How any formula which gives us $ \cos A/2 $ in terms of k should give us also the cosine of $ (2n \pi ± A) /2 $?

I'm learning Trigonometry right now with myself and at current about Trigonometry Ratios of multiple and Submultiple angles. I am not able to understand a statement given in the book. Please have a look at the image. The image is the explanation of why there is ambiguity when $ \cos A/2$ and $ \sin A/2$ are found from the value of $ \cos A$.
I can't understand the line any formula which gives us $ \cos A/2$ in terms of k should give us also the cosine of $(2n\pi ± A) /2$ .Please help. Thankyou in advance.

2 Answers

Here they are using the fact that:

$$\cos(\pi n + x) = \begin{cases}\cos(x) & n \equiv 0 \pmod{2} \\ -\cos(x) & n \equiv 1 \pmod{2}\end{cases}$$

The easiest way to see this is to consider the cosine graph:

If $cos(A/2)$ can be made to depend only on $cosA=k$ (which is not actually true), then so can $cos((2nπ+A)/2)$ be made to depend only on $cos(2nπ+A)=k$, by substitution, and hence it would be equal to $cos(A/2)$, a true implication between two false propositions.