Question 1160203
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc(x)\ =\ 7\ \Right\ \sin(x)\ =\ \frac{1}{7}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(x)\ =\ \frac{1}{49}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 49\sin^2(x)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 49\(1\ -\ \cos^2(x)\)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(x)\ =\ \frac{48}{49}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ =\ \pm\frac{4\sqrt{3}}{7}]


But *[tex \Large x\ \in\ \text{QII}\ \Right\ \cos(x)\ <\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ =\ -\frac{4\sqrt{3}}{7}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\(\frac{\varphi}{2}\)\ =\ \pm\sqrt{\frac{1\,+\,\cos(\varphi)}{2}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\(\frac{\varphi}{2}\)\ =\ \pm\sqrt{\frac{1\,-\,\cos(\varphi)}{2}}]


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


You can do the rest of the arithmetic yourself.  Hint: *[tex \Large \varphi\ \in\ \text{QII}\ \Right\ \frac{\varphi}{2}\ \in\ \text{QI}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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