Question 1167597
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Pythagorean Identity:


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


So if *[tex \Large \sin^2(t)\ =\ \frac{1}{5}] then *[tex \Large \cos^2(t)\ =\ \frac{4}{5}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(t)\ =\ \pm\sqrt{\frac{4}{5}}\ =\ \pm\frac{2\sqrt{5}}{5}]


But *[tex \Large \cos\varphi\ <\ 0\ \forall\ \varphi\ \in\ \text{QII}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(t)\ =\ -\frac{2\sqrt{5}}{5}]


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


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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