Basic Trigonometry - Clavius

Basic Trigonometry

1. Right Angle Triangle

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$\sin α = \frac{y}{r} = \frac{opposite}{hypotenuse}$

$\cos α = \frac{x}{r} = \frac{adjacent}{hypotenuse}$

$\tan α = \frac{y}{x} = \frac{opposite}{adjacent}$

$\sec α = \frac{1}{\cos α}$

$\csc α = \frac{1}{\sin α}$

$\cot α = \frac{1}{\tan α}$

2. Special Angles

	$0$	$30^{o}$	$45^{o}$	$60^{o}$	$90^{o}$
Sin	0	$\frac{1}{2}$	$\frac{1}{2} \sqrt{2}$	$\frac{1}{2} \sqrt{3}$	1
Cos	1	$\frac{1}{2} \sqrt{3}$	$\frac{1}{2} \sqrt{2}$	$\frac{1}{2}$	0
Tan	0	$\frac{1}{3} \sqrt{3}$	1	$\sqrt{3}$	∼

Example 01

Given a right angle triangle below:

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

Find the value of $r$

$\begin{aligned} x^{2} + 5^{2} & = r^{2} \\ 3^{2} + 4^{2} & = r^{2} \\ 9 + 16 & = r^{2} \\ 25 & = r^{2} \\ r & = \pm 5 \end{aligned}$

Use positive value, $r = 5$

$\sin α = \frac{opposite}{hypotenuse} = \frac{4}{5}$

$\cos α = \frac{adjacent}{hypotenuse} = \frac{3}{5}$

$\tan α = \frac{opposite}{adjacent} = \frac{4}{3}$

Example 02

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

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

$\sin A = \frac{opposite}{hypotenuse} = \frac{5}{13}$

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Find the value of $x$

$\begin{aligned} x^{2} + 5^{2} & = 13^{2} \\ x^{2} + 25 & = 169 \\ x^{2} & = 144 \\ x & = \pm 12 \end{aligned}$

Use positive value, $x = 12$

$\cos A = \frac{x}{r} = \frac{adjacent}{hypotenuse} = \frac{12}{13}$

$\tan A = \frac{y}{x} = \frac{opposite}{adjacent} = \frac{5}{12}$

Exercise

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