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I would not expect a student to memorize trig functions of easy angles. (Never memorized them myself.) I would expect a student to have enough understanding to be able to figure them out in seconds.

The trig functions for $30^\circ,45^\circ,60^\circ$ are based on two simple geometric figures: the square and the equilateral triangle.

The square has four sides of equal length, which we take to be $1$. It has four equal angles of $90^\circ.$

Next, cut the square along the diagonal, making two triangles. Either one of these triangles has angles of $90^\circ,45^\circ,45^\circ$; no need to memorize $45$, just divide $90$ by $2$. The triangle has two sides of length $1$; if you have memorized the theorem of Pythagoras, you can figure out that the length of the third side is $\sqrt2.$

Unfortunately, you have to memorize the definitions of the sine and tangent: $\sin=\text{opp }/\text{ hyp}$ and $\tan=\sin/\cos.$ The cosine is easier: cosine = complement's sine, so $\cos\theta=\sin(90^\circ-\theta).$

The point is that you can just read off the trig functions of $45^\circ$ from the $45^\circ$-$45^\circ$-$90^\circ$ triangle: $\sin45^\circ=1/\sqrt2,\ $ $\cos45^\circ=\sin(90^\circ-45^\circ)=\sin45^\circ=1/\sqrt2,$ and $\tan45^\circ=\sin45^\circ/\cos45^\circ=(1/\sqrt2)/(1/\sqrt2)=1.$

Next, take an equilateral triangle; each of the three sides has length $1,$ and each of the three angles is $60^\circ.$ (No need to memorize $60,$ just divide $180$ by $3.$) Cut the equilateral triangle in half by bisecting an angle and look at one of the resulting triangles. The angles are $30^\circ,\ 60^\circ,$ and $90^\circ$; the sides are $1$ and $1/2$ and (if you still have Pythagoras memorized) $\sqrt3/2.$ From this triangle you can read off the trig functions of $30^\circ$ and $60^\circ.$

Executive summary. Take the two simplest polygons, the equilateral triangle and the square. An angle bisector divides each of those figures into two congruent right triangles. The trigonometric functions of $30^\circ,60^\circ,$ and $45^\circ$ can be read off of those triangles.