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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.

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Do you mean that A is in the given quadrant, or that A/2 is in the given quadrant? –  Gabe Cunningham Nov 4 '10 at 18:54
    
They didn't provided any of such information, but I guess it's should be A/2. –  Quixotic Nov 4 '10 at 18:57
1  
It is possible to derive the following double identity that has no sign ambiguity: $\tan \dfrac{A}{2}=\dfrac{\sin A}{1+\cos A}=\dfrac{1-\cos A}{\sin A}$ –  Américo Tavares Nov 4 '10 at 23:03
    
Américo Tavares:This is interesting, I knew this formula before but never thought this way, thanks really appreciated :) –  Quixotic Nov 11 '10 at 15:29

2 Answers 2

up vote 4 down vote accepted

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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We have that $$\cos A=\cos \left(2\cdot\frac{A}{2}\right) = \cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}$$ and $$1 = \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}}.$$ Therefore \begin{align*} 1+\cos A & = \cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}} + \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}} \\ & = 2\cos^2{\frac{A}{2}} \\ \Rightarrow \frac{1+\cos A}{2} & = \cos^2{\frac{A}{2}} \\ \Rightarrow \cos\frac{A}{2} & = \sqrt{\frac{1+\cos A}{2}}. \end{align*} and \begin{align*} 1-\cos A & = \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}} - \cos^2{\frac{A}{2}}+\sin^2{\frac{A}{2}} \\ & = 2\sin^2{\frac{A}{2}} \\ \Rightarrow \frac{1-\cos A}{2} & = \sin^2{\frac{A}{2}} \\ \Rightarrow \sin\frac{A}{2} & = \sqrt{\frac{1-\cos A}{2}}. \end{align*}

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Sure... Try $A=3\pi$ for example. –  Did Jul 15 '12 at 6:29

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