Question 1112839
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \log_{\sin x}\(\cos x\)\ =\ \frac{1}{2}]


where *[tex \Large 0\ <\ x\ < \frac{\pi}{2}]


Start with the definition of the logarithm function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \log_b(x)\ =\ y\ \Leftrightarrow\ b^y\ =\ x]


Hence,


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


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin^2(x)\ +\ \sin(x)\ -\ 1\ =\ 0]


Let *[tex \Large u\ =\ \sin(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  u^2\ +\ u\ -\ 1\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  u\ =\ \frac{-1\ \pm\ \sqrt{1\ +\ 4}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  u\ =\ \frac{-1\ \pm\ \sqrt{5}}{2}]


Substitute back, recalling that the sine function is positive in the first quadrant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin(x)\ =\ \frac{\sqrt{5}\ -\ 1}{2}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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