An (intuitive) explanation for the emergence of $\pm$ in the expansion of $\cos\Big(\frac{x}{2}\Big)$ in terms of $\cos x$?

Question

For some angle $\alpha$, we have : $$\cos(2\alpha) = 2\cos^2\alpha - 1$$ $$\cos\Big(\dfrac{\alpha}{2}\Big) = \pm\sqrt{\dfrac{\cos\alpha + 1}{2}}$$ I want to gain a deeper and more intuitive understanding of why '$\pm$' emerges here.

At first, I thought that the emergence of $\pm$ implies that the answer is not unique, which is the case in most of the uses of $\pm$. For example, $\sqrt{4} = \pm 2$. This implies that the answer can be both $2$ and $-2$.

But, the cosine of any angle is always unique. So, in this particular case, I think that the use of $\pm$ implies that it is either $\Bigg(\sqrt{\dfrac{\cos\alpha + 1}{2}}\Bigg)$ or $\Bigg(-\sqrt{\dfrac{\cos\alpha + 1}{2}}\Bigg)$. From my point of view, this also implies that only the value of $\cos\alpha$ (even if the values of other trigonometric functions at $\alpha$ are provided) is not sufficient to evaluate the value of $\cos\Big(\dfrac{\alpha}{2}\Big)$ as there are infinitely many values of $\alpha$ for some given $\cos\alpha$ and for those possible values, overall, two values of $\cos\Big(\dfrac{\alpha}{2}\Big)$ emerge.

The reason that $\pm$ doesn't emerge in the expansion of $\cos(2\alpha)$ in my opinion, is that for some given value of $\cos\alpha$, the value for $\cos(2\alpha)$ is unique for all possible values of $\alpha$ (that are co-terminally related to each other).

I have mathematically proved that for all possible values of $\alpha$ for some given $\cos\alpha$, there is one and only one value that $\cos(2\alpha)$ corresponds to but rather two values that $\cos\Big(\dfrac{\alpha}{2}\Big)$ corresponds to.

So, I want to verify if the way I interpret the emergence of $\pm$ and especially, the statement that I have formatted with bold and italics are correct.

Thanks!

PS : I can add the proof that I reference in this post if that helps to make the question clearer. Let me know if I should do it.

Answer 1

I think it's worth pointing out that most of the time you can regard uses of the "$\pm$" sign like this as slightly informal: if you know $x^2 = k$ for some $k > 0$, then concluding that $x = \pm \sqrt{k}$ is really just a shorthand for saying "thus $x = \sqrt{k}$ or $x = - \sqrt{k}$, but I don't know which".

Written long-form like this, I think it becomes a lot clearer what you can and cannot conclude.

Also, using the example above, it might be the case in a given situation that only the case of $x = -\sqrt{k}$ (say) is possible due to additional information which places constraints on $x$. For example, you might know that $x$ is actually some function $f(y)$ of another variable $y$, and $f(y)$ only ever produces negative values. But in a vacuum, by only manipulating the equation $x^2 = y$ you won't be able to conclude this.

I think put this way, we can clearly evaluate the truth of what you've written in bold: it does not follow that knowledge of $\cos \alpha$ is not enough to evaluate $\cos \frac{\alpha}{2}$ directly from your manipulation---it might be possible, it's just that your equation doesn't tell you that it is. But as stated in other answers there are explicit examples that show that your statement in bold is true.

Thus, in retrospect with such examples in mind, there can't be any possible way to manipulate your first equation in order to obtain a unique possibility for $\cos \frac{\alpha}{2}$ (since this isn't actually true!).

Note on square roots in general: It is a common misconception that "$\sqrt{4} = \pm 2$" (using the usual notation common in mathematics). Formally the square root function $\sqrt{x}$ has only a single value, the so-called principal square root, which is just the positive one. That is, for example $\sqrt{4}$ is literally equal to $2$. The "$\pm$" creeps in when we are solving equations, since if we know that $x^2 = k$ (with $k > 0$) then $x = \sqrt{k}$ or $x = - \sqrt{k}$, but we don't know which. Often we write $x = \pm \sqrt{k}$ as a shorthand for this, but that is all that is going on.

Answer 2

Because $\sqrt{x^2}=|x|$ and since $$\cos^2\frac{\alpha}{2}=\frac{1+\cos\alpha}{2},$$ we obtain a right identity: $$\left|\cos\frac{\alpha}{2}\right|=\sqrt{\frac{1+\cos\alpha}{2}}.$$ Now,for $\cos\frac{\alpha}{2}\leq0$ we obtain: $$\cos\frac{\alpha}{2}=-\sqrt{\frac{1+\cos\alpha}{2}}$$ and for $\cos\frac{\alpha}{2}\geq0$ we obtain: $$\cos\frac{\alpha}{2}=\sqrt{\frac{1+\cos\alpha}{2}}$$

In my opinion the statement $$\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1+\cos\alpha}{2}}$$ without some adding words is wrong.

Also, in real numbers $\sqrt4=2$.

Answer 3

Knowing $\cos(2\alpha)$ does not determine what the sign of $\cos(\alpha)$ is. For example, $\alpha=0$ and $\alpha = \pi$ both have $\cos(2\alpha) = 1$, but one has $\cos(\alpha)=1$ and the other has $\cos(\alpha)=-1$.