An intuitive explanation for the emergence of the $\pm$ symbol in the expansion of sin, cos or tan of $\frac{\alpha}{2}$ in terms of $\cos\alpha$ For some angle $\theta$,
$$\cos(2\theta) = 2\cos^2\theta - 1 \implies \cos(x) = \cos\Big(2\cdot\dfrac{x}{2}\Big) = 2\cos^2\Big(\dfrac{x}{2}\Big)-1$$
$$\implies \cos^2\Big(\dfrac{x}{2}\Big) = \dfrac{1+\cos(x)}{2}$$
$$\implies \cos\Big(\dfrac{x}{2}\Big) = \pm \sqrt{\dfrac{1+\cos(x)}{2}}$$
The expansions for $\sin\Big(\dfrac{x}{2}\Big)$ and $\tan\Big(\dfrac{x}{2}\Big)$ in terms of $\cos(x)$ are mentioned below. I have not derived them for that would make this question unnecessarily lengthier.
$$\sin\Big(\dfrac{x}{2}\Big) = \pm \sqrt{\dfrac{1-\cos(x)}{2}}$$
$$\tan\Big(\dfrac{x}{2}\Big) = \pm \sqrt{\dfrac{1-\cos(x)}{1+\cos(x)}}$$

Now, I was looking for an intuitive explanation for the emergence of the $\pm$ symbol in these identities and also an explanation for the fact that the symbol does not appear in the expansion of $f(2\phi)$ in terms of $f(\phi)$ where $f$ is some trigonometric function.
I thought of taking the identity $\sin\varphi = \pm \sqrt{1 - \cos^2\varphi}$ as reference. In this identity, the cause of emergence of the $\pm$ symbol seems to be the fact that the value of $\cos\varphi$ alone is not enough information to determine the value of $\sin\varphi$. In other words, for a given value of $\cos\varphi$, there are multiple possible values of $\sin\varphi$ (i.e. the value of $\sin\varphi$ is not unique). For example, if $\cos\varphi = \dfrac{1}{2}$, then two possible values of $\varphi$ for $0 < \varphi \leq 2\pi$ are $\dfrac{\pi}{3}$ and $\dfrac{5\pi}{3}$ and hence, there are two possible values of $\sin\varphi$, specifically, $\dfrac{\sqrt{3}}{2}$ and $\dfrac{-\sqrt{3}}{2}$.
Now, my main question here is that when we talk about one of the three identities that I mentioned above, sy $\cos\Big(\dfrac{x}{2}\Big) = \pm \sqrt{\dfrac{1+\cos(x)}{2}}$, then what is it tha twe assume to be given to us? Is it just $\cos(x)$ that is given or are the values of all trigonometric functions at $x$ given?
I know that the obvious answers looks like the former and you might be wondering how the latter would even imply that the value of $\cos\Big(\dfrac{x}{2}\Big)$ is not unique if all the trigonometric ratios of $x$ are given. Let me elaborate.
Let's say that the values of all trigonometric functions at $\alpha$ for some angle $\alpha$ are given to us. Then, for $\alpha \in (0,2\pi]$, there is one and only one value of $\alpha$. Let us call that value $\lambda$. But, when we look past the previous restriction that $\alpha \in (0,2\pi]$ and we look for values of $\alpha$ for $-\infty < \alpha < \infty$ i.e. all possible values of $\alpha$, then there are infinite possible values of $\alpha$ and all of them are co-terminally related to $\lambda$. Now, if we take a look at all the possible values of $\alpha$ and for all of them, we evaluate $\cos(2\alpha)$, we get $\cos(2\lambda)$ in all the cases. This could explain why the $\pm$ symbol does not appear when we express $\cos(2\phi)$ in terms of $\cos\phi$ for some angle $\phi$.
But, I have observed (and mathematically proved) that for all possible values of $\alpha$ that are of the form $(2\pi)n + \lambda$, where $n$ is an odd, $\cos\Big(\dfrac{\alpha}{2}\Big) = -\cos\Big(\dfrac{\lambda}{2}\Big)$ and where $n$ is even, $\cos\Big(\dfrac{\alpha}{2}\Big) = \cos\Big(\dfrac{\lambda}{2}\Big)$.
Let me provide an example for the sake of clarity. Let $\cos\gamma = \dfrac{\sqrt{3}}{2}$ and $\sin\gamma = \dfrac{1}{2}$, then $\lambda = \dfrac{\pi}{6}$ and a co-terminal of $\lambda$ whose difference from $\lambda$ is an odd multiple of $2\pi$ is $\dfrac{13\pi}{6}$. Now, for both these two values, the corresponding values of cosine of half of these angles are : $\dfrac{\sqrt{3}+1}{2\sqrt{2}}$ and $-\dfrac{\sqrt{3}+1}{2\sqrt{2}}$ respectively.
So, we have established the fact that if we are provided with the values of all trigonometric functions at an angle, then there are infinitely many possible values of that angle and for all these values, the cosine of two times these angles is always a unique value but this is not the case with the cosine of half of these angles.
Now, this result is valid for sine instead of cosine as ell. So, we can also say that the value of sine of half of all the possible values of $\alpha$ is not unique either.
This is primarily due to the fact that $f(\pi + \delta) = -f(\delta)$ if $f(x)$ is either $\sin(x)$ or $\cos(x)$ for some angle $\delta$.

But, the period of the tangent function is $\pi$, unlike the sine and cosine functions, whose period is $2\pi$. So, if $\mu$ is some angle and we know the values of all trigonometric functions at $\mu$, then for all the possible values of $\mu$, the tangent of half of those values will always be unique.
So, this caused my previous assumption that the values of all trigonometric functions at the given angle are known to us while solving the problem to completely break down.
Conclusion : So, now I think that only the value of $\cos(\alpha)$ is known to us while solving the problem and that makes it way easier to deduce that the value of $f_x\Big(\dfrac{\alpha}{2}\Big)$ is not unique, where $f_x$ is either sine, cosine or tangent.

So, I want to know if the way that I finally interpret the cause of emergence of the $\pm$ symbol and how I think that only the value of $\cos(\alpha)$ is known to us in context of these identities is correct. Please let me know if I have made some conceptual error in this post. It was long one, so it was not possible for me to go through it one more time and hence, I would appreciate your help in making this post error-free.
Thanks!

PS : I previously asked this question which is similar to this question but I only asked about $\cos\Big(\dfrac{\alpha}{2}\Big)$ in that question and I got a pretty satisfying answer from Keeley Hoek but I had not mentioned about sine and tangent in that question and this question is more focused on whether the assumption that I made in the previous question was indeed right or wrong since it doesn't seem to work with $\tan\Big(\dfrac{\alpha}{2}\Big)$.
 A: The longer your question is (and with a yes/no question like "is this correct?") makes it hard to guess what sort of answer you're looking for, but I hope this is helpful.
Ways the sign can change
It is true that $\cos\left(\dfrac{\alpha+2\pi}{2}\right)=-\cos\left(\dfrac{\alpha}2\right)$ even though $\cos(\alpha+2\pi)=\cos(\alpha)$. So consideration of coterminal angles is enough to get a difference of sign when evaluating $\cos(\alpha/2)$. Note that you can also get a difference of sign with $\cos(2\pi-\alpha)=\cos(\alpha)$ in that $\cos\left(\dfrac{2\pi-\alpha}{2}\right)=-\cos\left(\dfrac{\alpha}2\right)$ (though $\sin(2\pi-\alpha)\ne\sin(\alpha)$, as you implicitly pointed out).
And it is true that $\tan\left(\dfrac{\alpha+2\pi}{2}\right)=\tan\left(\dfrac{\alpha}2\right)$. So consideration of coterminal angles is not enough to get a difference of sign when evaluating $\tan(\alpha/2)$. But you could still get a difference of sign with $\cos(2\pi-\alpha)=\cos(\alpha)$ in that $\tan\left(\dfrac{2\pi-\alpha}{2}\right)=-\tan\left(\dfrac{\alpha}2\right)$.
If I understood you correctly, this means "yes, you are correct about how the sign changes in these formulae."
Assumptions

Is it just $\cos(x)$ that is given or are the values of all trigonometric functions at $x$ given?

Technically, the formulas don't assume any knowledge at all. You don't need to know the value of $\cos(x)$ to know that either $\sin\left(\dfrac{x}2\right)=\sqrt{\dfrac{1-\cos(x)}2}$ or $\sin\left(\dfrac{x}2\right)=\sqrt{\dfrac{1-\cos(x)}2}$ is true. More relevantly, nothing in the formula encodes how you might use information about the other trig functions.
But if you wanted to, rather than writing "$\tan\left(\dfrac{x}2\right)=\pm\sqrt{\dfrac{1-\cos(x)}{1+\cos(x)}}$", you could write an improved formula like "When $\tan(x/2)$ is defined, we have:
$$\tan\left(\dfrac{x}2\right)=\begin{cases}\sqrt{\dfrac{1-\cos(x)}{1+\cos(x)}}\text{ if }\sin(x)\ge0\\-\sqrt{\dfrac{1-\cos(x)}{1+\cos(x)}}\text{ if }\sin(x)\le0\end{cases}\text{."}$$
That formula says more and uses the sign of sine.
A: I posted this question a while ago and I got the answer to it soon after that but the idea of posting an answer for this myself slipped from my mind. So, here it is.
My question was that when we are expressing $\tan(x/2)$ in terms of $\cos(x)$ i.e. $\tan(x/2) = \pm\sqrt{\dfrac{1-\cos(x)}{1+\cos(x)}}$, then in the context of this equation, do we only have the knowledge of $\cos(x)$ or do we have the knowledge of other trigonometric functions at $x$ too? The answer is the former.
It is pretty obvious to me now. We are expressing $\tan(x/2)$ in terms of $\cos(x)$ which means that for a given $\cos(x)$, we have an expression that gives the value of $\tan(x/2)$.
A little more info : Now that we've established that, we can have an intuitive explanation for the $\pm$ symbol. By the way, an explanation for these "ambiguities" is also given in Plane Trigonometry Part $1$ by SL Loney under the section "Explanation of ambiguities" in the multiples and sub multiples chapter. Now, when we have a given value of $\cos(x)$, we have a lot of possible values of $x$ and for all those possible values, there are many possible values of $x/2$. For all these possible values of $x/2$, there are two possible values of $\tan(x/2)$ and those values are additive inverses of each other. So, the $\pm$ symbol arises when we express $\tan(x/2)$ in terms of $\cos(x)$.
But, when we have a given value of $\sin(x)$ as well as $\cos(x)$, of the infinitely many values of $x/2$, the value of $\tan(x/2)$ is unique for them and so, the $\pm$ symbol does not arise when we express $\tan(x/2)$ in terms of $\sin(x)$ as well as $\cos(x)$ i.e. $\tan(x/2) = \dfrac{1-\cos(x)}{\sin(x)}$
