Question 334138
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t] in Quadrant II


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\pi}{2}\ \leq\ t\ \leq\ \pi]


In Quadrant II, *[tex \LARGE \sin(\varphi)\ \geq\ 0] and *[tex \LARGE \cos(\varphi)\ \leq\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc(\varphi)\ =\ \frac{1}{\sin(\varphi)]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc(t)\ =\ 2] and *[tex \LARGE \frac{\pi}{2}\ \leq\ t\ \leq\ \pi\ \Rightarrow\ \sin(t)\ =\ \frac{1}{2}]


Furthermore:


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


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(t)\ =\ \frac{1}{2}\ \Rightarrow\ \cos(t)\ =\ \pm\frac{\sqrt{3}}{2}]


But *[tex \LARGE \cos(\varphi)\ \leq\ 0] for Quadrant II so we need *[tex \LARGE \cos(t)\ =\ -\frac{\sqrt{3}}{2}]


Next:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \cot(\varphi)\ =\ \frac{\cos(\varphi)}{\sin(\varphi)}]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \cot(t)\ =\ \frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}\ =\ -\sqrt{3}]



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