Question 1130787
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Given:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\varphi\ =\ \frac{p}{q}]


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


(1) *[tex \LARGE \ \ \ \ \ \ q\sin\varphi\ =\ p\cos\varphi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ q^2\sin^2\varphi\ =\ p^2\cos^2\varphi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ q^2\sin^2\varphi\ =\ p^2\(1\,-\,\sin^2\varphi\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(q^2\ +\ p^2\)\sin^2\varphi\ = \ p^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\varphi\ =\ \frac{p^2}{q^2\,+\,p^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\varphi\ =\ \pm\sqrt{\frac{p^2}{q^2\,+\,p^2}}]


Then from (1):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\varphi\ =\ \pm\frac{q}{p}\sqrt{\frac{p^2}{q^2\,+\,p^2}}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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