<s>[INST] 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.
 [/INST] 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.
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