Question 1118725
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\beta\ =\ \sin\alpha\cos\alpha]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \cos^2\beta\ =\ \sin^2\beta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\beta\ =\ 1\ -\ \sin\alpha\cos\alpha]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ = 1\ -\ 2\sin\alpha\cos\alpha]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\(\frac{\pi}{4}\,+\,\alpha\)\ =\ \cos\(\frac{\pi}{4}\)\cos\alpha\ -\ \sin\(\frac{\pi}{4}\)\sin\alpha]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ \frac{\sqrt{2}}{2}\(\cos\alpha\ -\ \sin\alpha\)]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cos^2\(\frac{\pi}{4}\,+\,\alpha\)\ =\ 2\(\frac{1}{2}\)\(\cos^2\alpha\ -\ 2\sin\alpha\cos\alpha\ +\ \sin^2\alpha\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ 1\ -\ 2\sin\alpha\cos\alpha]

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