Projection of linear combination of real numbers Suppose a real number $\alpha$ is a rational linear combination of two linearly independent real numbers, such as $\pi$ and $\sqrt 3$. For instance, say $$\alpha = \tfrac12\pi + 5\sqrt 3.$$
By a simple linear algebra argument ($\{\pi,\sqrt3\}$ is a $\mathbb Q$-basis), it's clear that the coefficients of $\alpha$ are uniquely determined.

Question: Given $\alpha\in\operatorname{span}_\mathbb Q(\{\pi,\sqrt 3\})$, how do we determine the coefficients of $\pi$ and $\sqrt 3$?

It feels like there is enough information, but at the same time, I imagine that we can find different coefficients which give us numbers arbitrarily close to $\alpha$, so it is important that we have an exact expression for $\alpha$.
(In other words, if we only know that $\alpha\approx 10.23105$, say, I imagine it would be hopeless to expect that we can determine the coefficients, even if we restrict our view to the module $\operatorname{span}_{\mathbb Z}(\{\pi,\sqrt 3\})$.)

The reason I ask this question is because for each $n$, the integral $$I(n)=\int_0^{\pi/3}\sin^{2n}x\,dx$$ is a number of the form $a\pi+b\sqrt 3$. It's easy to obtain the recurrence relation
$$I(n) = \begin{cases} \hfil\frac\pi3\hfil &amp; \text{if $n=0$}\\[3pt]
(1-\tfrac1{2n})\,I(n-1)-\frac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1} &amp; \text{otherwise}
\end{cases}$$
for $I(n)$, but I want to be able to compute the coefficients of $\pi$ and $\sqrt 3$ separately (in exact form) using a computer program.

I appreciate any help with solving either of the two problems, I'm assuming a solution to one necessarily sheds light on the other, that's why I haven't posted these as two separate questions.
I will answer your specific question about $I_n$.
Let $J_n=\dfrac{2^{2n}(n!)^2}{(2n)!}I_n$. We have  $J_0=I_0=\dfrac{\pi}{3}$ and
$J_n=\dfrac{2^{2n}(n!)^2}{(2n)!}(\dfrac{2n-1}{2n}I_{n-1}-\dfrac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1})=\dfrac{2^{2n}(n!)^2}{(2n)!}(\dfrac{2n-1}{2n}\dfrac{(2n-2)!}{2^{2n-2}((n-1)!)^2}J_{n-1}-\dfrac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1})=\dfrac{2^{2n}(n!)^2}{(2n)!}(\dfrac{2n}{2n}\dfrac{2n-1}{2n}\dfrac{(2n-2)!}{2^{2n-2}((n-1)!)^2}J_{n-1}-\dfrac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1})=J_{n-1}-\dfrac{2^{2n}(n!)^2}{(2n)!}\dfrac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1})$,
thus
$J_n=J_{n-1}-\dfrac{1}{2n}\dfrac{3^{n}(n!)^2}{(2n)!}\dfrac{\sqrt{3}}{3}=J_{n-1}-\dfrac{3^{n-1}}{2n}\dfrac{(n!)^2}{(2n)!}\sqrt{3}$.
Summing relations yields $J_n=\dfrac{\pi}{3}-(\displaystyle\sum_{k=1}^n\dfrac{3^{k-1}}{2k}\dfrac{(k!)^2}{(2k)!})\sqrt{3}$.
Consequently, $I_n=\dfrac{(2n)!}{2^{2n}(n!)^2}\dfrac{\pi}{3}-\dfrac{(2n)!}{2^{2n}(n!)^2}(\displaystyle\sum_{k=1}^n\dfrac{3^{k-1}}{2k}\dfrac{(k!)^2}{(2k)!})\sqrt{3}$.
Here's easy way to determine them:
let $t \in [0..1]$, we define
$$ a_0 = t * \alpha $$
$$ a_1 = (1-t) * \alpha $$
Basically we have $a_0 + a_1 = \alpha$.
Now, after we have $a_0$ and $a_1$, we can calculate co-efficients:
$$c_0 = \frac{a_0}{\pi}$$
$$c_1 = \frac{a_1}{\sqrt{3}}$$.
I have a feeling that the $t$ cannot be calculated from the $\alpha$, given that they are both located in same dimension, basically you need some criteria how $\alpha \in R$ can be divided to two separate dimensions. Thus your end result is large number of possible alternative coefficients. Every choice of $t$ will have different coefficients. Limiting $c_0,c_1 \in R$ to $Q$ sounds kinda difficult.