02 — Unit Circle
Every point is \((\cos\theta,\;\sin\theta)\)
cos sin π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 7π/6 5π/4 4π/3 3π/2 5π/3 7π/4 11π/6
\((1,\; 0)\)
\(\left(\frac{\sqrt{3}}{2},\; \frac{1}{2}\right)\)
\(\left(\frac{1}{\sqrt{2}},\; \frac{1}{\sqrt{2}}\right)\)
\(\left(\frac{1}{2},\; \frac{\sqrt{3}}{2}\right)\)
\((0,\; 1)\)
\(\left(-\frac{1}{2},\; \frac{\sqrt{3}}{2}\right)\)
\(\left(-\frac{1}{\sqrt{2}},\; \frac{1}{\sqrt{2}}\right)\)
\(\left(-\frac{\sqrt{3}}{2},\; \frac{1}{2}\right)\)
\((-1,\; 0)\)
\(\left(-\frac{\sqrt{3}}{2},\; -\frac{1}{2}\right)\)
\(\left(-\frac{1}{\sqrt{2}},\; -\frac{1}{\sqrt{2}}\right)\)
\(\left(-\frac{1}{2},\; -\frac{\sqrt{3}}{2}\right)\)
\((0,\; -1)\)
\(\left(\frac{1}{2},\; -\frac{\sqrt{3}}{2}\right)\)
\(\left(\frac{1}{\sqrt{2}},\; -\frac{1}{\sqrt{2}}\right)\)
\(\left(\frac{\sqrt{3}}{2},\; -\frac{1}{2}\right)\)

How to read it

Every point on the unit circle is \((\cos\theta,\;\sin\theta)\). The x-coordinate is cosine, the y-coordinate is sine.

Example: at \(\frac{\pi}{4}\) (45°), the point is \(\left(\frac{1}{\sqrt{2}},\;\frac{1}{\sqrt{2}}\right)\), so \(\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\) and \(\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}\).

Degrees Radians \(\cos\theta\) \(\sin\theta\)
\(0°\)\(0\)\(1\)\(0\)
\(30°\)\(\dfrac{\pi}{6}\)\(\dfrac{\sqrt{3}}{2}\)\(\dfrac{1}{2}\)
\(45°\)\(\dfrac{\pi}{4}\)\(\dfrac{1}{\sqrt{2}}\)\(\dfrac{1}{\sqrt{2}}\)
\(60°\)\(\dfrac{\pi}{3}\)\(\dfrac{1}{2}\)\(\dfrac{\sqrt{3}}{2}\)
\(90°\)\(\dfrac{\pi}{2}\)\(0\)\(1\)
\(120°\)\(\dfrac{2\pi}{3}\)\(-\dfrac{1}{2}\)\(\dfrac{\sqrt{3}}{2}\)
\(135°\)\(\dfrac{3\pi}{4}\)\(-\dfrac{1}{\sqrt{2}}\)\(\dfrac{1}{\sqrt{2}}\)
\(150°\)\(\dfrac{5\pi}{6}\)\(-\dfrac{\sqrt{3}}{2}\)\(\dfrac{1}{2}\)
\(180°\)\(\pi\)\(-1\)\(0\)
\(210°\)\(\dfrac{7\pi}{6}\)\(-\dfrac{\sqrt{3}}{2}\)\(-\dfrac{1}{2}\)
\(225°\)\(\dfrac{5\pi}{4}\)\(-\dfrac{1}{\sqrt{2}}\)\(-\dfrac{1}{\sqrt{2}}\)
\(240°\)\(\dfrac{4\pi}{3}\)\(-\dfrac{1}{2}\)\(-\dfrac{\sqrt{3}}{2}\)
\(270°\)\(\dfrac{3\pi}{2}\)\(0\)\(-1\)
\(300°\)\(\dfrac{5\pi}{3}\)\(\dfrac{1}{2}\)\(-\dfrac{\sqrt{3}}{2}\)
\(315°\)\(\dfrac{7\pi}{4}\)\(\dfrac{1}{\sqrt{2}}\)\(-\dfrac{1}{\sqrt{2}}\)
\(330°\)\(\dfrac{11\pi}{6}\)\(\dfrac{\sqrt{3}}{2}\)\(-\dfrac{1}{2}\)
\(360°\)\(2\pi\)\(1\)\(0\)

The pattern — you don't need to memorize the table

For \(0°, 30°, 45°, 60°, 90°\) — cosine is:

$$\frac{\sqrt{4}}{2},\quad \frac{\sqrt{3}}{2},\quad \frac{\sqrt{2}}{2},\quad \frac{\sqrt{1}}{2},\quad \frac{\sqrt{0}}{2}$$

Just count down: 4, 3, 2, 1, 0 under the root. And \(\frac{\sqrt{4}}{2} = 1\), \(\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\), \(\frac{\sqrt{0}}{2} = 0\).

Sine is the same sequence counting up: \(\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}\)

Other quadrants: same values, just flip signs. Cosine is negative on the left half (Q2, Q3). Sine is negative on the bottom half (Q3, Q4).

For quantum — the ones you'll use constantly

\(\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \approx 0.707\) — the equal superposition value. Hadamard gate lives here.

\(\cos 0 = 1,\;\sin 0 = 0\) — the \(|0\rangle\) state (north pole of Bloch sphere).

\(\cos\frac{\pi}{2} = 0,\;\sin\frac{\pi}{2} = 1\) — but the Bloch formula uses \(\theta/2\), so \(\theta = \pi\) puts you at the south pole \((|1\rangle)\).