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.