Perbandingan Trigonometri

Segitiga Siku-siku
1. Segitiga Siku-siku

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\(\sin \alpha = \dfrac yr = \dfrac {\text{depan}}{\text{miring}}\)

\(\cos \alpha = \dfrac xr = \dfrac {\text{samping}}{\text{miring}}\)

\(\tan \alpha = \dfrac yx = \dfrac {\text{depan}}{\text{samping}}\)

 

 

\(\sec \alpha = \dfrac {1}{\cos \alpha} \)

\(\csc \alpha = \dfrac {1}{\sin \alpha} \)

\(\cot \alpha = \dfrac {1}{\tan \alpha} \)


2. Sudut Istimewa
\(0\) \(30^{\text{o}}\) \(45^{\text{o}}\) \(60^{\text{o}}\) \(90^{\text{o}}\) \(37^{\text{o}}\) \(53^{\text{o}}\)
Sin 0 \(\dfrac{1}{2}\) \(\dfrac{1}{2}\sqrt{2}\) \(\dfrac{1}{2}\sqrt{3}\) 1 \(0,6\) \(0,8\)
Cos 1 \(\dfrac{1}{2}\sqrt{3}\) \(\dfrac{1}{2}\sqrt{2}\) \(\dfrac{1}{2}\) 0 \(0,8\) \(0,6\)
Tan 0 \(\dfrac{1}{3} \sqrt{3}\) 1 \(\sqrt{3}\) \(\dfrac 34\) \(\dfrac 43\)

Contoh 01

Diketahui segitiga siku-siku di bawah ini:

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Tentukan nilai dari \(\sin \alpha\), \(\cos \alpha\) dan \(\tan \alpha\)

 

Menentukan nilai r

\begin{equation*} \begin{split} x^2 + 5^2 & = r^2 \\\\ 3^2 + 4^2 & = r^2 \\\\ 9 + 16  & = r^2 \\\\ 25 & = r^2 \\\\ r & = \pm 5 \end{split} \end{equation*}

Gunakan nilai positif untuk \(r = 5\)

\begin{equation*} \sin \alpha = \frac {\text{depan}}{\text{miring}} = \bbox[5px, border: 2px solid magenta] {\frac {4}{5}} \end{equation*}

 

\begin{equation*} \cos \alpha = \frac {\text{samping}}{\text{miring}} = \bbox[5px, border: 2px solid magenta] {\frac {3}{5}} \end{equation*}

 

\begin{equation*} \tan \alpha = \frac {\text{depan}}{\text{samping}} = \bbox[5px, border: 2px solid magenta] {\frac {4}{3}} \end{equation*}


Contoh 02

Pada sebuah segitiga siku-siku ABC, diketahui nilai dari \(\sin A = \dfrac {5}{13}\).

Tentukan nilai dari \(\cos A\) dan \(\tan A\)

 

\(\sin A = \dfrac yr = \dfrac {\text{depan}}{\text{miring}} = \dfrac {5}{13}\)

 

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Menentukan nilai \(x\)

\begin{equation*} \begin{split} x^2 + 5^2 & = 13^2 \\\\ x^2 + 25 & = 169 \\\\ x^2  & = 144 \\\\ x & = \pm 12 \end{split} \end{equation*}

Karena panjang AB harus bernilai positif, maka \(x = 12\)

 

\(\cos A = \dfrac xr = \dfrac {\text{samping}}{\text{miring}} = \bbox[5px, border: 2px solid magenta] {\dfrac {12}{13}}\)

\(\tan A = \dfrac yx = \dfrac {\text{depan}}{\text{samping}} = \bbox[5px, border: 2px solid magenta] {\dfrac {5}{12}}\)

SOAL LATIHAN

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