red}\displaystyle \frac{1}{\cos \alpha }\\\\ \tan \alpha =\displaystyle \frac{BC}{AC}\qquad\Leftrightarrow \quad \cot \alpha =\displaystyle \frac{AC}{BC}=\color{red}\displaystyle \frac{1}{\tan \alpha }  \end{matrix}$.

B. 2 Perbandingan Trigonometri untuk Sudut Istimewa.

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

B. 3. Tripel Pythagoras pada Segitiga Siku-Siku

$\begin{array}{|c|c|c|c|}\hline a&b&c=\sqrt{a^{2}+b^{2}}&\color{red}\textrm{Tripel}\\\hline 3&4&5&(3,4,5)\\\hline 5&12&13&(5,12,13)\\\hline 7&24&25&(7,24,25)\\\hline 8&15&17&(8,15,17)\\\hline 9&40&41&(9,40,41)\\\hline 11&60&61&(11,60,61)\\\hline 12&35&37&(12,35,37)\\\hline 15&112&113&(15,112,113)\\\hline 16&63&65&(16,63,65)\\\hline 17&144&145&(17,144,145)\\\hline 19&180&181&(19,180,181)\\\hline 20&21&29&(20,21,29)\\\hline 21&220&221&(21,220,221)\\\hline 29&420&421&(29,420,421)\\\hline 30&224&226&(30,224,226)\\\hline 31&480&481&(31,480,481)\\\hline 33&544&545&(33,544,545)\\\hline 35&612&613&(35,612,613)\\\hline 37&684&685&(37,684,685)\\\hline 39&760&761&(39,760,761)\\\hline 41&840&841&(41,840,841)\\\hline 43&924&925&(43,924,925)\\\hline 45&1012&1013&(45,1012,1013)\\\hline 47&1104&1105&(47,1104,1105)\\\hline 48&55&73&(48,55,73)\\\hline 49&1200&1201&(49,1200,1201)\\\hline 51&1300&1301&(51,1300,1301)\\\hline 60&63&87&(60,63,87)\\\hline \end{array}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai perbandingan}\: \:  \color{red}\sin \alpha \\ & \color{red}\cos \alpha , \tan \alpha  \: \: \color{black}\textrm{untuk segitiga berikut} \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Untuk sisi miringnya adalah}\\ &\begin{aligned}\textrm{Sisi  miring}&=\sqrt{3^{2}+4^{2}}\\ &=\sqrt{9+16}\\ &=\sqrt{25}\\ &=5 \end{aligned}\\ &\textrm{Sehingga}\\ &\color{red}\sin \alpha =\displaystyle \frac{3}{5},\: \: \:  \cos \alpha =\frac{4}{5}\: \:  \color{black}\textrm{dan}\: \color{red}\tan \alpha =\frac{3}{4}\\ &\textrm{Dengan langkah sama}\\ \end{aligned}\\&\begin{aligned}\textrm{b}.\quad&\textrm{Untuk sisi miringnya adalah}\\&\begin{aligned}\textrm{Sisi  miring}&=\sqrt{5^{2}+12^{2}}\\&=\sqrt{25+144}\\&=\sqrt{169}\\ &=13 \end{aligned}\\ &\textrm{Sehingga}\\ &\color{red}\sin \alpha =\frac{12}{13}\: \:  \cos \alpha =\frac{5}{13}\: \:  \color{black}\textrm{dan}\: \color{red}\tan \alpha =\frac{12}{5}. \end{aligned}  \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai dari}\\ &\begin{array}{llll} \textrm{a}.&\left ( \tan 30^{0}+\sin 30^{0} \right )\cos 30^{0}\\ \textrm{b}.&\left ( \tan 60^{0} \right )^{2}+4\left ( \sin 60^{0} \right )^{2}\\ \textrm{c}.&\tan 60^{0}-\sin 60^{0}-\tan 30^{0}\\ \textrm{d}.&\displaystyle \frac{1+\sin 30^{0}}{\sin 30^{0}}+\frac{\cos 30^{0}}{1+\sin 30^{0}}\\ \textrm{e}.&\displaystyle \frac{2\tan 30^{0}}{1+\tan ^{2}30^{0}} \end{array}\\\\ &\textbf{Jawab}:\\ \end{array}$.

$.\: \qquad\begin{array}{|l|l|}\hline \begin{array}{llll} \textrm{a}.&\left ( \tan 30^{0}+\sin 30^{0} \right )\cos 30^{0}\\ &=\displaystyle \left ( \frac{1}{3}\sqrt{3}+\frac{1}{2} \right ).\frac{1}{2}\sqrt{3}\\ &=\displaystyle \frac{1}{2}+\frac{1}{4}\sqrt{3} \end{array}&\begin{array}{llll} \textrm{b}.&\left ( \tan 60^{0} \right )^{2}+4\left ( \sin 60^{0} \right )^{2}\\ &\displaystyle =\left ( \sqrt{3} \right )^{2}+4\left ( \frac{1}{2}\sqrt{3} \right )^{2}\\ &\displaystyle =3+3\\ &=6 \end{array}\\\hline \begin{array}{llll} \textrm{c}.&\tan 60^{0}-\sin 60^{0}-\tan 30^{0}\\ &=\displaystyle \sqrt{3}-\frac{1}{2}\sqrt{3}-\frac{1}{3}\sqrt{3}\\ &=\displaystyle \sqrt{3}-\frac{5}{6}\sqrt{3}\\&=\displaystyle \frac{1}{6}\sqrt{3} \end{array}&\begin{array}{llll} \textrm{d}.&\displaystyle \frac{1+\sin 30^{0}}{\sin 30^{0}}+\frac{\cos 30^{0}}{1+\sin 30^{0}}\\ &=\displaystyle \frac{1+\frac{1}{2}}{\frac{1}{2}}+\frac{\frac{1}{2}\sqrt{3}}{1+\frac{1}{2}}\\ &=\displaystyle 3+\frac{1}{3}\sqrt{3} \end{array}.\\\hline \begin{array}{llll} \textrm{e}.&\displaystyle \frac{2\tan 30^{0}}{1+\tan ^{2}30^{0}}\\ &=\displaystyle \frac{2\times \frac{1}{3}\sqrt{3}}{1+\left ( \frac{1}{3}\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{\frac{2}{3}\sqrt{3}}{1+\frac{1}{3}}\\ &=\displaystyle \frac{\frac{2}{3}\sqrt{3}}{\frac{4}{3}}\\ &=\displaystyle \frac{1}{2}\sqrt{3} \end{array}.&\\\hline \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Perhatikanlah ilustrasi berikut}\\ \end{array}$

$.\: \qquad\begin{aligned}&\textrm{Jika Jarak antara kucing seorang pencatat}\\ &\textrm{dan kucing adalah 100 m, maka jarak}\\ &\textrm{Pencatat tersebut dengan seorang tentara}\\ &\textrm{sebagaimana gambar tersebut di atas adalah}?\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan gambar di atas dengan diberikan}\\ &\textrm{tambahan keterangan sebagai berikut} \end{aligned}$.

$.\: \qquad\begin{aligned}&\textrm{Ditanya berpakah  panjang jarak } \left ( x+100 \right ) ?\\ &\begin{aligned}\quad&\textrm{Pilih}\: :\qquad  \color{blue}y=y\\ &\Leftrightarrow x.\tan 60^{0}=\left ( x+100 \right ).\tan 30^{0}\\ &\Leftrightarrow x.\sqrt{3}=\left ( x+100 \right )\frac{1}{3}\sqrt{3}\\ &\Leftrightarrow 3x=x+100\\ &\Leftrightarrow 2x=100\\ &\Leftrightarrow x=50 \end{aligned}\\ &\textrm{Jadi}\: \:   x+100=50+100=\color{red}150 \: \: \color{black}\textrm{meter}. \end{aligned}$.

$\begin{array}{ll}\\ 4. &\textrm{Tentukanlah perbandingan trigonometri}\\ &\quad  \angle XOA\: \:   \textrm{jika}\: \:  A(3,5)\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.

$\: \qquad\begin{aligned}&\textrm{Dengan memandang ilustrasi gambar di atas}\\ &\textrm{kita mendapatkan}\: \:   \triangle OAA',\: \:  \textrm{dengan menggunakan}\\ &\textrm{teorema pythagoras kita mendapatkan}\\ &\begin{aligned}OA^{2}&=\left ( OA' \right )^{2}+\left ( AA' \right )^{2}\\ &=3^{2}+5^{2}\\ &=9+25\\ &=34\\ OA&=\sqrt{34} \end{aligned}.\\ &\textrm{Sehingga akan didapatkan}\\ &\begin{array}{llll} \textrm{a}.&\displaystyle \sin A'OA=\frac{5}{\sqrt{34}}=\frac{5}{34}\sqrt{34}\\ \textrm{b}.&\displaystyle \cos A'OA=\frac{3}{\sqrt{34}}=\frac{3}{34}\sqrt{34}\\ \textrm{c}.&\displaystyle \tan A'OA=\frac{5}{3} \end{array}. \end{aligned}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai dari}\\ &\begin{array}{llll} \textrm{a}.&\left ( \tan 60^{0}+\sin 45^{0} \right )\cos 0^{0}\\ \textrm{b}.&\left ( \tan 30^{0} \right )^{2}+4\left ( \sin 30^{0} \right )^{2}\\ \textrm{c}.&\tan 60^{0}+\sin 60^{0}+\tan 30^{0}\\ \textrm{d}.&\displaystyle \frac{1+\sin 45^{0}}{\sin 30^{0}}+\frac{\cos 30^{0}}{1+\sin 45^{0}}\\ \textrm{e}.&\displaystyle \frac{2\tan 60^{0}}{1+\tan ^{2}60^{0}} \end{array} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukan nilai perbandingan trigonometrinya}\\ &\textrm{dari gambar segitiga berikut} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: \tan \alpha =0,75.\: \: \textrm{Tentukan nilai}\\ &\textrm{perbandingan trigonometri berikut}\\ &\begin{array}{llll} \textrm{a}.&\sin \alpha \\ \textrm{b}.&\cos \alpha \\ \textrm{c}.&\csc \alpha \\ \textrm{d}.&\sec \alpha \\ \textrm{e}.&\cot \alpha \\ \textrm{f}.&\sin^{2} \alpha+\cos ^{2}\alpha  \\ \end{array} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: 1-\cos \beta  =\displaystyle \frac{9}{25(1+\cos \beta )}\: \: \textrm{Tentukan}\\ &\textrm{nilai dari}\: \: \: \tan \beta  \end{array}$.

DAFTAR PUSTAKA

1. Kurnianingsih, S., Kuntarti, Sulistiyono. 2007. Matematika SMA dan MA untuk Kelas X Semester 2 Standar Isi 2006. Jakarta: ESIS.
2. Marwanta, dkk. 2013. Matematika SMA Kelas X. Jakarta: YUDHISTIRA.
3. Yuana. R.A., Indriyastuti. 2017. Persepktif Matematika untuk Kelas X SMA dan MA Kelompok Matematika Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Trigonometri

 A. Ukuran Sudut

Sudut itu sendiri adalah suatu bangun yang dibatasi oleh dua buah sinar atau garis yang berpotongan di sekitar titik potongnya. Sebagai ilustrasinya perhatikanlah gambar berikut

Ada dua buah sinar yaitu  $\overrightarrow{OA}$  dan  $\overrightarrow{OB}$ yang bertemu atau berpotongan di titik $O$ den terbentuklah  $\angle AOB$.

Sedangkan berkaitan dengan pengukuran sudut nantinya minimal kita mengenal 3 jenis, yaitu: derajat, radian, dan gone atau grade. Berikut hubungan ketiga jenis ikuran sudut yang dimaksud beserta lambang/notasi penulisannya. Untuk anak ditingkat MA/SMA ukuran sudut yang dikenalkan biasanya derajat dan radian, akan tetapi untuk anak SMK ada satu lagi yaitu satuan gone atau grade.

$\begin{array}{|c|}\hline 1\: \: \textrm{keliling}\: \: \bigcirc =360^{0}=2\pi \: \: \textrm{radian}=400^{g}\\\hline \color{red}\textrm{atau}\\\hline \displaystyle \frac{1}{2}\: \: \textrm{keliling}\: \: \bigcirc =180^{0}=\pi \: \textrm{radian}=200^{9}\\\hline \color{red}\textrm{boleh juga}\\\hline \displaystyle \frac{1}{4}\: \: \textrm{keliling}\: \: \bigcirc =90^{0}=\displaystyle \frac{1}{2}\pi \: \textrm{radian}=100^{9}\\\hline \color{red}\textrm{Jika ingin diperkecil lagi}\\\hline \displaystyle \frac{1}{8}\: \: \textrm{keliling}\: \: \bigcirc =45^{0}=\displaystyle \frac{1}{4}\pi \: \textrm{radian}=50^{9}\\\hline   \end{array}$.

Perhatikan ilustrasi berikut

Sebagai catatan, terkadang sudut dalam satuan derajat dipresentasikan dalam bentuk seksagesimal, yaitu: diubah dalam satuan menit dan detik.

$\begin{array}{|l|}\hline \bullet \quad \displaystyle 1^{\circ}={60}'={3600}''\\\hline \bullet \quad {1}'={60}''\\\hline \end{array}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Ubahlah sudut berikut ke satuan}\\ & \textrm{yang diminta}\\ &\textrm{a}.\quad 1^{\circ}=....\: \textrm{rad}\\ &\textrm{b}.\quad 1\: \textrm{rad}=....\: ^{0}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}& \begin{aligned}\textrm{a}.\quad\qquad 360^{0}&=2\pi \: \textrm{rad}\\ 360\times 1^{0}&=2\pi \: \textrm{rad}\\ 1^{0}&=\displaystyle \frac{2\pi }{360}\\ &=\displaystyle \frac{\pi }{180}\: \textrm{rad} \end{aligned}\\&\begin{aligned}\textrm{b}.\quad\qquad 2\pi \: \textrm{rad}&=360^{0}\\ 2\pi \times 1\: \textrm{rad}&=360^{0}\\ 1\: \textrm{rad}&=\left ( \displaystyle \frac{360}{2\pi } \right )^{0}\\ &=\left ( \displaystyle \frac{180}{\pi } \right )^{0} \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Ubahlah sudut berikut ke satuan }\\ &\textrm{yang diminta}\\&\textrm{a}.\quad 53,24^{\circ}=....\: \textrm{(dalam sexagesimal)}\\ &\textrm{b}.\quad 23^{\circ}{12}'{24}''=....\: ^{0}\\\\&\textbf{Jawab}:\\&\begin{aligned}\textrm{a}.\quad53,24^{0}&=53^{0}+0,24^{0}\\ &=53^{0}+0,24\times 1^{0}\\ &=53^{0}+0,24\times {60}'\\ &=53^{0}+{14,4}'\\ &=53^{0}+{14}'+{0,4}'\\ &=53^{0}+{14}'+0,4\times {1}'\\ &=53^{0}+{14}'+0,4\times {60}''\\ &=53^{0}+{14}'+{24}''\\ 53,24^{0}&=53^{0}{14}'{24}'' \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad 23^{0}{12}'{24}''&=23^{0}+12\times {1}'+24\times {1}''\\ &=23^{0}+12\times \left ( \frac{1}{60} \right )^{0}+24\times \left ( \frac{1}{3600} \right )^{0}\\&=23^{0}+0,2^{0}+0,00\overline{666}^{0}\\ &=23,20\overline{666}^{0}\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Ubahlah ke dalam sudut-sudut berikut dalam radian}\\ &\begin{array}{llllllll}\\ \textrm{a}.&3^{0}&\textrm{e}.&90^{0}&\textrm{i}.&210^{0}&\textrm{m}.&300^{0}\\ \textrm{b}.&15^{0}&\textrm{f}.&120^{0}&\textrm{j}.&225^{0}&\textrm{n}.&315^{0}\\ \textrm{c}.&30^{0}&\textrm{g}.&135^{0}&\textrm{k}.&240^{0}&\textrm{0}.&12^{0}{24}'\\ \textrm{d}.&60^{0}&\textrm{h}.&150^{0}&\textrm{l}.&270^{0}&\textrm{p}.&30^{0}{12}'{30}'' \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bahwa}:\quad 180^{0}=180\times 1^{0}=\pi \: rad\Rightarrow 1^{0}=\displaystyle \frac{\pi }{180}\: rad \\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad 3^{0}&=\cdots \: \pi \: rad\\ 3^{0}&=\displaystyle \frac{3}{180}\pi \: rad\\ &=\displaystyle \frac{1}{60}\pi \: rad \end{aligned}&\begin{aligned}\textrm{m}.\quad 300^{0}&=\cdots \: \pi \: rad\\ 300^{0}&=\displaystyle \frac{300}{180}\pi \: rad\\ &=\displaystyle \frac{5}{3}\pi \: rad\\  \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad 30^{0}&=\cdots \: \pi \: rad\\ 30^{0}&=\displaystyle \frac{30}{180}\pi \: rad\\ &=\displaystyle \frac{1}{6}\pi \: rad\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{o}.\quad 12^{0}{24}'&=\cdots \: \pi \: rad\\ 12^{0}{24}'&=\displaystyle \frac{12+\left ( \displaystyle \frac{24}{60} \right )}{180}\pi \: rad\\ &=\displaystyle \frac{12+0,4}{180}\pi \: rad\\ &=\displaystyle \frac{12,4}{180}\pi \: rad\\ &=\displaystyle \frac{31}{450}\pi \: rad \end{aligned}\\\hline \begin{aligned}\textrm{f}.\quad 120^{0}&=\cdots \: \pi \: rad\\ 120^{0}&=\displaystyle \frac{120}{180}\pi \: rad\\ &=\displaystyle \frac{2}{3}\pi \: rad \end{aligned}&\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Ubahlah ke dalam sudut-sudut berikut dalam derajat}\\ &\begin{array}{llllllll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\pi \: rad&\textrm{e}.&\displaystyle \frac{4}{3}\pi \: rad\\\\ \textrm{b}.&\displaystyle \frac{3}{5}\pi \: rad&\textrm{f}.&2 \: rad\\\\ \textrm{c}.&\displaystyle \frac{2}{9}\pi \: rad&\textrm{g}.&12\: rad\\\\ \textrm{d}.&\displaystyle \frac{7}{12}\pi \: rad&\textrm{h}.&15\pi \: rad \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bahwa}:\quad \pi \: rad=180^{0}\Rightarrow 1\: rad=\displaystyle \frac{180^{0}}{\pi } \\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \displaystyle \frac{1}{2}\pi \: rad&=\displaystyle \frac{1}{2}\pi \times \displaystyle \frac{180^{0}}{\pi }\\ &=90^{0} \end{aligned}&\begin{aligned}\textrm{b}.\quad \displaystyle \frac{3}{5}\pi \: rad&=\displaystyle \frac{3}{5}\pi \times \displaystyle \frac{180^{0}}{\pi }\\ &=108^{0} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad \displaystyle \frac{7}{12}\pi \: rad&=\displaystyle \frac{7}{12}\pi \times \displaystyle \frac{180^{0}}{\pi }\\ &=105^{0}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{f}.\quad 2\: rad&=2\times \displaystyle \frac{180^{0}}{\pi }\\ &=2\times \displaystyle \frac{180^{0}}{\frac{22}{7}}\\ &=\displaystyle \frac{1260^{0}}{11}\\ &=114,\overline{54}\, ^{0} \end{aligned}\\\hline  \end{array} \end{array}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Ubahlah sudut berikut ke satuan }\\ &\textrm{yang diminta}\\ &\textrm{a}.\quad27^{0}=.....\: \textrm{rad}\\ &\textrm{b}.\quad 28\: \textrm{rad}=....^{0}\\  &\textrm{c}.\quad 31,35^{\circ}=....\: \textrm{(dalam sexagesimal)}\\ &\textrm{d}.\quad 30^{\circ}{24}'{12}''=....\: ^{0} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Ubahlah sudut berikut ke satuan }\\ &\textrm{yang diminta}\\ &\textrm{a}.\quad 135^{0}=.....\: \textrm{rad}\\ &\textrm{b}.\quad 6\: \textrm{rad}=....^{0}\\  &\textrm{c}.\quad 23,45^{\circ}=....\: \textrm{(dalam sexagesimal)}\\ &\textrm{d}.\quad 45^{\circ}{50}'{36}''=....\: ^{0} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Silahkan dicoba sendiri soal yang belum}\\ &\textrm{dibahas} \end{array}$.