I looked around quite a bit here on StackExchange and elsewhere and couldn't find an answer to my specific question, so here goes.
How can you algebraically calculate the exact value of a trig function applied to a non-transcendental angle value? I know that for certain angles (e.g. 75°), you can represent them as a sum/difference of two angles with nice, algebraic trig function results (in this case, 75° = 30° + 45°). But what about a less convenient angle, for example something like sin(63°)?
Some disclaimers/stipulations:
I'm working with degrees here because I know that you run into transcendental results when working with radians, e.g., sin(1 rad). I'm also trying to stay in the realm of trigonometric functions only, not that I can really see how complex numbers or exponentials would help here (aside from identities), though I suppose that is why I'm asking for help. I know some angles (e.g. 36°) have polygonal geometric proofs, but that's not really what I'm looking for either. I would prefer to formulate the problem purely in terms of equations. Finally, I am trying to use finite math only; I know I could use Taylor series expansions to do this, but that won't give me the nice closed-form solution I'm after.
I saw an example online where the angle in question is an integer factor of one of the more common angles (e.g. 18° which is $\frac{90°}{5}$), so I tried a similar approach for 63° and got nowhere. Here is (roughly) my attempt:
Calculating sin(63°):
$\mathrm{sin(63°)} = \mathrm{sin(3° + 60°)} = \mathrm{sin(3°)cos(60°) + sin(60°)cos(3°)}$
Calculating sin(3°):
Let $\theta = 3°$
$10\theta = 30°, 5\theta=30° - 5\theta, \mathrm{sin(5\theta)} = \mathrm{sin(30° - 5\theta)} = \mathrm{sin(30°)cos(5\theta) - sin(5\theta)cos(30°)}$
$\mathrm{sin(5\theta)} = \mathrm{\frac{1}{2}cos(5\theta) - \frac{\sqrt{3}}{2}sin(5\theta)}, \left(1 + \frac{\sqrt{3}}{2} \right) \mathrm{sin(5\theta)} = \frac{1}{2}\mathrm{cos(5\theta)}$
$\left(1 + \frac{\sqrt{3}}{2} \right) \mathrm{sin(2\theta + 3\theta)} = \frac{1}{2}\mathrm{cos(2\theta + 3\theta)}$
Use angle addition identities again:
$\mathrm{ \left(1 + \frac{\sqrt{3}}{2} \right) \left( \left( 2sin(\theta)cos(\theta) \right) \left( 4cos^{3}(\theta)-3cos(\theta) \right) + \left( 3sin(\theta) - 4sin^{3}(\theta) \right) \left( 1 - sin^{2}(\theta) \right) \right)} = $ ... [similarly expanded] ...
Want to solve for $\mathrm{sin(\theta)}$, at least initially. Convert all cosines possible to sines and simplify:
... [a lot of algebra, probably with mistakes] ...
$\mathrm{\left( 1 + \frac{\sqrt{3}}{2} \right)\left( 16sin^{5}(\theta) - 24sin^{3}(\theta) + 5sin(\theta) \right)} = \mathrm{\frac{1}{2}cos(\theta) \left( 4sin^{4}(\theta) - 2sin^{2}(\theta) + 1 \right)} = \mathrm{\frac{1}{2} \left(\pm \sqrt{1 - sin^{2}(\theta)} \right) \left( 4sin^{4}(\theta) - 2sin^{2}(\theta) + 1 \right)}$
And at this point, this is not analytically solvable for sin($\theta$) (as far as I can tell). I got the feeling soon into this derivation that this was not the right approach, but I wanted to see if this would work for any angle $\theta$.
I found a geometric proof for sin(3°) which also relies on a geometric proof for sin(18°), but I was hoping to do this without polygonal constructions.
TL;DR:
Did I make a mistake in my logic (other than assuming $\mathrm{sin(\theta)}$ was determinable this way)? How can I algebraically and exactly calculate the value of a trig function applied to any non-transcendental angle value?
To make the story short, we know that $$\sin (18 {}^{\circ})=\frac{\sqrt{5}-1}{4} \qquad \text{and} \qquad \cos (18 {}^{\circ})=\frac{\sqrt{10+2\sqrt 5}}{4} $$ $$\sin (15 {}^{\circ})=\frac{\sqrt{6}-\sqrt{2}}{4} \qquad \text{and} \qquad \cos (15 {}^{\circ})=\frac{\sqrt{6}+\sqrt{2}}{4}$$ $$\sin (3 {}^{\circ})=\sin ((18-15) {}^{\circ})=\sin (18 {}^{\circ})\cos (15 {}^{\circ})-\sin (15 {}^{\circ})\cos (18 {}^{\circ})$$ $$\sin (3 {}^{\circ})=\frac{\sqrt{2} \left(1+\sqrt{3}\right) \left(\sqrt{5}-1\right)-2 \left(\sqrt{3}-1\right) \sqrt{5+\sqrt{5}}}{16} $$
Do the same for the cosine.