Question 602425
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^4(\varphi)\ -\ \cos^4(\varphi)\ \equiv^?\ 2\sin^2(\varphi)\ -\ 1]


Factor the difference of two squares in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sin^2(\varphi)\ -\ \cos^2(\varphi)\right)\left(\sin^2(\varphi)\ +\ \cos^2(\varphi)\right)\ \equiv^?\ 2\sin^2(\varphi)\ -\ 1]


Note that the right-hand factor in the LHS is just the Pythagorean Identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sin^2(\varphi)\ -\ \cos^2(\varphi)\right)\left(1\right)\ \equiv^?\ 2\sin^2(\varphi)\ -\ 1]


Then, using the Pythagorean identity again in the form *[tex \LARGE \cos^2(\phi)\ =\ 1\ -\ \sin^2(\phi)], substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(\varphi)\ -\ \left(1\ -\ \sin^2(\varphi)\right)\ \equiv^?\ 2\sin^2(\varphi)\ -\ 1] 


Combine terms in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sin^2(\varphi)\ -\ 1\ \equiv\ 2\sin^2(\varphi)\ -\ 1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>