Trigonometry

Basic Trigonometry

1. Right Angle Triangle

$$\sin \alpha = \dfrac yr = \dfrac {\text{opposite}}{\text{hypotenuse}}$$

$$\cos \alpha = \dfrac xr = \dfrac {\text{adjacent}}{\text{hypotenuse}}$$

$$\tan \alpha = \dfrac yx = \dfrac {\text{opposite}}{\text{adjacent}}$$

$$\sec \alpha = \dfrac {1}{\cos \alpha}$$

$$\csc \alpha = \dfrac {1}{\sin \alpha}$$

$$\cot \alpha = \dfrac {1}{\tan \alpha}$$

2. Special Angles
 $$0$$ $$30^{\text{o}}$$ $$45^{\text{o}}$$ $$60^{\text{o}}$$ $$90^{\text{o}}$$ Sin 0 $$\dfrac{1}{2}$$ $$\dfrac{1}{2}\sqrt{2}$$ $$\dfrac{1}{2}\sqrt{3}$$ 1 Cos 1 $$\dfrac{1}{2}\sqrt{3}$$ $$\dfrac{1}{2}\sqrt{2}$$ $$\dfrac{1}{2}$$ 0 Tan 0 $$\dfrac{1}{3} \sqrt{3}$$ 1 $$\sqrt{3}$$ ∼

Example 01

Given a right angle triangle below:

Determine the value of $$\sin \alpha$$, $$\cos \alpha$$ and $$\tan \alpha$$

Find the value of $$r$$

\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*}

Use positive value, $$r = 5$$

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

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

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

Example 02

A right angle triangle ABC, given that $$\sin A = \dfrac {5}{13}$$.

Determine the value of $$\cos A$$ and $$\tan A$$

$$\sin A = \dfrac {\text{opposite}}{\text{hypotenuse}} = \dfrac {5}{13}$$

Find the value of $$x$$

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

Use positive value, $$x = 12$$

$$\cos A = \dfrac xr = \dfrac {\text{adjacent}}{\text{hypotenuse}} = \bbox[5px, border: 2px solid magenta] {\dfrac {12}{13}}$$

$$\tan A = \dfrac yx = \dfrac {\text{opposite}}{\text{adjacent}} = \bbox[5px, border: 2px solid magenta] {\dfrac {5}{12}}$$