Question 916840
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(\theta)\ =\ \frac{12}{13}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(\theta)\ =\ \frac{144}{169}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(\theta)\ =\ 1\ -\ \sin^2(\theta)\ =\ 1\ -\ \frac{144}{169}\ =\ \frac{25}{169}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(\theta)\ =\ \pm\frac{5}{13}]


But since *[tex \Large \theta] is obtuse *[tex \Large \Rightarrow\ \theta] is in the second quadrant where cosine is negative,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(\theta)\ =\ -\frac{5}{13}]


Since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\frac{\alpha}{2}\right)\ =\ \pm\sqrt{\frac{1\ -\ \cos(\alpha)}{2}}]


We have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\frac{\theta}{2}\right)\ =\ \pm\sqrt{\frac{1\ +\ \frac{5}{13}}{2}}\ =\ \pm\sqrt{\frac{9}{13}}\ =\ \pm\frac{3}{\sqrt{13}}]


Rationalizing the denominator


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\frac{\theta}{2}\right)\ =\ \pm\frac{3\sqrt{13}}{13}]


But since *[tex \Large \theta] is in quadrant 2, *[tex \Large \frac{\theta}{2}] must be in quadrant 1 (*[tex \Large x\ <\ \pi\ \Rightarrow\ \frac{x}{2}\ <\ \frac{\pi}{2}]).  And since the sine is positive in quadrant 1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\frac{\theta}{2}\right)\ =\ \frac{3\sqrt{13}}{13}] 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \Large \ \