Question 333365
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(\theta)\ =\ 0.577] on *[tex \LARGE 0\ <\ \theta\ <\ {\pi\over2}]


*[tex \LARGE \ \ \ \ \ \ \ ]Find *[tex \Large \cos(\theta)] and *[tex \Large \sin(\theta)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ +\ \tan^2(\varphi)\ =\ \sec^2(\varphi)]


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


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


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


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(\theta)\ =\ \sqrt{\frac{1}{1\ +\ (0.577)^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ sin(\theta)\ =\ \sqrt{1\ -\ \frac{1}{1\ +\ (0.577)^2}]


You can do your own arithmetic.


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