Sine and cosine exact values

Memorize three terminal points, all in the first quadrant:

1. $\theta = \pi/6$ (30°): point $(\sqrt{3}/2,\; 1/2)$. So $\cos(\pi/6) = \sqrt{3}/2$ and $\sin(\pi/6) = 1/2$.
2. $\theta = \pi/4$ (45°): point $(\sqrt{2}/2,\; \sqrt{2}/2)$. Sine and cosine are equal here, both $\sqrt{2}/2$.
3. $\theta = \pi/3$ (60°): point $(1/2,\; \sqrt{3}/2)$. The $\pi/6$ values with the coordinates traded.

The values come in one ladder: $1/2 < \sqrt{2}/2 < \sqrt{3}/2$ (about 0.5, 0.71, 0.87). A size check catches most slips: as the angle climbs from 0 toward $\pi/2$, the point rises, so sine grows and cosine shrinks. At the small angle $\pi/6$, sine is the small value $1/2$; at the big angle $\pi/3$, sine is the big value $\sqrt{3}/2$. If your claimed value moves the wrong way, you traded the coordinates.

The reference angle of $\theta$ is the acute angle between the terminal ray and the x-axis (never the y-axis). Angles sharing a reference angle have terminal points that are mirror images of each other, so their sines and cosines match in absolute value and differ only in sign.

Finding it is quadrant bookkeeping: in QII the reference angle is $\pi - \theta$; in QIII it is $\theta - \pi$; in QIV it is $2\pi - \theta$. So $5\pi/6$ folds to $\pi/6$, $4\pi/3$ folds to $\pi/3$, and $7\pi/4$ folds to $\pi/4$. Evaluate the reference angle from the family of three, then attach the quadrant's signs.

No mnemonic needed: cosine is the x-coordinate, so it is positive on the right half of the circle (QI and QIV) and negative on the left. Sine is the y-coordinate, so it is positive on the top half (QI and QII) and negative on the bottom. Read the quadrant, read the signs.

Worked example: $\cos(5\pi/6)$. The angle is in QII (between $\pi/2$ and $\pi$), reference angle $\pi/6$, so the magnitude is $\sqrt{3}/2$. QII is the left half, x negative, so $\cos(5\pi/6) = -\sqrt{3}/2$. Meanwhile $\sin(5\pi/6) = +1/2$: top half, y positive. Two decisions, made separately, every time.

At $\theta = 0,\ \pi/2,\ \pi,\ 3\pi/2$ the terminal ray lies on an axis and the terminal point is one of $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$. Read the values straight off the coordinates:

1. $\theta = 0$: point $(1, 0)$, so $\cos = 1$, $\sin = 0$.
2. $\theta = \pi/2$: point $(0, 1)$, so $\cos = 0$, $\sin = 1$.
3. $\theta = \pi$: point $(-1, 0)$, so $\cos = -1$, $\sin = 0$.
4. $\theta = 3\pi/2$: point $(0, -1)$, so $\cos = 0$, $\sin = -1$.

Nothing to memorize once the picture is up: half a turn from $(1,0)$ is the far side of the circle, $(-1, 0)$, which is why $\sin(\pi) = 0$ and not 1. The top of the circle belongs to $\pi/2$, not to $\pi$.