How to find trigonometric values that = radical values?

I have always hoped to rearrange a trigonometric expression into another expression without trigonometric functions. For example, it is common known that $\sin 60^\circ =\dfrac{\sqrt{3}}{2}$, but what about other ones?
I have found some already (below) and I wish to find more as this may be helpful when I am to simplify an expression.

instructions

If you find any typos or improvements comment under my comment that reads: report
The angles I am looking for are integers in range $[0^\circ ,180^\circ]$ .
Please do not comment straight away as things will get messy.
If you have found a new solution, comment under my comment below this note in format:
<equation>
<proof———>

The Transformation table

Resolution found
Formulae used:
$(1)~~~\sin (\pi -A)=\sin A,~~\cos (\pi -A)=-\cos A,~~\tan (\pi -A)=-\tan A.$
$(2)~~~\sin ^2 A+\cos ^2 A=1,~~\tan A=\dfrac{\sin A}{\cos A}.$
$(3)~~~\sin (\dfrac{\pi}{2}+A)=\cos A,~~\cos (\dfrac{\pi}{2}+A)=-\sin A.$
$(4)~~~\sin 2A=2\sin A\cos A.$
$(5)~~~\cos 2A=\cos ^2 A-\sin ^2 A=2\cos ^2 A-1 =1-2\sin ^2 A.$
$(6)~~~\tan 2A=\dfrac{2\tan A}{1-\tan ^2 A}.$
Table 1: integer values in degrees:

angle $\theta$ in degrees	$\sin \theta$	$\cos \theta$	$\tan \theta$	Proof
$15^\circ$	$\dfrac{\sqrt{6}-\sqrt{2}}{4}$	$\dfrac{\sqrt{6}+\sqrt{2}}{4}$	$2-\sqrt{3}$	The figure explains everything. In the figure, note that $\triangle ACD$ is isosceles, so $\angle ADB =15^\circ .$
$18^\circ$	$\dfrac{\sqrt{5}-1}{4}$	$\dfrac{\sqrt{10+2\sqrt{5}}}{4}$	$\dfrac{\sqrt{25-10\sqrt{5}}}{5}$	We can draw a regular pentagon as follows: Let $AB=1$ . From symmetry and some algebra, we get $AC=BC=\dfrac{\sqrt{5}+1}{4}=\phi$ and $AD=BD=1$ . Then apply the Pythagorean Theorem on the triangle. Calculations should be tedious but we can arrive at the result. Or we can do this: $\begin{aligned} \because \cos 18^\circ \gt 0, \\ \therefore \cos 18^\circ &=\sqrt{1-\sin ^2 18^\circ}\\ &=\sqrt{1-\frac{(5-2\sqrt{5} +1)}{16}}~~~\text{the complete square formula helps} \\ &=\sqrt{\dfrac{10+2\sqrt{5}}{16}} \\ &=\dfrac{\sqrt{10+2\sqrt{5}}}{4}.\end{aligned}$ It is still a tedious job to calculate $\tan 18^\circ$ though.
$20^\circ$				We must use triple angle identities to find the value of $\sin 20^\circ$ and $\cos 20^\circ$ . This shall be separated into two sections. $\huge \text{COSINE}$ Perhaps we can try $\cos 20^\circ$ first as $\cos (3\times 20^\circ )=\dfrac{1}{2}$ . Let $\cos 20^\circ =x$ , then $4x^3-3x=\dfrac{1}{2}.$ Therefore $x^3-\frac{3}{4}x-\dfrac{1}{8}=0.$ So we can use the cubic formula to solve this equation. If the function has real coefficients $x^3+ax^2+bx+c=0$ , then one real root would be: $\sqrt[3]{-q+\sqrt{r}}+\sqrt[3]{-q-\sqrt{r}}-\frac{a}{3},$ Where $q=\dfrac{1}{27}a^3-\dfrac{1}{6}ab+\dfrac{1}{2}c$ , $r=q^2+\dfrac{1}{27}(b-\dfrac{1}{3}a^2)^3$ . Plug in $\color{#D61F06} a=0, ~~b=\dfrac{3}{4}, ~~c=\dfrac{1}{8}$ to get the result: $q=\dfrac{1}{16},~~r=\frac{5}{256}.$ $\sqrt{r}=\dfrac{\sqrt5}{16}.$ $\sqrt[3]{-q+\sqrt{r}}+\sqrt[3]{-q-\sqrt{r}}-\frac{a}{3}=\sqrt[3]{\dfrac{\sqrt5-1}{16}}+\sqrt[3]{\dfrac{-\sqrt5-1}{16}}=\dfrac{\sqrt[3]{\sqrt5-1}+\sqrt[3]{-\sqrt5-1}}{2\sqrt[3]{2}}=-0.161092677...$ Check if this is the expected root. $\cos 20^\circ =0.93969262...$ , therefore we must try another root but things get complicated. $\huge \text{SINE}$ Let $y=\sin 20^\circ$ and do the same thing. This time $3y-4y^3=\dfrac{\sqrt3}{2}.$ Therefore $y^3-\dfrac{3}{4}y+\dfrac{\sqrt3}{2}=0.$ Again plug in the equation.
$30^\circ$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{3}}{3}$	N/A
$36^\circ$	$\dfrac{\sqrt{10-2\sqrt{5}}}{4}$	$\dfrac{\sqrt{5}+1}{4}$	$\sqrt{5-2\sqrt5}$	Use formulae $(4),(5),(6)$ on the values of $18^\circ$ to get the values.
$45^\circ$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{2}}{2}$	$1$	N/A
$54^\circ$	$\dfrac{\sqrt5+1}{4}$
$60^\circ$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\sqrt{3}$	N/A
$72^\circ$		$\dfrac{\sqrt{5}-1}{4}$
$75^\circ$	$\dfrac{\sqrt{6}+\sqrt{2}}{4}$	$\dfrac{\sqrt{6}-\sqrt{2}}{4}$	$2+\sqrt{3}$	Apply formula $(3)$ and let $A=15^\circ$ to get the sine and cosine. Then we can find tangent easily.
$90^\circ$	$1$	$0$	undefined	N/A
$105^\circ$	$\dfrac{\sqrt6 +\sqrt2}{4}$	$\dfrac{\sqrt2 -\sqrt6}{4}$	$-2-\sqrt3$	Use formula $(1)$ on the values of $75^\circ$ to get the values.
$108^\circ$		$\dfrac{1-\sqrt{5}}{4}$
$120^\circ$	$\dfrac{\sqrt3}{2}$	$-\dfrac{1}{2}$	$-\sqrt3$	Use formula $(1)$ on the values of $60^\circ$ to get the values.
$126^\circ$
$135^\circ$	$\dfrac{\sqrt{2}}{2}$	$-\dfrac{\sqrt{2}}{2}$	$-1$	Use formula $(1)$ on the values of $45^\circ$ to get the values.
$144^\circ$	$\dfrac{\sqrt{10-2\sqrt{5}}}{4}$	$\dfrac{-\sqrt{5}-1}{4}$	$-\sqrt{5-2\sqrt5}$	Use formula $(1)$ on the values of $36^\circ$ to get the values.
$150^\circ$	$\dfrac{1}{2}$	$-\dfrac{\sqrt3}{2}$	$-\dfrac{\sqrt3}{3}$	Use formula $(1)$ on the values of $30^\circ$ to get the values.
$162^\circ$	$\dfrac{\sqrt5-1}{4}$	$-\dfrac{\sqrt{10+2\sqrt{5}}}{4}$	$-\dfrac{\sqrt{25-10\sqrt{5}}}{5}$	Use formula $(1)$ on the values of $18^\circ$ to get the values.
$165^\circ$	$\dfrac{\sqrt6-\sqrt2}{4}$	$\dfrac{-\sqrt6-\sqrt2}{4}$	$\sqrt3-2$	Use formula $(1)$ on the values of $15^\circ$ to get the values.

Table 2: fraction values in radians:

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

How to find trigonometric values that = radical values?

instructions

The Transformation table

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