Trigonometric ratios of special angles

Trigonometric Ratios of Special Angles

Determination of the Sine, Cosine and Tangent of 30°, 45° and 60° without using Mathematical Tables

The special angles we are going to deal with are 30°, 45°, 60°, 90°. Let us see how to get the Tangent, Sine and Cosine of each angle as follows:

Deriving values for 30° and 60°

First, consider an equilateral triangle ABC below, the altitude from C bisects at D.

From Pythagoras Theorem; $(AD)^2 + (CD)^2 = (AC)^2$

$\begin{aligned} 1^2 + (CD)^2 &= 2^2 \\ (CD)^2 &= 4 - 1 \\ (CD)^2 &= 3 \end{aligned}$

Squaring both sides, we get

$(CD) = \sqrt{3}$

$\begin{aligned} \sin 60^\circ &= \frac{CD}{AC} = \frac{\sqrt{3}}{2} \\ \tan 60^\circ &= \frac{CD}{AD} = \frac{\sqrt{3}}{1} = \sqrt{3} \\ \cos 60^\circ &= \frac{AD}{AC} = \frac{1}{2} \\ \sin 30^\circ &= \frac{AD}{AC} = \frac{1}{2} \\ \cos 30^\circ &= \frac{CD}{AC} = \frac{\sqrt{3}}{2} \end{aligned}$

Deriving values for 45°

Secondly, consider the isosceles triangle ABC below, with base angles 45° and $\overline{AC}=\overline{BC}=1$.

The side $\overline{AB}$ (Hypotenuse side) = $\sqrt{1^2 + 1^2} = \sqrt{2}$ (by Pythagoras Theorem). So,

$\tan 45^\circ = \frac{AC}{BC} = \frac{1}{1} = 1$

$\sin 45^\circ = \frac{AC}{AB} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

$\cos 45^\circ = \frac{BC}{AB} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Summary of results

The results above can be summarized in table as here below:

θ	sin θ	cos θ	tan θ
30°	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$
45°	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	1
60°	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
90°	1	0	undefined

Note: $\tan \theta = \frac{\sin \theta}{\cos \theta}$

Simple Trigonometric Problems Related to Special Angles

Example 3

Find the value of $x$ if $\cos x^\circ = \frac{1}{2}$

Solution

Recalling the special angles, $\cos 60^\circ = \frac{1}{2}$

Therefore, the value of $x = 60^\circ$