Question 1011260
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Use the sine and cosine of the difference of angles to assure yourself that *[tex \Large \sin\left(\frac{\pi}{2}\ -\ x\right)\ =\ \cos x] and *[tex \Large \cos\left(\frac{\pi}{2}\ -\ x\right)\ =\ \sin x].  Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left\[\left(\frac{\pi}{2}\ -\ x\right)\ +\ y\right\]\ =\ \sin\left(\frac{\pi}{2}\ -\ x\right)\cos y\ +\ \cos\left(\frac{\pi}{2}\ -\ x\right)\sin y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,=\ \cos x\cos y\ +\ \sin x\sin y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,=\ \cos(x\ -\ y)]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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