Hello, premedtim!

You're expected to know the trig values for 30°, 45°, and 60°.

I'll add diagrams to what Captain Black said.

For the 45° angle, we have "half a square". Code:

+ - - - - - *
: * *
: x * *
: * * 1
: * *
: * 45° *
* * * * * * *
1

If the sides of the square are of length 1,

. . we use Pythagorus to find the length of the diagonal.

Hence:. $\displaystyle x^2\,=\,1^2+1^2\quad\Rightarrow\quad x = \sqrt{2}$

For a 45° angle, the sides of the right triangle are: .$\displaystyle \{1,\,1,\,\sqrt{2}\}$

If you memorize that, you can write all the trig values for 45°.

. . $\displaystyle \begin{array}{cccc}\sin45^o \:= \\ \cos45^o\:= \\ \tan45^o\:= \\ \vdots \end{array}

\begin{array}{cccc}\frac{1}{\sqrt{2}}\:= \\ \frac{1}{\sqrt{2}}\:= \\ \frac{1}{1} \:\:=\\ \vdots\end{array}

\begin{array}{cccc}\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \\ 1 \\ \vdots\end{array}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For the 30° and 60° angles, we have an equilateral triangle.

Draw an altitude, bisecting the vertex angle and the base. Code:

*
*:*
* : *
* : *
1 *30*: *
* : *
* : _ *
* :½√3 *
* : *
* 60* : *
* * * * * * * * * * *
½

We have a 30-60 right triangle with $\displaystyle base \,= \,\frac{1}{2},\:hyp\,=\,1$

Pythagorus says that: $\displaystyle height\,=\,\frac{\sqrt{3}}{2}$

The sides are: .$\displaystyle \frac{1}{2},\:\frac{\sqrt{3}}{2},\:1$

Multiply by 2: .$\displaystyle \underbrace{1}_\uparrow\:\underbrace{\sqrt{3}}_\up arrow \:\underbrace{2}_\uparrow$

. . . . . . . . . opp 30° . opp 60° .hyp

If you memorize that, you can recreate these triangles any time. Code:

*
2 * *
* * 1
* 30° *
* * * * * * * * *
√3
*
**
* *
2 * * _
* *√3
* *
* *
* 60° *
* * * * *
1