Persamaan Trigonometri

 $\Large\textrm{A. 1  Identitas Trigonometri}$.

A. 1. 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

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

A. 1. 2  Identitas Trigonometri Dasar

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

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

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

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

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

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

$\color{purple}\textrm{d. Aturan sinus pada segitiga sebarang}$.

$\displaystyle \frac{BC}{\sin \angle A}=\displaystyle \frac{AC}{\sin \angle B}=\displaystyle \frac{AB}{\sin \angle C}$

$\color{purple}\textrm{e. Aturan cosinus pada segitiga sebarang}$.

Perhatikanlah gmabar pada poin e di atas, aturan cosinusnya adalah:

$\begin{aligned}\bullet \: \: &\cos \angle A=\displaystyle \frac{b^{2}+c^{2}-a^{2}}{2bc}\\ \bullet \: \: &\cos \angle B=\displaystyle \frac{a^{2}+c^{2}-a^{2}}{2ac}\\ \bullet \: \: &\cos \angle C=\displaystyle \frac{a^{2}+b^{2}-a^{2}}{2ab} \end{aligned}$.

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

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

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

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

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

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

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

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

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