Special Angles on Unit Circle

The special angles on the unit circle refer to the angles that have corresponding coordinates which can be solved with the Pythagorean Theorem. Each of these angles are measured from the positive $x$-axis as the initial side, and the terminal side is the segment connecting the origin to the terminal point on the unit circle.

$$\text{The sixteen special angles (measured in radians) on the unit circle, each labeled at the terminal point.}$$

These angles are commonly given as an argument of a trigonometric function such as the sine or cosine functions. When this is the case, one does not need a calculator to compute the value of these functions; the value is easily memorized by pattern. The derivation of the main values is below; once they are obtained the unit circle can be created by reflecting the lengths.

First consider a right triangle with a $45^\circ$ angle and a hypotenuse of 1. The other angle must be $180^\circ - 90^\circ - 45^\circ = 45^\circ ,$ and since there are two equal angles the triangle is isosceles.

The congruent sides are marked with $x$ on the diagram. Now apply the Pythagorean Theorem:

$$\begin{aligned} x^2 + x^2 &= 1^2 \\ 2x^2 &= 1 \\ x^2 &= \frac{1}{2} \\ x &= \sqrt{\frac{1}{2}} \end{aligned}$$

While $\sqrt{\frac{1}{2}}$ is correct, typically some rationalization is done: $\sqrt{\frac{1}{2}} = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1\sqrt{2}}{\sqrt{2}\sqrt{2}} = \frac{\sqrt{2}}{2} .$

Now consider a right triangle with a $30^\circ$ angle and a hypotenuse of 1. The other angle must be $180^\circ - 90^\circ - 30^\circ = 60^\circ .$ Reflect a copy of the triangle such that the adjoining triangles form an equilateral triangle as shown.

Note that since $\overline{BD} \cong \overline{DC}$ and triangle $ABC$ is equilateral, $m\overline{BD} = m\overline{DC} = \frac{1}{2} .$

$m\overline{AD}$ can now be worked out via Pythagorean Theorem:

$$\begin{aligned} (m\overline{AD})^2 + {\frac{1}{2}}^2 &= 1^2 \\ (m\overline{AD})^2 + \frac{1}{4} &= 1 \\ (m\overline{AD})^2 &= 1 - \frac{1}{4} \\ (m\overline{AD})^2 &= \frac{3}{4} \\ m\overline{AD} &= \sqrt{\frac{3}{4}} \end{aligned}$$

Usually $\sqrt{\frac{3}{4}}$ is simplified: $\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2} .$