Question 632146
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First thing to note is that since *[tex \LARGE 0\ <\ 2a\ <\ \frac{\pi}{2}], a simple division operation shows that *[tex \LARGE 0\ <\ a\ <\ \frac{\pi}{4}], forcing *[tex \LARGE a] to be a first quadrant angle, hence all six basic trigonometric functions will be positive.


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


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(a)\ -\ \sin^2(a)\ =\ -\frac{17}{18}]


Since we know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(\varphi)\ +\ \sin^2(\varphi)\ \equiv\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ 2\sin^2(a)\ =\ -\frac{17}{18}]


And then a little algebra music, Sammy:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(a)\ =\ \pm\frac{\sqrt{35}}{6}]


From which we discard the negative root because of the interval analysis earlier to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(a)\ =\ \frac{\sqrt{35}}{6}]



And likewise:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cos^2(a)\ -\ 1\ =\ -\frac{17}{18}]


From which we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(a)\ =\ \frac{1}{6}]


Use the following identities to get the other four functions:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot(\varphi)\ \equiv\ \frac{1}{\tan{\varphi}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec(\varphi)\ \equiv\ \frac{1}{\cos{\varphi}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc(\varphi)\ \equiv\ \frac{1}{\sin{\varphi}]


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