How to find sin of any fraction-angles, and how do you find them in fraction forms and not in decimal forms? Ok so on doing a whole lot of Geometry Problems, since I am weak at Trigonometry, I am now focused on $2$ main questions :-
$1)$ How to calculate the $\sin,\cos,\tan$ of any angle?
Some Information :- This site :- https://www.intmath.com/blog/mathematics/how-do-you-find-exact-values-for-the-sine-of-all-angles-6212 , produces a clear understanding and a detailed approach of finding the $\sin$ of any angle from $1$ to $90^\circ$ , and I found it very interesting. But now the Questions arise :-
Can you find the $\sin$, $\cos$ or $\tan$ of any fraction angles, like $39.67$? 
Can you find the $\sin$, $\cos$ or $\tan$ of recurring fractions like $\frac{47}{9}$? 
Can you find the $\sin$, $\cos$ or $\tan$ of irrationals, like $\sqrt{2}?$
Since I am a bit new to Trigonometry, I will be asking if there is a formula to find the $\sin$ of fractions, or even recurring fractions. I can use the calculator to find them obviously, but I have another Question :-
$2)$ How to calculate the trigonometric ratios of every angle in fractional form?
We all know $\sin 45^\circ = \frac{1}{\sqrt{2}}$ , but what will be $\sin 46^\circ$ in fractions? I can use a calculator to calculate the decimal of it, but it is hard to deduce the fraction out of the value, especially because the decimal will be irrational. I know how to convert recurring decimals to fractions, but this is not the case. Right now I am focused on a particular problem, which asks me to find the $\sin$ of a recurring fraction, in a fraction form. I am struggling to do this unless I clear up the ideas.
Edit: My problem is to find the $\sin$ of $\frac{143}{3}^\circ$ . I do not have any specific formula to find this, and I am mainly stuck here. I need a formula which shows how this can be done.
Can anyone help me? Thank You.
 A: This will be my attempt at answering your question about finding $\sin(\frac{143°}{3})=\sin(\frac x3)$.
Let θ=$\frac x3$ and using this website
$\sin(3θ)=3\sinθ-4\sin^3θ$ Therefore, we need to solve this equation for sin(θ):
$4\sin^3θ-3\sinθ+ \sin(3θ)=0⇔ \sin^3θ -\frac 34 \sinθ+ \frac 14 \sin(3θ)=0$
Using Cardano’s Depressed Cubic Formula gets us the first root of the cubic equation gets us:
$\sqrt[3]{ \frac {-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$+ $\sqrt[3]{ \frac {-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$ where q=$\frac 14 \sin(3θ)$= $\frac 14 \sin(x)$= $\frac 14 \sin(143°)$= $\frac 14 \sin(37°)$ which is the constant value of the equation and p=-$\frac 34$ which is the coefficient of the linear value of degree one in the equation above.
Plug these into the formula and simplify to get an answer. We will use the fact that $\sin143°=\sin37°$:
$\sqrt[3]{ \frac {-(1/4)\sin37°}{2}+\sqrt{\frac{((1/4)\sin37°)^2}{4}+\frac{(-3/4)^3}{27}}}$+ $\sqrt[3]{ \frac {-1/4)\sin37°}{2}-\sqrt{\frac{((1/4)\sin37°)^2}{4}+\frac{(-3/4)^3}{27}}}$=
$\sqrt[3]{ \frac {-\sin37°}{8}+\frac 18 \sqrt{\sin^2 37°-1}}$+$\sqrt[3]{ \frac {-\sin37°}{8}-\frac 18 \sqrt{\sin^2 37°-1}}$=$\frac 12 (\sqrt[3]{i\cos37°- \sin37°}-\sqrt[3]{\sin37°+i\cos37°})$.
Unfortunately, $\sin37°=\cos53°$ and $\cos37°=\sin53°$ do not have easily solvable forms, but this website has the exact values for sine. However, $\sin37°=\sin143°=\sin\frac{37π}{180}$ so here are the steps for finding this value:
1.Use the same technique but sin(5θ)=sinx,x=π. Then, $θ=\frac {x}{5}$ and using the multiple angle formulas for sin in the above website to get sinπ=0=$5y-20y^3-16y^6$,and solve for  $y=\sinθ=\sin\frac π5$
2.Use the cubic technique on $sin\frac π5$ to get $\sin\frac{π}{15}$
3.Use the half angle formula twice to get $\sin\frac{π}{60}$
4.Use the cubic technique again on $sin\frac π{60}$ to get $\sin\frac{π}{180}$
5.Finally use the multiple angle formula for $\sin(37a)=\sin\frac{37π}{180}$
6.Evaluate $\sqrt{1-\sin^2\frac{37π}{180}}$= $\cos\frac{37π}{180}$
This means the final answer is:
$\sin\frac{143°}{3}$= $\frac 12 \bigg(\sqrt[3]{\bigg[}\bigg($$\bigg)i-\bigg($
$\bigg)-\sqrt[3]{\bigg[}$$+\bigg($$\bigg)i\bigg]\bigg)$
Here is proof of my answer. Please correct me if I am wrong or give me feedback!
A: For any value angle I would turn it into radians in order to use the power series of sine.
$sin(x)=\sum_{k=0}^{\infty}{(-1)^k\frac{x^{2k+1}}{(2k+1)!}}$, $x\epsilon\mathbb R$
$\theta=\frac{143^o}{3}\Rightarrow x=\frac{\pi}{180}\cdot \theta$
The result will always be a rational number that is an approximation of a transcendental number since both x and sin(x) are transcendental numbers related to $\pi$.
