Question 1024808
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Start with the Pythagorean Identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\varphi\ +\ \sin^2\varphi\ =\ 1]


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


For your problem, the angle is restricted to Quadrant III, where both cosine and sine are negative.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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