Unit circle trigonometry

The unit circle is a circle of radius $1$ centered at the origin.
Draw an angle $\theta$ from the positive $x$-axis; where its ray meets the circle is a point $A$.
That point's coordinates are exactly $(\cos\theta, \sin\theta)$.

This gives a simple rule: the $x$-coordinate of the point is $\cos\theta$, and the $y$-coordinate is $\sin\theta$.
So if you know where the point is, you can read off both values with no triangle needed.
And $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{y}{x}$.

Because $x$ and $y$ change sign in different quadrants, so do $\cos$ and $\sin$.
In Quadrant II, $x < 0$ and $y > 0$, so $\cos\theta$ is negative and $\sin\theta$ is positive.
If you know one coordinate, the other comes from $x^2 + y^2 = 1$, choosing the sign that fits the quadrant.

A few key angles come up again and again.
At $45^\circ$, the point is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.
Angles are often measured in radians, where $\frac{\pi}{6} = 30^\circ$, $\frac{\pi}{4} = 45^\circ$, and $\frac{\pi}{3} = 60^\circ$.

For angles past $90^\circ$, use symmetry.
The point for $\frac{7\pi}{6}$ sits in Quadrant III directly opposite the point for $\frac{\pi}{6}$, so both coordinates simply flip sign.
This lets you find $\cos$ and $\sin$ of large angles from the familiar small ones.

A point on the unit circle at angle $\theta$ has coordinates $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. What is $\cos\theta$?
On the unit circle, $\cos\theta$ is the $x$-coordinate of the point.
So $\cos\theta = -\frac{1}{2}$.

An angle $\theta$ is in Quadrant II with $\sin\theta = \frac{5}{13}$. What is $\cos\theta$?
Here $\sin\theta = y = \frac{5}{13}$, so use $x^2 + y^2 = 1$: $x^2 = 1 - \frac{25}{169} = \frac{144}{169}$.
So $x = \pm\frac{12}{13}$, and since Quadrant II has $x < 0$, $\cos\theta = -\frac{12}{13}$.

The terminal side of an angle meets the unit circle at $45^\circ$. What is $\cos(45^\circ)$?
At $45^\circ$ the point is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, and $\cos$ is the $x$-coordinate.
So $\cos(45^\circ) = \frac{\sqrt{2}}{2}$.