Consider an equilateral triangle ABC of side 2 units.

In $\triangle ABC$ of Figure 8.8, the altitude AD bisects line segment BC. Using the Pythagorean theorem:

$AB^2 = BD^2 + AD^2$

$2^2 = 1^2 + AD^2$

$AD^2 = 4 - 1 = 3$

$AD = \sqrt{3}$

Hence,

$\sin 60° = \cos 30° = \frac{AD}{AB} = \frac{\sqrt{3}}{2}$

$\cos 60° = \sin 30° = \frac{BD}{AB} = \frac{1}{2}$

$\tan 60° = \frac{AD}{BD} = \frac{\sqrt{3}}{1} = \sqrt{3}$

$\tan 30° = \frac{BD}{AD} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Consider an isosceles triangle ABC.

Let $AB = BC = 1$ unit. By the Pythagorean theorem:

$AC^2 = AB^2 + BC^2$

$AC^2 = 1^2 + 1^2 = 2$

$AC = \sqrt{2}$

Then,

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

$\cos 45° = \frac{BC}{AC} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

$\tan 45° = \frac{AB}{BC} = \frac{1}{1} = 1$

The sine, cosine, and tangent of special angles are summarized in Table 8.1.

Angles	0°	30°	45°	60°	90°	180°	270°	360°
sin	0	1/2	√2/2	√3/2	1	0	-1	0
cos	1	√3/2	√2/2	1/2	0	-1	0	1
tan	0	√3/3	1	√3	∞	0	∞	0

Example 2

Evaluate $2\cos 60° \sin 45° + 3\tan 60° \cos 30°$, leaving the answer in surd form.

Solution:

$2\cos 60° \sin 45° + 3\tan 60° \cos 30° = 2\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + 3(\sqrt{3})\left(\frac{\sqrt{3}}{2}\right)$

$= \frac{\sqrt{2}}{2} + \frac{9}{2}$

$= \frac{\sqrt{2} + 9}{2}$