Question 1131102
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\frac{\varphi}{2}\ =\ \pm\sqrt{\frac{1\,+\,\cos\varphi}{2}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\frac{\varphi}{2}\ =\ \pm\sqrt{\frac{1\,-\,\cos\varphi}{2}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\frac{\varphi}{2}\ =\ \pm\frac{1\,-\,\cos\varphi}{\sin\varphi}]


If your given angle is in the range *[tex \Large \pi\ <\ x\ <\ \frac{3\pi}{2}], then the half angle must be in the range *[tex \Large \frac{\pi}{2}\ <\ x\ <\ \frac{3\pi}{4}], hence the cosine and tangent will be negative and the sine positive.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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