Question 1171626
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For an angle in QII:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(a)\ =\ -\sqrt{1\,-\,\cos^2(a)}]


For an angle in QII:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(b)\ =\ \sqrt{1\,-\,\sin^2(b)}]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(a\,+\,b)\ =\ \cos(a)\cos(b)\ -\ \sin(a)\sin(b)]


You get to do your own arithmetic.

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