trigonometry - Expressions of $sin frac{A}{2}$ and $cos frac{A}{2}$ in terms of $sin A$

I am trying to understand the interpretation of $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ in terms of $\sin A$ from my book, here is how it is given :

We have $ \bigl( \sin \frac{A}{2} + \cos \frac{A}{2} \bigr)^{2} = 1 + \sin A $ and $ \bigl( \sin \frac{A}{2} - \cos \frac{A}{2} \bigr)^{2} = 1 - \sin A $

By adding and subtracting we have $ 2 \cdot \sin \frac{A}{2} = \pm \sqrt{ 1 + \sin A } \pm \sqrt{ 1 - \sin A } $ ---- (1) and $ 2 \cdot \cos \frac{A}{2} = \mp \sqrt{ 1 + \sin A } \mp \sqrt{ 1 - \sin A } $ ---(2)

I have understood upto this far well,

Now they have broke the them into quadrants :

In 1st quadrant :

$$ 2 \cdot \sin \frac{A}{2} = \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$

$$ 2 \cdot \cos \frac{A}{2} = \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$

In 2nd quadrant :

$$ 2 \cdot \sin \frac{A}{2} = \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$

In 3rd quadrant :

$$ 2 \cdot \sin \frac{A}{2} = \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$

$$ 2 \cdot \cos \frac{A}{2} = \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$

In 4th quadrant :

$$ 2 \cdot \sin \frac{A}{2} = - \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = - \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$

Now, In knew the ALL-SINE-TAN-COSINE rule but still I am not able to figure out how the respective signs are computed in these (above) cases.

Answer

The easiest way of computing the signs is to make them match; we know that sin x > 0 if 0 < x < π and that cos x > 0 if -π/2 < x < π/2. Knowing whether sin A is greater than 0 or less than zero tells you whether $\sqrt{1-\mathrm{sin} A}$ is greater or less than $\sqrt{1+\mathrm{sin} A}$; that in turn lets you figure out what the overall sign on all of the right-hand terms is, and each quadrant corresponds to one of the four positive/negative pairs on the right-hand terms.