In calculus classes when you are asked to evaluate a trig function at a specific angle, it’s 99.9% of the time at one of the so-called special angles we use in our chart. Since you are likely to have learned degrees first I’ll include degree angles in the first chart, but after that, it’s going to be radian only.
Begin by setting up the table on scratch paper as follows:
$$\begin{array}{ l| |c|c|c|c|c } \theta & 0 = 0º & \tfrac{\pi}{6} = 30º & \tfrac{\pi}{4}=45º & \tfrac{\pi}{3}=60º & \tfrac{\pi}{2}=90º \\ \hline\hline \sin\theta & & & & & \\ \hline \cos\theta & & & & & \\ \hline \tan\theta & & & & & \\ \hline \end{array} $$
Then remember ##\sin\theta## starts at zero, fill in the pattern
$$\begin{array}{ l| |c|c|c|c|c } \theta & 0 & \tfrac{\pi}{6} & \tfrac{\pi}{4} & \tfrac{\pi}{3} & \tfrac{\pi}{2} \\ \hline\hline\sin\theta & \tfrac{\sqrt{0}}{2} & \tfrac{\sqrt{1}}{2} & \tfrac{\sqrt{2}}{2} & \tfrac{\sqrt{3}}{2} & \tfrac{\sqrt{4}}{2} \\...
##\theta~\rightarrow## | 0° = 0 | 30° = ##\dfrac{\pi}{6}## | 45° = ##\dfrac{\pi}{4}## | 60° = ##\dfrac{\pi}{3}## | 90° = ##\dfrac{\pi}{2}## |
##\sin \theta## | ##\sqrt{\dfrac{0}{4}}## | ##\sqrt{\dfrac{1}{4}}## | ##\sqrt{\dfrac{2}{4}}## | ##\sqrt{\dfrac{3}{4}}## | ##\sqrt{\dfrac{4}{4}}## |
##\cos \theta## | ##\sqrt{\dfrac{4}{4}}## | ##\sqrt{\dfrac{3}{4}}## | ##\sqrt{\dfrac{2}{4}}## | ##\sqrt{\dfrac{1}{4}}## | ##\sqrt{\dfrac{0}{4}}## |
##\tan \theta## | ##\sqrt{\dfrac{0}{4 - 0}}## | ##\sqrt{\dfrac{1}{4 - 1}}## | ##\sqrt{\dfrac{2}{4 - 2}}## | ##\sqrt{\dfrac{3}{4 - 3}}## | ##\sqrt{\dfrac{4}{4 - 4}}## |