Question 1176263
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin^2x\ -\ \cos^2x}{\sin x\ +\ \cos x}\ \equiv\ \sin x\ -\ \varphi(x)]


Factor the difference of two squares in the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\(\sin x\ +\ \cos x\)\(\sin x\ -\ \cos x)}{\sin x\ +\ \cos x}\ \equiv\ \sin x\ -\ \varphi(x)]


Eliminate common factors from numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin x\ -\ cos x\ \equiv\ \sin x\ -\ \varphi(x)]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi(x)\ =\ \cos x]


And the identity is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin^2x\ -\ \cos^2x}{\sin x\ +\ \cos x}\ \equiv\ \sin x\ -\ \cos x]


Note: You must exclude *[tex \LARGE x\ =\ \frac{3\pi}{4}\ +\ k\pi\ \forall\ k\ \in\ \mathbb{Z}] so that the denominator in the left hand expression is never zero.  Hence, the original expression is not precisely an identity without having made that exclusion in the first place.

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