Question 1158076
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Find *[tex \Large \tan\theta] given *[tex \Large \sin\theta\ =\ \frac{1}{2}] and *[tex \Large \theta\ \in\ QII]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ \frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ \frac{1}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta\ =\ 1\ -\ \sin^2\theta\ =\ \frac{3}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ \pm\frac{sqrt{3}}{2}]


But cosine in QII is negative so reject the positive value.


Recall that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}]


Hence,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\ =\ \frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}\ =\ -\frac{1}{\sqrt{3}}\ =\ -\frac{\sqrt{3}}{3}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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