Question 333375
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(\varphi)\ =\ \frac{1}{\cot(\varphi)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(\varphi)\ =\ \frac{1}{1\ +\ \tan^2(\varphi)}]


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



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(\theta)\ =\ -\frac{1}{4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(\theta)\ =\ \frac{1}{\frac{17}{16}}\ =\ \frac{16}{17}\ \ \Rightarrow\ \ \cos(\theta)\ =\ \pm\frac{4\sqrt{17}}{17}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(\theta)\ =\ 1\ -\ \frac{16}{17}\ =\ \frac{1}{17}\ \ \Rightarrow\ \ \sin(\theta)\ =\ \pm\frac{\sqrt{17}}{17}]


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