Question 1177237
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Since *[tex \Large u] is in the fourth quadrant, *[tex \Large \sin(u)\,<\,0], and *[tex \Large \cos(u)\,>\,0].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot(u)\ =\ -3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\cos(u)}{\sin(u)}\ =\ -3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(u)\ =\ -3\sin(u)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(u)\ =\ 9\sin^2(u)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \sin^2(u)\ =\ 9\sin^2(u)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\sin^2(u)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(u)\ =\ \frac{1}{10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(u)\ =\ \pm\frac{1}{\sqrt{10}}]


But since *[tex \Large \sin(u)\,<\,0\ \forall\ u\ \in \(\frac{3\pi}{2},2\pi\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(u)\ =\ -\frac{1}{\sqrt{10}}]


Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(u)\ =\ 9\sin^2(u)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(u)\ =\ 9\ -\ 9\cos^2(u)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\cos^2(u)\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(u)\ =\ \frac{9}{10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(u)\ =\ \pm\frac{3}{\sqrt{10}}]


But since *[tex \Large \cos(u)\,>\,0\ \forall\ u\ \in \(\frac{3\pi}{2},2\pi\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(u)\ =\ \frac{3}{\sqrt{10}}]


Next:


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


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


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


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


Or you could just use


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


to get the same answer.


All you have to do is plug in the numbers and do the arithmetic.

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

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