Question 340898
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot(\varphi)\ =\ \frac{sin(\varphi)}{cos(\varphi)}]


So, in order for *[tex \LARGE \cot(\varphi)\ =\ -1], *[tex \LARGE sin(\varphi)\ =\ -cos(\varphi)]


Look at the unit circle and find the two places where *[tex \LARGE sin(\varphi)\ =\ -cos(\varphi)]


<img src="http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png">


But you didn't restrict your solution to any particular set of angles, so your complete answer is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot^{-1}(-1)\ =\ \frac{3\pi}{4}\ +\ 2k\pi]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot^{-1}(-1)\ =\ \frac{7\pi}{4}\ +\ 2k\pi]


which can be simplified to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot^{-1}(-1)\ =\ \frac{3\pi}{4}\ +\ k\pi]


where


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ \in\ \mathbb{Z}]



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