# 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: $\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2} .$

**Cite as:**Special Angles on Unit Circle.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/special-angles-on-unit-circle/