Persamaan Trigonometri

 $\text{A. 1 Identitas Trigonometri}$.

A. 1. 1  Nilai Trigonometri Sudut

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

Perhatikanlah ilustrasi sebuah segitiga siku-siku sama kaki berikut

Diketahui pula bahwa :

$\begin{array}{c}\bullet \sin {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}}\\ \bullet \cos {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}}\\ \bullet \tan {45}^{\circ }=1\end{array}$.

$\begin{array}{c}\bullet \csc {45}^{\circ }={\displaystyle \sqrt{2}}\\ \bullet \sec {45}^{\circ }={\displaystyle \sqrt{2}}\\ \bullet \cot {45}^{\circ }=1\end{array}$.

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

$\begin{array}{|cc|}\hline \begin{array}{c}\bullet \sin {30}^{\circ }={\displaystyle \frac{1}{2}}\\ \bullet \cos {30}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \bullet \tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}}\\ \bullet \sin {60}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \bullet \cos {60}^{\circ }={\displaystyle \frac{1}{2}}\\ \bullet \tan {30}^{\circ }={\displaystyle \sqrt{3}}\end{array}& \begin{array}{c}\bullet \csc {30}^{\circ }={\displaystyle 2}\\ \bullet \sec {30}^{\circ }={\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}\\ \bullet \cot {30}^{\circ }={\displaystyle \sqrt{3}}\\ \bullet \csc {60}^{\circ }={\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}}\\ \bullet \sec {60}^{\circ }={\displaystyle 2}\\ \bullet \cot {30}^{\circ }={\displaystyle \frac{1}{3}\sqrt{3}}\end{array}\\ \hline\end{array}$

Perhatikan segitiga ABC siku-siku di C berikut

Perhatikanlah segitiga OAB berikut

$\begin{array}{rl}\text{a}.& \sin \alpha ={\displaystyle \frac{y}{r}}\\ \text{b}.& \cos \alpha ={\displaystyle \frac{x}{r}}\\ \text{c}.& \tan \alpha ={\displaystyle \frac{y}{x}}\\ \text{d}.& \csc \alpha ={\displaystyle \frac{r}{y}}\\ \text{e}.& \sec \alpha ={\displaystyle \frac{r}{x}}\\ \text{f}.& \cot \alpha ={\displaystyle \frac{x}{y}}\end{array}$.

A. 1. 2  Identitas Trigonometri Dasar

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

$\begin{array}{c}\\ \begin{array}{rl}& \text{Dalil/rumus Pythagoras}\\ & a^2+b^2=c^2\\ & \text{atau}\\ & c=\sqrt{a^2+b^2}\end{array}& \begin{array}{rl}& \sin \mathrm{\angle }ACB={\displaystyle \frac{a}{c}}\\ & \cos \mathrm{\angle }ACB={\displaystyle \frac{b}{c}}\\ & \tan \mathrm{\angle }ACB={\displaystyle \frac{a}{b}={\displaystyle \frac{\sin \mathrm{\angle }ACB}{\cos \mathrm{\angle }ACB}}}\\ & \csc \mathrm{\angle }ACB={\displaystyle \frac{c}{a}}\\ & \sec \mathrm{\angle }ACB={\displaystyle \frac{c}{b}}\\ & \cot \mathrm{\angle }ACB={\displaystyle \frac{b}{a}={\displaystyle \frac{\cos \mathrm{\angle }ACB}{\sin \mathrm{\angle }ACB}}}\end{array}\end{array}$

$\text{b. Identitas trigonometri pada segitiga siku-siku}$.

$\begin{array}{rl}& \text{Dalil/rumus Pythagoras}\\ & a^2+b^2=c^2\\ & \text{Perhatikan lagi gambar di poin c di atas}\\ & \begin{array}{|cl|}\hline 1.& \text{Rumus saat dibagi dengan}{c}^2\\ & {\displaystyle \frac{a^2}{c^2}+{\displaystyle \frac{b^2}{c^2}={\displaystyle \frac{c^2}{c^2}\iff {\displaystyle \frac{a^2}{c^2}+{\displaystyle \frac{b^2}{c^2}=1}}}}}\\ & \\ & \text{menjadi}{\left({\displaystyle \frac{a}{c}}\right)}^2+{\left({\displaystyle \frac{b}{c}}\right)}^2=1\\ & {\sin }^2\mathrm{\angle }ACB+{\cos }^2\mathrm{\angle }ACB=1\\ 2& \text{Rumus saat dibagi dengan}{b}^2\\ & {\displaystyle \frac{a^2}{b^2}+{\displaystyle \frac{b^2}{b^2}={\displaystyle \frac{c^2}{b^2}\iff {\displaystyle \frac{a^2}{b^2}+1={\displaystyle \frac{c^2}{b^2}}}}}}\\ & \\ & \text{menjadi}{\left({\displaystyle \frac{a}{b}}\right)}^2+1={\left({\displaystyle \frac{c}{b}}\right)}^2\\ & {\tan }^2\mathrm{\angle }ACB+1={\sec }^2\mathrm{\angle }ACB\\ 3& \text{Rumus saat dibagi dengan}{a}^2\\ & {\displaystyle \frac{a^2}{a^2}+{\displaystyle \frac{b^2}{a^2}={\displaystyle \frac{c^2}{a^2}\iff 1+{\displaystyle \frac{b^2}{a^2}={\displaystyle \frac{c^2}{a^2}}}}}}\\ & \\ & \text{menjadi}1+{\left({\displaystyle \frac{b}{a}}\right)}^2={\left({\displaystyle \frac{c}{a}}\right)}^2\\ & 1+{\cot }^2\mathrm{\angle }ACB={\csc }^2\mathrm{\angle }ACB\\ \hline\end{array}\end{array}$

$\text{c. Tabel trigonometri nilai sudut istimewa}$.

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

$\text{d. Aturan sinus pada segitiga sebarang}$.

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

$\text{e. Aturan cosinus pada segitiga sebarang}$.

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

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

$\begin{array}{rl}& \text{Macam-Macam Identitas Trigonometri Dasar}\\ & 1.\csc \alpha ={\displaystyle \frac{1}{\sin \alpha }5.\tan \alpha ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}}\\ & 2.\sec \alpha ={\displaystyle \frac{1}{\cos \alpha }6.{\tan }^2\alpha +1={\sec }^2\alpha }\\ & 3.\cot \alpha ={\displaystyle \frac{1}{\tan \alpha }7.{\cot }^2\alpha +1={\csc }^2\alpha }\\ & 4.\cot \alpha ={\displaystyle \frac{\cos \alpha }{\sin \alpha }8.{\sin }^2\alpha +{\cos }^2=1}\end{array}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa}\\ & \tan \alpha ={\displaystyle \frac{\sin \alpha \cos \alpha }{1-{\sin }^2\alpha }}\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}\tan \alpha & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}\\ & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }\times \frac{\cos \alpha }{\cos \alpha }}\\ & ={\displaystyle \frac{\sin \alpha \cos \alpha }{{\cos }^2\alpha }}\\ & ={\displaystyle \frac{\sin \alpha \cos \alpha }{1-{\sin }^2\alpha }◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1}{\sqrt{{\tan }^2\beta }}\times \sin \beta =\cos \beta }\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{\sqrt{{\tan }^2\beta }}\times \sin \beta }& ={\displaystyle \frac{1}{\tan \beta }\times \sin \beta }\\ & ={\displaystyle \frac{\cos \beta }{\sin \beta }\times \sin \beta }\\ & =\cos \beta ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{{\cos }^2\gamma }{1-\sin \gamma }=1+\sin \gamma }\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{{\cos }^2\gamma }{1-\sin \gamma }}& ={\displaystyle \frac{1-{\sin }^2\gamma }{1-\sin \gamma }}\\ & ={\displaystyle \frac{(1-\sin \gamma )(1+\sin \gamma )}{1-\sin \gamma }}\\ & =1+\sin \gamma ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }={\cos }^2\theta -{\sin }^2\theta }\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }}& ={\displaystyle \frac{1-{\tan }^2\theta }{{\sec }^2\theta }={\displaystyle \frac{1-{\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }}}{{\displaystyle \frac{1}{{\cos }^2\theta }}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta }}}{{\displaystyle \frac{1}{{\cos }^2\theta }}}}\\ & ={\cos }^2\theta -{\sin }^2\theta ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tunjukkan bahwa}\\ & {\cos }^4\alpha -{\sin }^4\alpha =1-2{\sin }^2\alpha \\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\cos }^4\alpha -{\sin }^4\alpha & ={({\cos }^2\alpha )}^2-{({\sin }^2\alpha )}^2\\ & =({\cos }^2\alpha -{\sin }^2\alpha )({\cos }^2\alpha +{\sin }^2\alpha )\\ & =({\cos }^2\alpha -{\sin }^2\alpha )\times 1\\ & ={\cos }^2\alpha -{\sin }^2\alpha \\ & =(1-{\sin }^2\alpha )-{\sin }^2\alpha \\ & =1-2{\sin }^2\alpha ◼\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tunjukkan bahwa}\\ & {\displaystyle \frac{\sin \beta \sec \beta }{{\sin }^2\beta -{\tan }^2\beta }=-{\displaystyle \frac{\cos \beta }{{\sin }^3\beta }}}\\ \\ & \text{Bukti}:\\ & \begin{array}{rl}{\displaystyle \frac{\sin \beta \sec \beta }{{\sin }^2\beta -{\tan }^2\beta }}& ={\displaystyle \frac{\sin \beta \left({\displaystyle \frac{1}{\cos \beta }}\right)}{{\sin }^2\beta -{\displaystyle \frac{{\sin }^2\beta }{{\cos }^2\beta }}}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{\sin \beta }{\cos \beta }}\right)}{{\sin }^2\beta (1-{\displaystyle \frac{1}{{\cos }^2\beta }})}\times \frac{{\cos }^2\beta }{{\cos }^2\beta }}\\ & ={\displaystyle \frac{\sin \beta \cos \beta }{{\sin }^2\beta ({\cos }^2\beta -1)}}\\ & ={\displaystyle \frac{\cos \beta }{\sin \beta (-{\sin }^2\beta )}}\\ & =-{\displaystyle \frac{\cos \beta }{{\sin }^3\beta }◼}\end{array}\end{array}$.

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