Question 1194018
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You misstated your problem: You want to prove *[tex \Large \cos\(x\,+\,\frac{\pi}{2}\)\ =\ -\sin(x)]


Use the addition formulas:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(\alpha\,+\,\beta)\ =\ \sin(\alpha)\cos(\beta)\ +\ \cos(\alpha)\sin(\beta)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(\alpha\,+\,\beta)\ =\ \cos(\alpha)\cos(\beta)\ -\ \sin(\alpha)\sin(\beta)]


To prove both *[tex \Large \sin\(x\,+\,\frac{\pi}{2}\)\ =\ \cos(x)] and *[tex \Large \cos\(x\,+\,\frac{\pi}{2}\)\ =\ -\sin(x)]


Then *[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\,\cos(x)\ =\ \frac{d}{dx}\,\sin\(x\,+\,\frac{\pi}{2}\)]


Use the chain rule:  Let *[tex \Large u\ =\ x\,+\,\frac{\pi}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{du}\,\sin(u)\ =\ \cos(u)]


from the assumption, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(u)\ =\ \cos\(x\,+\,\frac{\pi}{2}\)\ =\ -\sin(x)]

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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